Pre-U STPM Physics Term 1 CC039242a

294 11 Physics Term 1 STPM Chapter 11 Heat Transfer 1. Fibreglass, which consists of a very large number of intertwined fine glass fibres is used as a thermal insulator, because A glassfibre does not absorb much heat. B thermal conductivity of glass is very low. C fibreglass prevents the faster-moving air molecules from escaping. D fibreglass reduces air movement. 2. In Lee’s method to determine the thermal conductivity of poor conductors, the sample is in the form of a thin disc A to reduce heat loss from the large surface area. B so that the rate of heat flow can be measured. C so that the temperature gradient is large. D to produce a small difference in temperature. 3. (a) Explain why good electric conductors are also good thermal conductors. (b) In the experimental determination of the thermal conductivity of a good conductor, the sample used is long and its sides are lagged. For poor conductors, the sample is rather thin and is not lagged. Explain why. 4. A vacuum flask has a cylindrical cork of diameter 10.0 cm and thickness 2.5 cm. It contains 1.00 kg of ice at 0°C. When the temperature of the externalsurface of the cork is 20°C, the ice in the flask takes 30 hours to melt. What is the mean temperature of the inner surface of the cork if heat flow is only by conduction through the cork. (Specific latent heat of fusion of ice = 3.35 × 105 J kg–1, thermal conductivity of cork = 0.67 W m–1 K–1) 5. The eskimos make igloos from blocks of compact snow of thickness 0.50 m each. What isthe thickness of expanded polystyrene which would give the same thermal insulation, which is the rate of heat loss for the same temperature difference across the wall. (Thermal conductivity : compact ice, 0.20 W m–1 K–1; expanded polystyrene, 0.012 W m–1 K–1) 6. The thermal conductivity of glassfor a window is 0.80 W m–1 K–1. Calculate the rate of heat flow through a glass window of area 2.0 m2 and thickness 4.00 mm if the temperature inside is 20°C and the temperature outside is 35°C. The calculated value for the heat flow is much greater than the actual value. Explain why. Thermal Resistance, Rθ 1. From the microscopic viewpoint, thermal conduction and electric conduction are similar. Both conductions are by the translational motion of free electrons. 2. From the macroscopic viewpoint, the two conductions are also similar. In thermal conduction, rate of heat flow, dQ dt = –kA dθ dx = kA Length × (Temperature difference) ............... a In electric conduction, rate of flow of electric charge, I = V, potential difference R, resistance ............... b Comparing equation a and b, the following quantitiesin thermal conduction and electrical conduction are analogous. Thermal conduction Electrical conduction Heat Electric charge Temperature Electric potential Temperature difference Potential difference Thermal resistance, Rθ Electrical resistance, R Rθ = l kA l = length R = l σA σ = electrical conductivity Quick Check 3

295 11 Physics Term 1 STPM Chapter 11 Heat Transfer Jointed Rods 0 Temperature gradient x l 2 l 1 (c) First rod Second rod Lagging x l 1 l 2 k1 k1 < k2 k2 θ1 θ’ θ2 < θ1 (a) 0 Temperature x l 2 l 1 θ1 θ’ θ2 (b) Figure 11.8 1. Figure 11.8(a) shows two uniform rods of the same cross-sectional area A but having different lengths l 1 and l 2 , different thermal conductivities k1 and k2 (k1 < k2 ) joined in series. The joined rods are well-lagged. 2. Figure 11.8(b) shows the variation of temperature with distance from the hot end. Since the rods are lagged, there is no heat loss from the sides and the rate of heat flow along both rods are the same. Then dQ dt = –k1 A( dθ dx )1 = –k2 ( dθ dx )2 Since k1 < k2 , then ( dθ dx )1 > ( dθ dx )2 . Magnitude of the gradient of the graph for the first rod is greater than that of the second graph (Figure 11.8 (b) and (c)). 3. If θ′ = temperature at the junction of the two rods, then using ( dQ dt )1 = ( dQ dt )2 –k1 A( dθ dx )1 = –k2 A( dθ dx )2 k1 A (θ1 – θ ′) l 1 = k2 A(θ ′ – θ2 ) l 2 Exam Tips When rods are joined in series and insulated, temperature gradient, dθ dx ∝ 1 k

296 11 Physics Term 1 STPM Chapter 11 Heat Transfer θ′ ( k1 l 1 + k2 l 2 ) = k1 θ1 l 1 + k2 θ2 l 2 θ ′ = ( k1 θ1 l 1 + k2 θ2 l 2 ) ( k1 l 1 + k2 l 2 ) 4. From dQ dt = k1 A (θ1 – θ′) l 1 θ1 – θ ′ = l 1 k1 A ( dQ dt ) .............................. c and from dQ dt = k2 A (θ′ – θ2 ) l 2 θ ′ – θ2 = l 2 k2 A ( dQ dt ) ..............................d Equation c + d: θ1 – θ2 = ( l 1 k1 A + l 2 k2 A ) × ( dQ dt ) dQ dt = (θ1 – θ2 ) ( l 1 k1 A + l 2 k2 A ) = Temperature difference (Total thermal resistance) 5. Figure 11.9 The equation Rate of heat flow, dQ dt = Temperature difference Total thermal resistance can be extended for any number of rods joined in series as shown in Figure 11.9. The rate of heat flow, dQ dt = (θ1 – θ2 ) Σ ( l i ki A ) i = 1, 2, 3, 4 Example 9 The total surface area of the walls of a room is 15 m2 . The wall is made of brick of thickness 0.25 m and thermal conductivity 0.50 W m–1 K–1. The temperature inside the room is 20°C and outside the room is 30°C. (a) Calculate the rate of heat conducted through the walls.

297 11 Physics Term 1 STPM Chapter 11 Heat Transfer (b) What is the rate of heat conducted through the walls if the inner wall is covered with an insulator of thermal conductivity 0.04 W m–1 K–1, and thickness 20 mm? Solution: (a) Rate of heat conducted through the walls, dQ dt = –kA dθ dx = (0.50)(15.0)(30 – 20) 0.25 = 300 W (b) Rate of heat flow through the walls and insulator in series, dQ dt = Temperature difference Total thermal resistance = (θ1 – θ2 ) ( l 1 k1 A + l 2 k2 A ) = (30 – 20) × 15.0 ( 0.25 0.50 + 0.020 0.04 ) = 150 W Example 10 X 0°C 100°C Y 0°C X 100°C Y (a) (b) Figure (a) shows two metal bars X and Y which have the same cross-section and length joined in series. The thermal conductivity of Y is three times of X. (a) If the rods are well lagged, find the temperature at the common junction. (b) The rods are then joined in parallel as shown in figure (b). Compare the rate of heat flow in figure (b) with that in figure (a) if the rods are well-lagged in both cases. Solution: (a) The rates of heat flow in X and Y are the same. Hence, using dQ dt = –k1 A( dθ dx )1 = –k2 A( dθ dx )2 kA (θ – 0) l = (3k)A( 100 – θ l ) θ = 3(100 – θ ) θ = 75°C (b) In figure (a), rate of heat flow ( dθ dt )1 = Temperature difference Total thermal resistance = (100 – 0) ( l kA + l 3kA) = 300 kA 4l

298 11 Physics Term 1 STPM Chapter 11 Heat Transfer In figure (b), rate of heat flow ( dQ dt )2 = ( dQ dt )x + ( dQ dt )y = kA (100 – 0) l + 3kA (100 – 0) l = 400 kA l ( dQ dt )2 ( dQ dt )1 = 400 kA l 300 kA 4l = 16 3 Example 11 Three metal rods made of aluminium, brass and copper are joined in series. The length of each rod is 50.0 cm and their diameter is 2.0 cm. The free end of the aluminium rod is kept at 100 °C and the copper rod is 0 °C. The rods are lagged and the thermal conductivity of copper is twice that of aluminium and four times that of brass. Find the equilibrium temperature at the aluminium-brass, and brass-copper interface. Solution: Aluminium Brass Copper 2k θ1 θ2 100°C k 4k 0°C If k = thermal conductivity of brass 2k = thermal conductivity of aluminium 4k = thermal conductivity of copper The rate of heat flow through all the rods are the same, dQ dt = Temperature difference Total thermal resistance = 100 – 0 l (2k)A + l kA + l (4k)A = 400kA 7l If θ1 = temperature at the aluminium-brass interface, for the aluminium rod, dQ dt = (2k)A (100 – θ1 ) l 400kA 7l = 2kA l (100 – θ1 ) 200 = 7 (100 – θ1 ) θ1 = 71.4°C For the copper rod, 400kA 7l = (4k) A (θ2 – 0) l θ2 = 14.3°C

299 11 Physics Term 1 STPM Chapter 11 Heat Transfer Example 12 The wall of a refrigerator consists of two aluminium plates of the same thickness separated by a layer of air as shown on the right. The temperature outside the refrigerator is 30 °C and inside the refrigerator is 0°C. (a) Sketch a graph to show the variation of temperature across the wall of the refrigerator. State any assumptions you make. (b) If the thickness of each aluminium plate is 2.00 mm, and the thickness of the air layer is 1.60 cm, calculate the rate of heat conducted through a unit surface area of refrigerator wall. (Thermal conductivity : aluminium = 210 W m–1 K–1, air = 0.024 W m–1 K–1). (c) Explain the effect on the rate of heat conducted through the refrigerator wall (i) if the separation between the two aluminium plates is reduced. (ii) if the aluminium plates are replaced with steel plates, the separation remains at 1.60 cm. (d) The separation between the aluminium plates remains at 1.60 cm, but the space between the plates are filled with a piece of expanded polystyrene with a thermal conductivity which is almost the same value as that of air. It is observed that the rate of the heat flow through the walls is very much reduced compared to when there is a layer of air between the aluminium plates. Why? Solution: (a) Assumption : Heat lost from the sides of the aluminium plates and air layer is negligible. (b) Using rate of heat flow dQ dt = Temperature difference Total thermal resistance = (30 – 0) x k1 A + l k2 A + x k1 A Rate of heat flow per unit area, 1 A dQ dt = 30 2(2 ×10–3) 210 + 0.016 0.024 = 45.0 W m–2 (c) (i) The rate of heat conducted through the wall increases because the thermal resistance of the air layer decreases when its thickness decreases. (ii) The change in the rate of heat conducted is negligible as the thermal resistance of the metal plates, aluminium or steel is negligible compared to the thermal resistance of the air layer. (d) With a layer of air between the aluminium plates, heat flow through the air layer is mainly by convection which is the main method of heat transfer in air, not by conduction. With expanded polystyrene between the aluminium plates, heat is transferred by conduction only. Air is trapped in the cellular structure of the expanded polystyrene. The trapped air is unable to transfer heat by convection. Hence, rate of heat flow is reduced. 30°C 0°C Aluminium Air 30 0 x l x Temperature /°C Distance

300 11 Physics Term 1 STPM Chapter 11 Heat Transfer Example 13 A layer of ice of thickness x is formed on the surface of a lake. The temperature of the air above the ice is –θ °C. (a) Derive an expression for the rate of increase in the thickness of ice in terms of k the thermal conductivity of ice, ρ its density, L the specific latent heat of fusion of ice, and θ. (b) Calculate the time taken for the thickness of ice in a lake to increase from 10 cm to 11 cm when the temperature above the ice is –15°C. (Density of ice = 920 kg m–3, specific latent heat of fusion of ice = 3.35 × 105 J kg–1, thermal conductivity of ice = 1.7 W m–1 K–1). Solution: Ice Water x 0°C – θ°C (a) Consider an area A of the ice on the lake. Mass of ice of surface area A and thickness x m = ρAx ρ = density of ice Heat lost by water to the air above it when the mass m of ice is formed is Q = mL = ρAxL L = specific latent heat of fusion of ice Rate of heat loss, dQ dt = ρAL dx dt ......................... a Rate of heat conducted through the ice, dQ dt = –kA dθ dx = –kA (–θ – 0) x = kAθ x ......................... b Equation a = b: ρAL dx dt = kAθ x Rate of increase of the thickness of ice dx dt = kθ ρLx (b) Using the equation dx dt = kθ ρLx ∫ 0. 0. 10 11 x dx = ∫ t1 0 kθ ρL dt [ 1 2 x2 ]0. 0. 11 10 = kθ ρL [ t ]t 1 0 Time taken for the thickness of ice to increase from 0.10 m to 0.11 m, t 1 = 1 2 (0.112 – 0.102 ) × ρL kθ = 1 2 (0.112 – 0.102 ) × 920 × (3.35 × 105 ) 1.7 × 15 = 1.27 × 104 s

301 11 Physics Term 1 STPM Chapter 11 Heat Transfer 1. Ice is kept in a box of total surface area 1.20 m2 . The temperature in the box is 0°C and the room temperature is 30°C. The thickness of the wall is 2.50 cm and its thermal conductivity is 0.22 W m–1 K–1. How much heat is conducted through the walls of the box in an hour? A 1.90 kJ B 4.75 kJ C 1.14 MJ D 2.35 MJ 2. Two lagged metal rods P and Q of the same length are place with the ends at temperatures of 100°C and 0°C asshown in the diagram. The diameter of P is twice that of Q. The thermal conductivity of P is half that of Q. P Lagging 100º C 0º C Q What is the value of the ratio of the rate of heat flow along P to that of Q? A 0.50 C 2.0 B 0.25 D 4.0 3. An aluminium rod of length a and a copper rod of length b have the same cross sectional area and are perfectly lagged. The ends of the rods are maintained at the same temperature T1 and T2 (T1 > T2 ). The rates of heat flow in both the rods are the same. What is the value of a/b? [Thermalconductivityofcopper=400Wm–1K–1, aluminium = 240 W m–1 K–1] A 0.36 C 1.7 B 0.60 D 2.8 4. Two metal rods, X and Y, of lengths 40 cm and 20 cm, with the same cross-section are joined end-to-end as shown. The joined rod is lagged and the opposite ends are kept at 100°C and 0°C. The thermal conductivity of metal Y is twice that of X. Metal X 40 cm 100°C θ°C 0°C 20 cm Metal Y What is the temperature at the junction? A 20°C C 50°C B 40°C D 80°C 5. The figure shows a joined rod made of different metals, P and Q. The ends of the joined rod are kept at 0°C and 100°C. The temperature at the junction is 20°C. P Q 0°C 20°C 100°C Lagging If the jointed rod is lagged, what is the ratio of thermal conductivity of P thermal conductivity of Q ? A 1 4 C 2 B 1 2 D 4 6. Two metal bars of lengths a and b have the same cross-sectional area, but different thermal conductivities k1 and k2 respectively. The opposite ends of the composite bar are kept at temperatures Th and Tc , whereas the temperature of the junction is Tj (Th > Tc ). Th k1 k2 Tj Tc a b If there is no heat loss from the sides of the bars, which of the following gives the ratio of k1 k2 ? A k1 k2 = (Th – Tj )b (Tj – Tc )a C k1 k2 = (Th – Tj )a (Tj – Tc )b B k1 k2 = (Tj – Tc )b (Th – Tj )a D k1 k2 = (Tj – Tc )a (Th – Tj )b 7. Two different metal rods of the same length and cross-section are joined in series and lagged. When steady state is attained, the temperatures at the opposite ends of the composite rods are 100°C and 20°C, and the temperature gradient across the rod P is three times that across the rod Q. Quick Check 4

302 11 Physics Term 1 STPM Chapter 11 Heat Transfer 100°C P θ°C Q 20°C Lagging What is the temperature θ °C of the junction? A 40°C B 50°C C 60°C D 80°C 8. Three rods X, Y and Z are of lengths l, l, and 2l, thermal conductivities k, 2k and 3k. The rods have the same cross-section and are joined in series. The composite rod is welllagged. θ1 X θ Y Z 2 l l 2l The temperatures of the ends of the composite rod are θ1 and θ2 (θ1 > θ2 ). At steady state, the temperature difference across rod X is Δθ. What are the temperature differences across rod Y and rod Z? Rod Y Rod Z A 1 2 Δθ 4 3 Δθ B 1 2 Δθ 2 3 Δθ C 2Δθ 2 3 Δθ D 2Δθ 3Δθ 9. A composite rod consists of two rods X and Y of the same length and cross-section. The composite rod is well-lagged and the temperature at the ends are 100°C and 0°C. The thermal conductivity of X is twice that of Y. Which of the following graphs shows the variation of temperature t with distance x along the rod in the steady state? X x l l 100°C Y 0°C Lagging A C 100 0 x 2ll t / °C 100 0 x 2ll t / °C B D 100 0 x 2ll t / °C 100 0 x 2ll t / °C 10. The figure shows a composite rod consisting of two rods P and Q of thermal conductivity k and k′ (k < k′) respectively. The ends of the rods P and Q are kept at 100°C and 30°C respectively. Only rod P is perfectly lagged. X Y P x 100°C Q 30°C Z Which of the following graphs best represent the variation of temperature θ along the composite rod? A C X Y Z 100 30 0 x θ/°C X Y Z 100 30 0 x θ/°C B D X Y Z 100 30 0 x θ/°C X Y Z 100 30 0 x θ/°C

303 11 Physics Term 1 STPM Chapter 11 Heat Transfer 11. Two brassrods are joined together using a type of glue which is a poor thermal conductor. The joined rods are lagged and their ends maintained at temperatures T1 and T2 (T1 > T2 ) as shown in the figure. Lagging Brass Glue Brass T1 T2 Which of the graphs shows correctly the variation of temperature T with the distance x from the hot end? A C 0 d x T T1 T2 0 d x T T1 T2 B D 0 d x T T1 T2 0 d x T T1 T2 12. For a certain temperature difference across the base of a metal pot of thermal conductivity λ, the rate of heat flow across the base of the pot is k. An insulating scale of thickness equal to 1% of the thickness of the base is deposited on the inside surface of the base. If the thermal conductivity of the scale is λ 100, what is the rate of heat flow through the base forthe same temperature difference? A k 2 C k 100 B k 4 D 2k 13. Two blocks which are kept at different temperatures are connected in two different ways using two identical metal rods P and Q which are lagged as shown in figure (a) and (b). The thermal conductivity of P is twice that of Q. Rod P Rod P Rod Q Hot block Hot block Cold block Figure (a) Rod P Rod Q Rod P Rod Q Hot block Cold block Hot block Cold block Figure (b) The ratio of the rate of heat flow between the two blocks in figure (a) to that in figure (b) is A 6 : 1 C 9 : 2 B 3 : 1 D 3 : 2 14. A copper rod of length 50.0 cm is joined end-to-end with an iron rod of length l. The cross-sectional area of both rods is 5.0 cm2 and the composite rod is lagged. At steady state, the end of the copper rod is at 100°C and the end of the iron rod is at 0°C. The temperature at the junction is 60°C. The thermal conductivity of copper and iron is 385 W m–1 K–1 and 50 W m–1 K–1. Calculate (a) the rate of heat flow in the copper rod. (b) the length l of the iron rod. 15. A copper rod of length 18 cm is joined to an aluminium rod of length 50 cm using a layer of glue of thickness 0.5 cm. The cross-sectional area is the same for the copper rod, glue and aluminium rod. (a) Calculate the temperature of the (i) copper rod, and (ii) the aluminium rod at the joint, if the temperatures at the opposite ends of the composite rod are 100°C and 0°C. (b) Hence, sketch a graph to show the variation of temperature along the joined rod. (Thermal conductivity: copper = 380Wm–1 K–1, aluminium=210Wm–1 K–1, glue = 1.3 W m–1 K–1)

304 11 Physics Term 1 STPM Chapter 11 Heat Transfer 11.2 Convection Students should be able to: • describe heat transfer by convection • distinguish between natural and forced convection Learning Outcomes 1. Convection is the transfer of heat by a heated material that moves, carrying heat with it. Convection takes place only in fluids, that is gas or liquids. Warm water moves up Cold water moves down Heat Figure 11.10 2. Figure 11.10 shows water in a beaker being heated. • Water that is heated expands and its density decreases. • Warm water rises carrying heat with it. • Cold water which is denser moves down and in turn gets heated. 3. There are two types of convection, natural convection and forced convection. 4. In natural convection, the convection currents are set up due to gravity. Cold fluid with higher density sinks because its weight is greater than the buoyant force. Hence natural convection does not occur in weightless environment. 5. In forced convention, movement of the fluid is caused by external forces such as a fan or pump. Forced convection increases the rate of heat transfer. 6. When a body cools in still air, still-cooling occurs and it is natural convection. When the body cools in a draught (example a fan blowing), it is forced convention. 7. Newton’s law of cooling states that in a draught, the rate of heat loss is proportional to the temperature excess which is the difference between the body and the surroundings temperature. Rate of heat loss ∝ temperature excess Example 14 Expanded polystyrene are used as insulation for refrigerator walls. Explain how heat transfer through the walls is reduced. Solution: Expanded polystyrene consists of air cells. No heat transfer by convection between the outer and inner surfaces of the walls. Heat transfer by conduction is negligible because polystyrene is a poor thermal conductor. 2017/P1/Q14

305 11 Physics Term 1 STPM Chapter 11 Heat Transfer Quick Check 5 1. Which arrangement would not increase heat transfer by convection? A Open windows of a room. B Ice that floats in a glass of water. C A heater is fixed close to the ceiling of a room to warm up the room. D An air conditioner is fixed close to the ceiling of a room to cool down the room. 2. (a) An electric fire is to be fixed on the wall of a room. Mark with the letter F in the figure below, a suitable position for the electric fire. Explain your answer. Wall (b) Mark with the letter A in the figure, a suitable position for an air-conditioner unit. Explain your answer. 3. With the aid of diagrams, explain the formation of (a) sea breeze during the day, and (b) land breeze during the night. 11.3 Radiation Students should be able to: • describe heat transfer by radiation • use Stefan-Boltzmann equation dQ dt = eσAT4 • define a black body Learning Outcomes 1. Radiation is the transfer of thermal energy from a body to its surrounding in the form of electromagnetic radiations. No medium is required for the transfer of thermal energy. 2. Thermal energy from the Sun is transfer to its surroundings, including the Earth by radiation. 3. Thermal radiations is emitted from all bodies whose temperature is above 0 K. Bodies of low temperatures emit electromagnetic radiations of long wavelength, infrared radiations. The human body emits thermal radiations in the infrared region of the electromagnetic spectrum. 4. Visible light and ultra violet radiations are emitted by high-temperature bodies, example the Sun. 5. The human skin is quite sensitive to thermalradiations. Instruments used to detect thermalradiations include a mercury thermometer with its bulb blackened, a bolometer and a thermopile. Blackened platinum strip Thermal radiation Figure 11.11

306 11 Physics Term 1 STPM Chapter 11 Heat Transfer 6. A bolometer consists of a narrow strip of platinum with its surface blackened. The blackened surface is a good absorber of thermal radiations. Thermal radiations falling on the bolometer causes its temperature and resistance to increase. The increase in resistance is measured by an electrical circuit. Thermal radiations Hot junctions Antimony Cold junctions Bismuth mV Figure 11.12 7. The thermopile is composed of several thermocouples connected in series. Each thermocouple consists of a strip of bismuth and a strip of antimony. One junction of each thermocouple is exposed to the incident thermal radiations. The temperature of this junction increases. The temperature difference between the hot junctions and the cold junction produces an emf which is measured by a millivoltmeter. Stefan-Boltzmann Law 1. The energy radiated by a body per second, or the radiated power is proportional to the surface area A, the nature of the surface and to the fourth power of its temperature T in kelvin. Stefan-Boltzmann law states that the radiated power P is given by P = eσAT4 where e is the emissivity of the surface. 0 ≤ e ≤1, e = 1 for a perfect radiator. σ (pronounced as sigma) is known as Stefan-Boltzmann constant. σ = 5.67 × 10–8 W m–2 K–4 2. A body absorbs radiation from its surrounding and at the same time it radiates radiation. If the temperature of the body is T, and T0 is the temperature of the surroundings, then the net power radiated by the body Pnet = eσA(T4 – T4 0 ) 3. If the initial temperature of the body is higher than the temperature of the surroundings T0 , it is a net emitter. As it loses energy, it temperature decreases until its temperature equals to T0 , the temperature of the surroundings. Thus, thermal equilibrium is said to occur. 4. Similarly, if the initial temperature of the body is lower than the temperature of the surroundings T0 , the body is a net absorbed. As it absorbs radiations, its temperature rises until it equals the temperature T0 of the surroundings. 5. The same value of e is applies for both emission and absorption of radiation. A perfect emitter (e = 1) is also a perfect absorber. Such a body is known as a black body. A black body is defined as a body that absorbs all electromagnetic radiations. 6. For an ideal reflector, e = 0. It does not absorb any radiation. 7. The emissivity e of a dull black surface is close to 1. Hence it is both a good radiator and absorber of radiation. The value of e for a highly polished surface is close to zero. It is a poor thermal radiator and absorber.

307 11 Physics Term 1 STPM Chapter 11 Heat Transfer Temperature of the Sun 1. The solar constant S is the average intensity of radiation from the Sun on the Earth’s surface. The value of the solar constant S is 1.34 × 103 W m–2. Using the value of S and Stefan-Boltzmann law, the surface temperature of the Sun can be estimated. S r R Earth Sun Figure 11.13 2. Let E = energy emitted per second per unit surface area of the Sun. R = radius of the Sun r = average distance of the Earth from the Sun Then total energy radiated per second from the Sun’s surface P = (4πR2 )E This energy would fall on a sphere of radius r, and S is the intensity at a distance r Hence P = (4πR2 )E = (4πr2 )S E = (r2 R2 )S Assuming that the Sun is a black body with surface temperature T, using Stefan-Boltzmann law, P = σ(4πR2 )T4 = (4πR2 ) (r2 R2 )S Surface temperature of the Sun, T = [ r2 R2 ( S σ ) ] 1 4 3. Average distance of the Earth from the Sun, r = 1.50 × 1011 m. Radius of the Sun, R = 7.00 × 108 m. Surface temperature of the Sun, T = [ r2 R2 ( S σ ) ] 1 4 = [ (1.50 × 1011)2 (7.00 × 108 )2 ( 1.34 × 103 5.67 × 10–8) ] 1 4 K = 5 700 K

308 11 Physics Term 1 STPM Chapter 11 Heat Transfer Example 15 A metal sphere has a radius of 5.0 cm and at a temperature of 310 K. (a) Calculate the radiated power if its emissivity e is (i) 0.8 and (ii) 1. (b) What is the net radiated power if the temperature of the surroundings is 290 K, if e = 1? Solution: (a) Apply Stefan-Boltzmann law, Radiated power, P = eσAT4 (i) If e = 0.8, P = (0.8)( 5.67 × 10–8)4π(0.050)2 (310)4W = 13.2 W (ii) If e = 1, P = (1)( 5.67 × 10–8)4π(0.050)2 (310)4W = 16.5 W (b) e = 1, Net radiated power, Pnet = σA(T4 – T4 0 ) = (5.67 × 10–8)4π(0.050)2 (3104 – 2904 ) W = 3.85 W Example 16 The solar constant S = 1.34 × 103 W m–2 is the intensity of radiation from the Sun on the Earth’s surface. The Earth achieves its equilibrium temperature when its radiated power equals the rate of absorption of radiation from the Sun. Assuming that the Earth is a black body, estimate the equilibrium temperature of the Earth. Solution: R Area of Earth normal to radiation from the Sun, A1 = πR2 Earth Radiation from the Sun Let T = equilibrium temperature of the Earth Using Stefan-Boltzmann law, radiated power of the Earth P = σAT4   (surface area of Earth, A = 4πR2 ) = σ(4πR2 )T4 Rate of absorption of radiation from the Sun = (A1 )S = (πR2 )S When the Earth is in thermal equilibrium, radiated power = rate of absorption σ(4πR2 )T4 = (πR2 )S Equilibrium temperature, T = ( S 4σ ) 1 4 = ( 1.34 × 103 4 × 5.67 × 10–8 ) 1 4 K = 277 K

309 11 Physics Term 1 STPM Chapter 11 Heat Transfer Quick Check 6 1. Three cans X, Y and Z are identical except that the surface of can X is covered with lamp soot, the surface of can Y is highly polished and the surface of can Z is painted white. The same amount of water at 90o C is poured into the can. After 10 minutes, the temperature of water in which can (i) is the lowest? (ii) is the highest? Give reasons for your answer. 2. A metal sphere of radius 2.50 cm is heated and then allowed to cool in a room. The room temperature is 30o C. The emissitivity of the sphere’s surface is 0.67. (a) Calculate the net rate of thermal energy radiated from the sphere when its temperature is (i) 330o C, and (ii) 165o C. (b) Discuss what would be the final temperature of the sphere? 3. A spherical bulb israted 120 W and has a radius of 5.00 cm. The emissitivity of the surface of the bulb is 0.90. Estimate the temperature on the surface of the bulb. 11.4 Global Warming Students should be able to: • explain the greenhouse effect and thermal pollution • suggest ways to reduce global warming Learning Outcomes 1. The Earth receives energy from the Sun mainly in the form of ultraviolet, visible light and short wavelength infrared due to the high temperature of the Sun. 2. The surface temperature of the Earth (average of about 300 K) being lower than the surface temperature of the Sun, emits the longer wavelength infrared. Water vapour and CO2 in the atmosphere Earth Short wavelengths from the Sun Long wavelength infrared Figure 11.14 3. These long wavelength infrared radiations are absorbed by water vapour, carbon dioxide, methane and ozone (known as greenhouse gases) in the Earth’s atmosphere. Most of these radiations are radiated back to the Earth by the atmosphere (Figure 11.14). This causes the temperature of the Earth to be higher than it would be otherwise. This phenomenon is known as greenhouse effect. 2016/P1/Q215, 2017/P1/Q15

310 11 Physics Term 1 STPM Chapter 11 Heat Transfer Short wavelengths from the Sun Long wavelength infrared trapped in greenhouse Glass Figure 11.15 4. The situation illustrated in Figure 11.14 is identical to that of a greenhouse shown in Figure 11.15. Radiations of short wavelengths are able to pass through the glass and are absorbed by the plants and soil. Glass is opaque to the long wavelength infrared emitted by objects in the greenhouse. The trapped infrared radiation and air warmed by the hot interior causes the temperature in the greenhouse to rise. 5. Emission of carbon dioxide from power stations and motor vehicles due to the burning of fossil fuels increases the amount of carbon dioxide in the atmosphere. This causes a rise in the temperature of the Earth due to greenhouse effect. Ways to Reduce Global Warming 1. The effects of global warming are: • Rising sea levels • Decrease in snow in the Northern hemisphere • Expansion of deserts • More frequent occurrence of extreme weather conditions such as droughts, heavy rainfall and snowfall. • Reduce in crop yield • Extinction of plant and animal species due to extreme climate • Human health – example increase in dengue and malaria cases 2. Global warming is mainly due to the increase of greenhouse gases in the atmosphere. To reduce global warming effort must be made to reduce the emission of greenhouse gases. 3. Natural processes that remove greenhouse gases include • Condensation and precipitation of water vapour • Mixing of atmospheric gases into the ocean • Chemical reactions in the atmosphere, example methane is oxidized by hydroxyl radical, OH4. Governments around the world have agreed under the Kyoto Protocol to lower the emission of greenhouse gases. Efforts made include • Generation of electricity using renewable energy sources such as wind turbines and solar panels • Promoting carbon capture and storage • Promote the use of electric vehicles

311 11 Physics Term 1 STPM Chapter 11 Heat Transfer Quick Check 7 1. What can you do to fight climate change? 2. Which radiation is absorbed by carbon dioxide gas in the atmosphere? A Visible light C Ultraviolet B Infrared D Microwaves Important Formulae 1. dQ dt = –kA dθ dx 2. When rods are joined in series and insulated, temperature gradient, dθ dx ∝ 1 k 3. Thermal resistance, Rθ = l kA 4. Stefan – Boltzman law: P = eσAT4 Pnet = eσA(T4 – T4 0 ) STPM PRACTICE 11 1. A wooden box of total surface area 1.2 m2 is used to keep 6.0 kg ice. The initial temperature of the ice is 0o C. The thickness of the wall is 1.5 cm. The temperature of the room is 32o C. How long does it takes for all the ice in the box to melt? (Thermal conductivity of wood = 0.52 W m–1 K–1, specific latent heat of fusion of ice = 3.3 × 105 J kg-1) A 510 s B 1100 s C 3900 s D 7800 s 2. A uniform metal rod of length L is insulated except for the middle section as shown in the figure. Heat flow through the rod. At the steady state, the temperature T1 > T2 . T1 T2 Which graph correctly shows the variation of temperature T with distance x from the hot end? A T L x T1 T2 0 C T L x T1 T2 0 B T L x T1 T2 0 D T L x T1 T2 0 3. A copper rod and a silver rod have the same mass but different cross-sectional areas, A1 and A2 respectively. The rods are insulated. When the temperature differences acrossthe rods are the same, the rates of heat flow in the rod are equal. The densities of copper and silver are ρ1 and ρ2 . The thermal conductivities of copper and silver are k1 and k2 respectively. What is the value of k1 /k2 ?

312 11 Physics Term 1 STPM Chapter 11 Heat Transfer A A2 2 ρ2 A1 2 ρ1 C A2 ρ1 A1 ρ2 B A2 ρ2 A1 ρ1 D A1 ρ2 A2 ρ1 4. A composite rod consists of two different metals P and Q of the same length joined in series. The rod is perfectly insulated and the temperatures of the end are 100o C and 30o C when steady state is attained. The variation of temperature θ with distance x is as shown in the figure below. θ / °C 100 80 30 0 x Insulation P Q Which of the following deduction is correct? A Thermal conductivity of P is greater than that of Q B Rate of heat flow in P is less than in Q C Temperature gradient along P equal to temperature gradient along Q D Rate of heat loss from P is greater than from Q 5. A covered styrofoam cup is filled with ice at 0°C. The thickness of the cup is 1.5 × 10–3 m, its effective conducting surface area is 0.02 m2 , and the temperature outside the cup is 30°C. If conduction is the only way to transfer the heat, what is the mass of ice that will melt in 200 seconds? [Thermal conductivity of styrofoam = 0.010 W m–1 K–1; specific latent heat of fusion of ice = 3.35 × 105 J kg–1.] A 1.2 × 10–5 kg B 1.2 × 10–4 kg C 2.4 × 10–3 kg D 2.4 × 10–2 kg 6. Which of the following does not reduce global warming? A Use of fossil fuel B Increase use of public transport C Photovoltaic power station D Use of electric power vehicles 7. The figure below shows Searle’s apparatus for the measurement of thermal conductivity of a good conductor. T4 x Water Metal rod Thermometer T3 T2 T1 N Lagging Electric heater Cooling coil Constant pressure head By keeping other variables constant, the temperature gradient along the metal rod is increased if A the distance x between thermometers T3 and T4 is increased. B the current through the electric heater is reduced. C the height of the constant pressure head is increased. D the distance x between thermometers T3 and T4 is decreased. 8. Two parallel-sided slabs P and Q are of different materials. The thermal conductivity of P is k, and of Q is 2k. P Heat θ1 θ2 θ1 > θ2 Q In steady state, opposite faces of the slabs are at θ1 °C and θ2 °C (θ1 > θ2 ). What is the fraction of the total rate of heat flow through the composite slab passing through P? A 1 4 B 1 3 C 1 2 D 2 3

313 11 Physics Term 1 STPM Chapter 11 Heat Transfer 9. The graph below shows the variation of temperature along a composite rod which consists of three rods, X, Y and Z joined in series and of thermal conductivities 400 W m–1 K–1, 100 W m–1 K–1 and 200 W m–1 K–1. The composite rod is welllagged and the ends are kept at 100°C and 30°C. Temperature /°C (i) (ii) (iii) Distance 100 60 50 30 0 100°C 30°C The arrangement of the three rods is (i) (ii) (iii) A X Y Z B Y Z X C Z Y X D Y X Z 10. A sphere P of radius r is maintained at a temperature of T, and another sphere Q of radius 2r is maintained at a temperature of 2T. Assuming that the spheres have the same emissivity, what is the value of the ratio radiate power of P : radiate power of Q? A 1 : 4 C 1 : 16 B 1 : 8 D 1 : 64 11. A black body has a constant temperature of T and its surface area is A. The energy radiated by the black body in time (∆t) is directly proportional to A AT2 (∆t) B AT3 (∆t) C AT4 (∆t) D AT(∆t) 12. A hot metal sphere is placed in a draught and allowed to cool by forced convection. The temperature of the surroundings is 30o C. The rate of heat loss from the sphere when its temperature is 100o C is R1 and at 50o C is R2 . What is the value of R1 R2 ? A 2.0 C 12.3 B 3.5 D 150 13. (a) Give one condition for the conduction of heat in a medium. (b) Discuss how energy consumption efficiency of a cooking utensil is affected by the values of (i) its thermal conductivity, and (ii) heat capacity. (c) The base of a steel cooking pot has an area of 0.032 m2 and thickness 2.5 mm. Calculate the rate of heat flow through the base of the pot for a temperature difference of 500o C. (Thermal conductivity of steel is 67 W m–1 K–1). 14. (a) Distinguish between the terms thermal energy and heat. (b) Describe the main mechanism of thermal conduction in (i) metals (ii) non-metal solids. (c) The figure below showsthe apparatus used to determine the thermal conductivity of brass. T3 T4 T1 T2 Cold water Warm water Brass rod Steam Steam d x Insulation One end of the brass rod is heated by steam, and heat conducted through the rod is removed from the other end using water flowing in cooling coils. The following readings of the thermometers are recorded at the steady state. T1 = 82.5 o C   T2 = 60.0 o C T3 = 37.0 o C   T4 = 30.0 o C Mass of water collected in the beaker in 5.00 minutes = 120 g The diameter of the rod d = 2.80 cm, and the distance x between the thermometers T1 and T2 is 12.5 cm. Calculate (i) rate of heat flow along the brass rod, (ii) thermal conductivity of brass. (Specific heat capacity of water = 4200 J kg–1 K–1)

314 11 Physics Term 1 STPM Chapter 11 Heat Transfer 15. (a) Describe in molecular terms, thermal conduction in a non-metallic solid. (b) In Lee’s experimental determination of the thermal conductivity of a poor conductor, the sample used has special features. Describe and explain the features of the sample. (c) An air-conditioned room has a brick wall of thickness 0.20 m with a layer of insulation of thickness 0.025 m on the inside wall. The total effective surface area of the walls is 36.0 m2 . The temperature outside the room is 30°C and inside the room is 23°C. Calculate (i) rate of heat conducted through the walls, if there is no insulation, (ii) rate of heat conducted through the walls with insulation as a fraction of the answer in (i), (iii) the temperature at the brick-insulation interface. [Thermal conductivity of brick = 0.84 W m–1 K–1, insulation layer = 0.04 W m–1 K–1] 16. (a) Describe themechanismofheat conduction in a metal. (b) Two metal rods P and Q of the same cross sectional area, A are joined in series. The length of rod P is 2L and the length of rod Q is L. The temperatures at the end of rod P is 100°C, at the interface is 25°C and at the end of rod Q is 0°C as shown in the figure. The rods are perfectly insulated. The thermal conductivity of rods P and Q are kP and kQ respectively. 100°C P Q 0°C 25°C Insulator (i) Write an expression for the rate of heat flow in the joint rod. State two conditions for the expression to be valid. (ii) Skectch a graph to show the temperature gradient along the joint rod. Discuss which rod is a better thermal conductor. (iii) Determine the value of the ratio of kP to kQ. 17. Two metal rods, A and B are of the same length but the diameter of A is twice that of B. One end of each rod is kept at a temperature θ°C > 0 °C and the other end in contact with melting ice. The rods are lagged and in the same time duration, the mass of ice melted by A and B are 10 g and 20 g respectively. Calculate the value of the ratio of the thermal conductivity of A to that of B. 18. The Earth has a crust of thermal conductivity 2 W m–1 K–1. On average, the temperature of the Earth’s crust increases by 30°C for each kilometer of thickness. Estimate the rate of heat lossfrom the Earth’ssurface through heat conducted through its crust. (Radius of Earth = 6 × 106 m). 19. There are similarities between thermal conduction and electrical conduction. (a) Write the equations for (i) rate of heat flow dQ dt across a slab of cross-sectional area A, thermal conductivity σ when the temperature gradient is dT dx . (ii) the electric current in a conductor of electrical conductivity σ, crosssectional area A and electric potential gradient dV dx . (b) By comparing the two equations, name the electric quantities which are analoguous to (i) heat (ii) temperature. 20. (a) A bolometer is used to detect thermal radiation from a hot body. With the aid of a labelled diagram, describe the principles used. (b) The reading registered by a bolometer is proportional to the intensity of the incident thermalradiation. A metalsphere is at a constant temperature. The reading of the bolometer when it is at distances of r from the sphere is X, and its reading is Y when the distance is 2r. Find the ratio of X Y .

315 11 Physics Term 1 STPM Chapter 11 Heat Transfer 21. (a) (i) Write an equation for the current I in a conductor of resistance R when the potential difference across the conductor is V. (ii) A lagged metal rod of length l and cross-sectional area A has thermal conductivity k. The temperature at the ends are θ1 and θ2 , (θ1 > θ2 ). Write an equation for the rate of heat flow. (iii) By comparing the equationsin (i) and (ii) above, deduce an expression for the thermal resistance Rθ of a rod. (b) The base of an aluminium kettle is 2.4 mm thick and of cross-sectional area 0.020 m2 . On the inside surface, it is deposited a layer of scale of thickness 0.50 mm. When water in the kettle boils at 100°C, heat is conducted at a rate of 2.0 kW through the base. Calculate the temperature on the outer surface of the base. (Thermal conductivity : aluminium = 240 W m–1 K–1, scale = 1.0 W m–1 K–1). 22. (a) Two metalrods, P and Q of the same length and cross-sectional area but different thermal conductivity are joined end-toend and lagged. The left end of the rod P is kept at a highertemperature than the right end of the rod Q. When atsteady state, the temperature, θ along the composite rod is as shown below. P θ 0 x Q Compare the following quantities for the two rods, P and Q. (i) Magnitude of the temperature gradient (ii) Rate of heat flow (iii) Thermal conductivity (b) A refrigerator wall has a total surface area of 4.0 m2 and thickness 50 mm. The cooling mechanism absorbs heat at a rate of 100 W. If the room temperature is 25°C, what is the lowest temperature achieved inside the refrigerator? (Thermal conductivity of refrigerator wall = 0.039 W m–1K–1) 23. (a) What is meant by thermal conductivity? (b) A metal rod of length L is well-lagged and at an initial temperature T0. The temperature of one end of the rod is maintained at T1 > T0 . The other end is kept at T0 . On the same axes, sketch a series of graphs to show the variation of temperature along the rod at different times until steady state is achieved. (c) Show that for a composite wall which consists of 3 layers, the rate of heat flow per unit area at the steady state is given by q = T0 – T3 3 Σi = 1 ( xi ki ) where xi and ki is the thickness and thermal conductivity of the i th layer, T0 and T3 the steady temperatures of the opposite faces of the composite wall. (d) A copperrod and a steelrod each of length 60 cm and cross-sectional area 5.0 cm2 are joined end-to-end. The exposed end of the copper rod is kept at 100°C, and that of the steel rod at 0°C. The sides of the composite rod is lagged.When steady state is achieved, calculate (i) temperature at the interface (ii) the total heat flow in one minute across a cross-section of the rod. (Thermal conductivity : Copper = 380 W m–1 K–1, steel = 46 W m–1 K–1) 24. (a) Explain what is meant by a black body. (b) The heating element of an electric fire has a diameter of 0.80 cm and length of 25.0 cm. It is rated as 800 W. (i) Assuming that the heating element radiates as an isolated black body, find the equilibrium temperature of the heating element. (ii) The heating element actual does not radiate as a black body. Discuss how this affects the equilibrium temperature. (iii) The calculation neglect the heat absorbed by the heating element

316 11 Physics Term 1 STPM Chapter 11 Heat Transfer from the room. How would the heat absorbed affect the equilibrium temperature? 25. (a) Explain what is meant by greenhouse effect. (b) The Sun may be assumed to be a black body with surface temperature of 6000 K. The radius of the Sun R = 7.00 × 108 m. The Earth is at a distance r = 1.50 × 1011 m from the Sun. Calculate (i) the rate of thermal energy radiated from the Sun’s surface per second. (ii) the intensity ofradiation from the Sun on the Earth’s surface. (iii) the equilibrium temperature of the Earth. (c) Give one reason why the temperature of the Earth is higher than the equilibrium temperature calculated above. ANSWERS 1 1. D 2. C 3. C 4. B 5. C 6. C 7. B 8. D 9. C 10. C 2 1. B 2. B 3. C 4. A 5. B 6. Long, >100 mm apart: temperature difference bigger. Lagged: dQ dt uniform along rod, dθ dx constant. 3 1. D 2. B Refer to page 289 3. (a) Many free electrons. Electrons responsible for electric and thermal conduction. 4. 5.3°C dQ dt = mL t 5. 0.030 m dQ dt = –kpolyA( Dq Dx ) = –kiceA( Dq 0.50) 6. 6.00 × 103 W Use dQ dt = –kA( dq dx ) Temperature gradient < the assumed value. Actual temperature of the outer surface < 35°C, and actual temperature of the inner surface > 20°C. 4 1. B: dQ dt = –kA dθ dx Q = (0.22)(1.20)( 30 – 0 0.025 )(3600) J = 1.14 MJ 2. C: dQ dt = –kA dθ dx , same dθ dx 3. B: dQ dt = –kA dθ dx , same dQ dx a b = 240 400 = 0.60 4. A 5. D 6. D 7. A 8. B 9. D 10. B 11. C 12. A 13. C 14. (a) 15.4 W dQ dt = –k1 A( dθ dx )1 (b) 0.0974 m dQ dt = –k2 A( dθ dx )2 15. (a) 92.9°C, 35.3°C dQ dt = Temperature difference Total thermal resistance and dQ dt = –kA dθ dx for copper; then Al. (b) 0 100 92.9 35.3 1818.5 68.5 x / cm °C 5 1. C Cold air or water is more dense and move down.

317 11 Physics Term 1 STPM Chapter 11 Heat Transfer 2. (a) Air above the electric fire is heated and rises. Cool air moves in and in turn gets heated. Convection currents set up as shown in the figure. Cool air Hot air A F (b) Cooled air is denser and moves down from A. Warm air that moves in get cooled. Convection current set up as shown. 3. Specific heat capacity of land is lower than that of water. (a) Warm air Cool air During the day Water Sea breeze (cool air) Land During the day, the land heats up faster, and is hotter than the sea. The warm land heats the air above it. Hot air rises. Cooler air above the sea flows in. Sea breeze is formed. (b) Cool air Warm air At night Water Sea breeze (cool air) Land At night, the land cools faster than the sea. The warm water in the sea heats the air above it. Warm air rises, cooler air above the land moves towards the sea. Land breeze is formed. 6 1. (i) Can X. Reason: its surface which is covered with lamp soot is dull black. Dull black surface is the best radiator. (ii) Can Y: Highly polished surface is the worst radiator. 2. (a) (i) T = (330 + 273) K = 603 K, T0 = (30 + 273) K = 303 K Net radiated power, Pnet = eσA(T4 – T4 0 ) = (0.67)(5.67 × 10–8)4π(2.50 × 10–2)2 (6034 – 3034 ) W = 36.9 W (ii) T = (165 + 273) K = 438 K, T0 = (30 + 273) K = 303 K Pnet = eσA(T4 – T4 0 ) = 8.47 W (b) Final temperature = 30°C, temperature of the room. Reason: When T = T0 , Pnet = eσA(T4 – T4 0 ) = 0 3. Use P = eσAT4 , e = 0.90, A = 4πR2 T = ( P eσ(4πR2 )) 1 4 = 500 K 7 1. • save energy use at home • drive less • recycle • save water • plant a tree • reduce food intake • reduce waste. 2. B Refer to page 310 STPM Practice 11 1. C : Rate of heat flow through the wall = kA dθ dx = (0.52)(1.2)( 32 – 0 0.015 ) W = 512 W Time taken = ml 512 = (6.0)(3.3 × 105 ) 512 s = 3900 s 2. D : Curve for non-insulated section. 3. A : Mass, m = A1 L1 ρ1 = A2 L2 ρ2 L1 L2 = A2 ρ2 A1 ρ1 k1 A1 ∆θ L1 = k2 A2 ∆θ L2 k1 k2 = ( A2 L1 A1 L2 ) = A2 2 ρ2 A1 2 ρ1 4. A : Rate of heat flow is the same in both P and Q. Thermal conductivity inversely proportional to gradient. 5. C : ml t = dQ dt = –kAdθ dx m = (0.010)(0.02)(30/1.5 × 10–3)(200) 3.35 × 105 kg = 2.4 × 10–3 kg 6. A : Burning of fossil fuel emits CO2 . 7. C : When the pressure head is increased, rate of heat removed from the rod increases.

318 11 Physics Term 1 STPM Chapter 11 Heat Transfer 8. B : Rate of heat flow ∝ k 9. D : Gradient of graph ∝ 1 k 10. D : P1 = eσAT4 , P2 = eσ(4A)(2T)4 P1 : P2 = 1 : 64 11. C : P = σAT4 Energy = P(∆t) 12. B : Rate of heat loss ∝ (T – T0 ) R1 R2 = (100 – 30) (50 – 30) = 3.5 13. (a) Condition: There must be a temperature difference across the medium. (b) (i) High value of thermal conductivity so that heat is transferred quickly through the wall of the utensil. (ii) Low value of heat capacity – less heat is absorbed by the utensil for every degree rise in temperature. (c) Rate of heat flow = kAdθ dx = (67)(0.032)( 500 0.0025 ) W s–1 = 4.3 × 105 W 14. (a) Thermal energy is the internal energy of an object due to the kinetic energy of the atoms or molecules. Heat is the energy transferred between two points in a medium due to a temperature difference between the points. (b) (i) In metals: When one end of the metal is heated, free electrons in the metal gain kinetic energy and diffuse from the hot end to the cold end. Heat is transferred when these free electrons collide with ions of the metal. (ii) In non-metals vibration of atoms at the hot end increases. Waves are set up and propagate across the solid. Thermal energy is transferred. (c) (i) Rate of heat flow = ( m t )c(∆θ) = ( 0.120 5.0 × 60) 4200(37.0 – 30.0)W = 11.8 W (ii) Rate of heat flow = kAdθ dx Thermal conductivity, k = 11.8 p(0.0142 )(82.5 – 60.0)/0.125 W m–1 K–1 = 106 W m–1 K–1 15. (a) Refer to page 279 (b) Special feature: thin disc which is not lagged and with large diameter. Thin: in order for the steady state to be achieve in a short time. Not lagging requires because surface area of the sides is negligible compared to cross sectional area (large diameter). Heat loss from the sides is negligible. (c) (i) dQ dt = –kA dθ dx = (0.84)(36.0)( 30 – 23 0.20 ) W = 1.06 kW (ii) dQ dt = (∆θ) x1 + x2 k1 A k2 A = (30 – 23)(36.0) 0.20 + 0.025 0.84 0.040 W = 292 W = ( 292 1060 ) = 0.275 of rate without insulation. 16. (a) Diffusion of free electrons from the hot end to the cold end. Vibration of metal atoms set up waves which propagate through the metal. (b) (i) dQ dt = –kP A 25 – 100 2L = 75kP A 2L , or dQ dt = –kQA 0 – 25 L = 25kQA L Conditions: • Linear flow of heat • no loss of heat form the sides of the rod (ii) Temperature gradient (°C m–1) x 2L P Q 0 3L k ∝ 1 temperature gradient , hence kQ > kP Q is the better thermal conductor (iii) kP kQ = 63 94 = 0.67 17. k1 k2 = 1 8 dQ dt = –kA dθ dx = mL t 18. 2.7 × 1013 W dQ dt = –kA dθ dx

319 11 Physics Term 1 STPM Chapter 11 Heat Transfer 19. (a) (i) dQ dt = –σ A dT dx (ii) I = σ A dV dx , I = V R , R = l σA (b) (i) Electric charge (ii) Electric potential 20. (a) Blackened platinum strip Thermal radiation Principles used: Black surface – good absorber Increase in resistance when its temperature increases. (b) Intensity ∝ 1 r2 X Y = (2r)2 r2 = 4 21. (a) (i) I = V R (ii) dQ dt = –kA (θ2 – θ 1 ) l (iii) Rθ = l kA (b) 151°C dQ dt = (θ – 100) l 1 + l 2 k1 A k2 A 22. (a) (i) ( dθ dx )P > ( dθ dx )Q , Temperature gradient of P is greater. (ii) dQ dt , same rate of heat flow. (iii) kP > kQ (b) –7.1°C 23. (a) Refer to page 280 (b) t 1 < t 2 < t 3 t 1 t 2 t 3 T1 L T0 Temperature Steady state Distance 0 (c) Use dQ dt = Temperature difference Total thermal resistance (d) (i) 89.2 °C (ii) 205 J Heat = dQ dt × time 24. (a) Refer to page 306 (b) (i) At equilibrium temperature, Radiate power = power supplied P = σAT4 = 800 W σ(2πrl)T4 = 800 W T = 1220 K (ii) Power radiated < 800 W T < 1220 K (iii) Room temperature is about 300 K Pnet = σA(12204 – 3004 ) ≈ σA(12204 ) Hence effect on T is negligible 25. (a) Refer to page 310 (b) (i) P = σAT4 = 4.52 × 1026 W (ii) Intensity on the Earth’s surface S = P 4πr2 = 4.52 × 1026 4π(1.50 × 1011)2 W m–2 = 1.60 × 103 W m–2 (iii) At equilibrium temperature T, radiated power = absorption rate σ(4πa2 )T4 = S(πa2 ) (a = radius of Earth) T = 290 K (c) Greenhouse effect.

320 Paper 1 Kertas 1 (1 hour 30 minutes) (1 jam 30 minit) Section A [15 marks] Bahagian A [15 markah] Instructions There are fifteen questions in Section A. Each question is followed by four choices of answer. Select the best answer. Answer all questions. Marks will not be deducted for wrong answers. Arahan Ada lima belas soalan dalam Bahagian A. Setiap soalan diikuti dengan empat pilihan jawapan. Pilih jawapan yang terbaik. Jawab semua soalan. Markah tidak akan ditolak bagi jawapan yang salah. 1. The following measurements are obtained in an experiment. L = (10.0 ± 0.1) cm and H = (2.0 ± 0.1) cm. Which expression has the greatest percentage uncertainty? Bacaan berikut diperolehi dalam suatu eksperimen. L = (10.0 ± 0.1) cm dan H = (2.0 ± 0.1) cm. Ungkapan yang manakah mempunyai peratus ketakpastian tertinggi? A (L – H) C LH B (L + H) D L H 2. A particle X moves from the point O to a point S via the point Q at constant speed. Another particle Y moves from point O to the point S via the point P with the same speed as shown in the diagram. Suatu zarah X bergerak dari titik O ke titik S melalui titik Q pada laju seragam. Suatu zarah yang lain Y bergerak dari titik O ke titik S melalui titik P pada laju yang sama seperti ditunjukkan dalam gambar rajah. Q S O P Which statement is correct about the magnitude and direction of the average acceleration of the particles X and Y? Pernyataan yang manakah yang betul tentang magnitud dan arah pecutan purata zarah X dan Y? Magnitude of acceleration Direction of acceleration Magnitud pecutan Arah pecutan A Same / Sama Same / Sama B Same / Sama Different / Berbeza C Different / Berbeza Same / Sama D Different / Berbeza Different / Berbeza STPM Model Paper (960/1)

321 Physics Term 1 STPM STPM Model Paper (960/2) 3. The displacement-time graph of a body is shown below. Graf sesaran-masa suatu jasad ditunjukkan di bawah. 0 t 1 t 2 t 3 Displacement Sesaran Time, t Masa When is the body (i) stationary, and (ii) accelerating? Bilakah jasad itu (i) pegun, dan (ii) memecut? Stationary Accelerating Pegun Memecut A t = 0 and t 3 0 t < t 1 B t = 0 and t 3 t 2 t < t 3 C t 1 t < t 2 0 t < t 1 D t 1 t < t 2 t 2 t < t 3 4. An object is dropped from height and as it falls it experiences air resistance. Which statement is correct about the acceleration of the object? Suatu objek dilepaskan dari suatu ketinggian dan semasa jatuh objek itu mengalami rintangan udara. Pernyataan yang manakah yang betul tentang pecutan objek itu? A Acceleration is constant but less than the acceleration of free fall. Pecutannya malar tetapi kurang daripada pecutan jatuh bebas. B Acceleration equals the acceleration of free fall and air resistance is constant. Pecutannya sama dengan pecutan jatuh bebas dan rintangan udara adalah malar. C Acceleration decreases because the weight of the object increases. Pecutannya semakin berkurang sebab berat objek semakin bertambah. D Acceleration decreases because air resistance increases. Pecutannya semakin berkurang sebab rintangan udara bertambah. 5. The force F on a body which is initially at rest on a smooth horizontal surface varies with time t as shown in the graph below. Suatu jasad yang pada awalnya pegun di atas permukaan ufuk yang licin dikenakan daya F yang berubah dengan masa t seperti di tunjukkan dalam graf di bawah. 0 t F Which graph correctly shows the variation of the velocity v of the body with time t? Graf yang manakah menunjukkan dengan betul ubahan halaju v jasad itu dengan masa t? A 0 t v B 0 t v C 0 t v D 0 t v

322 Physics Term 1 STPM STPM Model Paper (960/2) 6. A load is dropped from rest from a height and falls onto a pile in the ground. The pile is driven a distance into the ground before it stops. Suatu pemberat dilepaskan dari suatu ketinggian dan jatuh di atas sebatang kayu yang tercecak pada tanah. Kayu itu bergerak suatu jarak tertentu ke dalam tanah sebelum berhenti. Pile Kayu Load Pemberat Ground Tanah Which statement is correct? Pernyataan yang manakah adalah betul? A Collision between the load and pile is elastic. Perlanggaran antara pemberat dan kayu adalah perlanggaran kenyal. B Momentum of load and pile after collision equals momentum of pile before collision. Jumlah momentum pemberat dan kayu selepas perlanggaran adalah sama dengan momentum pemberat sebelum perlanggaran C Kinetic energy of pile immediately after collision equals loss of potential energy of load. Tenaga kinetik kayu sebaik sahaja selepas perlanggaran adalah sama dengan kehilangan tenaga keupayaan pemberat. D Work done against resistance of the ground equals kinetic energy of load and pile immediately after collision. Kerja yang dilakukan menentang rintangan tanah sama dengan tenaga pemberat dan kayu baik sahaja selepas perlanggaran. 7. Two loads of 80 N and 50 N hang from a rope that passes over a pulley which is attached to a motor. The pulley has a diameter of 0.20 m and when it is being rotated at 5.0 rad s–1 the loads are stationary. Pemberat 80 N dan 50 N tergantung pada hujung tali yang melalui satu takal yang diputarkan oleh sebuah motor. Diameter takal ialah 0.20 m dan apabila takal berputar pada 5.0 rad s-1, pemeberat-pemberat adalah pegun. 5.0 rad s–1 50 N 80 N What is the output power of the motor? Berapakah kuasa output motor? A 15 W C 40 W B 30 W D 45 W

323 Physics Term 1 STPM STPM Model Paper (960/2) 8. A uniform beam of weight W rests on a smooth peg P. The normal reaction of the peg P on the beam is N. The lower end of the beam is on a rough floor and the force on this end is F. Which diagram correctly shows the force W, N and F? Sebatang rod yang beratnya W terletak di atas kayu P yang licin. Tindak balas normal kayu P ke atas rod itu ialah N. Hujung bawah rod itu adalah di atas lantai yang kasar dan daya pada hujung bawah itu ialah F. Gambar rajah yang manakah menunujukkan dengan betul daya W, N dan F? A N W Floor Lantai P F C N W P F Floor Lantai B N W P F Floor Lantai D W P F N Floor Lantai 9 A stone of mass m at the end of a string of length l is projected from the point P with a velocity u as shown in the diagram. The stone then moves in a vertical circle. Seketul batu berjisim m diikat pada hujung seutas tali yang panjangnya l. Batu itu dilancarkan dari titik P dengan halaju u seperti ditunjukkan dalam gambar rajah di sebelah. Batu itu kemudian bergerak dalam bulatan tegak. Q P T u Which expression is correct for the tension T in the string when the stone is at the lowest point Q? Ungkapan yang manakah adalah betul bagi tegangan T dalam tali apabila batu di titik terendah Q? A mu2 l + mgl B mu2 l + 2mg C mu2 l + 3mg D mu2 l + 4mgl 10. The mass of the Earth is M and its radius is R. If G is the universal gravitational constant, which expression gives the gravitation field strength at a height h = 2R above the Earth? Jisim Bumi ialah M dan jejarinya R. Jika G ialah pemalar semesta kegravitian, yang manakah adalah ungkapan bagi kekuatan medan graviti pada tinggi h = 2R dari permukaan Bumi? A 2GMR2 B GM 9R2 C GM 3R2 D GM 2R2 11. A copper wire is of length L and diameter d. A steel wire is of length 2L and diameter 2d. The Young’s modulus of steel is twice that of copper. Loads of the same mass hang from the lower ends of the wires. The extension of the copper wire is x, what is the extension of the steel wire? Panjang seutas dawai kuprum ialah L dan diameternya ialah d. Panjang seutas dawai keluli ialah 2L dan diameternya 2d. Modulus Young keluli adalah dua kali nilai modulus Young kuprum. Pemberat-pemberat yang sama digantung daripada hujung bawah kedua-dua dawai itu. Pemanjangan dawai kuprum ialah x, berapakah pemanjangan dawai keluli? A 1 4 x B x C 2x D 4x

324 Physics Term 1 STPM STPM Model Paper (960/2) 12. An ideal gas in a container of volume 2.0 × 10–2 m3 is at a temperature of 300 K and pressure of 1.10 × 105 Pa. How many moles of the gas is in the container? Suhu suatu gas unggul di dalam suatu bekas berisi padu 2.0 × 10–2 m3 ialah 300 K dan tekanan gas ialah 1.10 × 105 Pa. Berapakah bilangan mol gas di dalam bekas itu? A 0.44 moles B 0.88 moles C 1.13 moles D 2.27 moles 13. At room temperature, the molar heat capacity of a gas at constant volume, CV,m is xR. If Cp,m is the molar heat capacity at constant pressure, what is the value of the ratio Cp,m Cv,m ? Pada suhu bilik, muatan haba molar pada isi padu malar CV,m suatu gas ialah xR. Jika Cp,m ialah muatan haba molar pada tekanan malar, berapakah nilai nisbah Cp,m Cv,m ? A x – 1 x B x + 1 x C x x + 1 D x x – 1 14. Two different metal rods P and Q of the same cross sectional area of 5.0 cm2 are joined in series and perfectly lagged. The lengths of the rods are 20 cm and 10 cm respectively. The thermal conductivity of rod P is 200 W m-1 K-1 and the thermal conductivity of rod Q is 400 W m-1 K-1. The temperatures at the ends of the jointed rod are 100o C and 30o C. Luas keratan rentas dua batang logam yang berlainan P dan Q adalah sama, 5.0 cm2 . Dua batang logam itu disambungkan secara siri dan ditebat dengan sempurna. Panjang P dan Q adalah masing-masing 20 cm dan 10 cm. Kekonduksian termal P dan Q adalah masing-masing 200 W m-1 K-1 dan 400 W m-1 K-1. Suhu di hujung-hujung rod gabungan adalah 100o C dan 30o C. 100°C P Q 30°C 20 cm 10 cm What is the rate of heat flow along the joint rod? Berapakah kadar aliran haba sepanjang rod gabungan itu? A 4.7 W B 26.7 W C 28.0 W D 33.4 W 15. Material medium is not required if heat is transferred by Bahan antara tidak diperlukan jika haba dipindah melalui A conduction / kekonduksian B convection / perolakan C radiation / sinaran D convection and radiation / perolakan dan sinaran Section B (15 marks) Bahagian B [15 markah] Answer all the questions in this section. Jawab semua soalan dalam bahagian ini. 16. (a) An object is projected at an angle of 45o to the horizontal. Find the value of the ratio of the maximum height reached H to the range R of the projectile. Suatu objek dilancarkan pada sudut 45o dari ufukan. Tentukan nilai nisbah tinggi maksimum H kepada julat R luncuran itu. [4 marks / 4 markah]

325 Physics Term 1 STPM STPM Model Paper (960/2) (b) A ball is kicked from a distance of 6.00 m from a wall of height 3.00 m. Sebiji bola ditendang pada jarak 6.00 m dari suatu dinding yang tingginya 3.00 m. (i) State the angle of projection so that the ball just clear the wall at the maximum point of its trajectory. [1 mark] Nyatakan sudut luncuran bola supaya bola itu melepasi dinding pada tinggi maksimum lintasan bola. [1 markah] (ii) Find the speed of projection of the ball. [3 marks] Cari laju bola diluncurkan. [3 markah] 17. (a) (i) Define ideal gas. [1 mark] Takrifkan gas unggul. [1 markah] (ii) State the quantities that determine the state of an ideal gas. [1 mark] Nyatakan kuantiti-kuantiti yang menentukan keadaan suatu gas unggul. [1 markah] (b) On the same axes, sketch graphs to show how the pressure p of an ideal gas varies with the volume V when the gas expands each time to twice its initial volume from the same initial state. Pada paksi-paksi yang sama, lakarkan graf untuk menunjukkan perubahan tekanan p suatu gas unggul dengan isi padunya V apabila gas itu mengembang kepada dua kali isi padu asal daripada keadaan asal yang sama. (i) in an isothermal expansion [2 marks] secara pengembangan isotermal [2 markah] (ii) in an isobaric expansion, and [1 mark] secara pengembangan isobarik, dan [1 markah] (iii) an adiabatic expansion. [2 marks] secara pengembangan adiabatik. [2 markah] Section C (30 marks) Bahagian C [30 markah] Answer any two questions in this section. Jawab mana-mana dua soalan dalam bahagian ini. 18. A car accelerates from rest to a speed of 20 m s-1 in 5.0 s. The total mass of the car and driver is 1920 kg. Sebuah kereta memecut daripada pegun kepada laju 20 m s-1 dalam 5.0 s. Jumlah jisim kereta dan pemandu ialah 1920 kg. (a) Name two forces that resist the motion of the car, and state how each force varies when the speed of the car increases. [3 marks] Nyatakan dua daya yang menentang pergerakan kereta, dan nyatakan bagaimana setiap satu daya itu berubah apabila laju kereta bertambah. [3 markah] (b) (i) Calculate the acceleration of the car. [2 marks] Hitungkan pecutan kereta. [2 markah]

326 Physics Term 1 STPM STPM Model Paper (960/2) (ii) What is the resultant force on the car during the acceleration? [2 marks] Berapakan daya paduan pada kereta semasa kereta memecut? [2 markah] (iii) State and explain how the horizontal forward force on the car changes when it accelerates. [3 marks] Nyatakan dan jelaskan bagaimana daya mengufuk ke depan pada kereta berubah apabila kereta memecut. [3 markah] (iv) Draw a labelled diagram to show the directions of the horizontal forces on the car during its motion. [2 marks] Lukiskan gambar rajah berlabel untuk menunjukkan arah daya-daya mengufuk pada kereta semasa kereta bergerak. [2 markah] (c) When the car is travelling at a constant speed of 20 m s-1, its power output is 25 kW. Find the total resistive force. [3 marks] Apabila laju kereta adalah tetap 20 m s-2, kuasa output ialah 25 kW. Tentukan jumlah daya rintangan pada kereta itu. [3 markah] 19. (a) (i) Define Young’s modulus of a material. [1 mark] Takrifkan modulus Young suatu bahan. [1 markah] (ii) A wire is of length L and its cross sectional area is A. Derive an expression that relate the Young’s modulus Y of the material to the force constant k of the wire. [2 marks] Panjang seutas dawai ialah L dan luas keratan rentasnya A. Terbitkan ungkapan yang mengaitkan modulus Young Y bahan dawai itu dengan pemalar daya k dawai. [2 markah] (b) The natural length of a vertical wire is 3.00 m and its diameter is 1.20 mm. Loads are slowly added to the lower end. The extension-force graph of the wire during the loading process up to the point Q is shown below. Panjang asal seutas dawai mencancang ialah 3.00 m dan diameternya 1.20 mm. Beban ditambahkan dengan perlahan di hujung bawah dawai. Graf pemanjangan-daya dawai itu semasa beban ditambah sehingga titik Q ditunjukkan di bawah. 0 Force Daya (N) Extension (mm) Pemanjangan 1.0 8.0 16.0 24.0 2.0 3.0 O P Q (i) State the types of deformation experienced by the wire along OP, and along PQ. Describe in atomic terms the difference in deformations along OP and PQ. [4 marks] Nyatakan jenis canggaan yang dialami dawai itu sepanjang OP, dan PQ. Huraikan dalam sebutan atomatom perbezaan canggaan sepanjang OP, dan PQ. [4 markah] (ii) Calculate the Young’s modulus of the material of the wire. [3 marks] Hitungkan modulus Young bahan dawai itu. [3 markah] (iii) Copy the graph and draw a line to show the variation of extension with force when the force is reduced from the point Q to zero. Hence estimate the permanent extension in the wire. [2 marks] Salin graf di atas dan lukiskan satu garis untuk menunjukkan ubahan pemanjangan dengan daya apabila daya pada dawai dikurangkan dari titik Q ke sifar. Seterusnys anggarkan pemanjangan kekal pada dawai. [2 markah]

327 Physics Term 1 STPM STPM Model Paper (960/2) (iv) Estimate the energy dissipated from the wire in the loading and unloading process. [3 marks] Anggarkan tenaga yang terlesap daripada dawai semasa daya ditambahkan dan daya dikurangkan. [3 markah] 20. (a) (i) Write an equation to represent the first law of thermodynamics. Explain clearly the terms in the equation. [2 marks] Tuliskan persamaan yang mewakili hukum termodinamik pertama. Jelaskan sebutan-sebutan dalam persamaan. [2 markah] (ii) Use the first law of thermodynamics to explain that in an isothermal expansion, although heat is supplied to the gas, the temperature of the gas remains unchanged. What happens to the heat supplied? [2 marks] Gunakan hukum termodinamik pertama untuk menjelaskan bahawa dalam pengembangan isotermal, walaupun haba dibekalkan kepada gas, suhu gas kekal tidak berubah. [2 markah] (b) An ideal gas has a volume of V0 when its temperature is T0 and pressure is p0 . The gas undergoes the following processes: Pada suhu T0 dan tekanan p0 , isi padu suatu gas unggul ialah V0 . Gas itu melakukan proses berikut: I: The pressure of the gas is increased to three times its initial pressure at constant volume. I: Tekanan gas ditambahkan sehingga tiga kali tekanan asal pada isi padu malar. II: The temperature of the gas is then reduced back to T0 at constant pressure. II: Suhu gas kemudian dikurangkan sehingga T0 pada tekanan malar. III: The gas expands isothermally until its volume is V0 and its pressure is p0 . III: Gas itu mengembang secara isotermal sehingga isi padunya V0 , dan tekanannya p0 . Determine in terms of p0 , V0 and T0 , Tentukan dalam sebutan p0 , V0 dan T0 , (i) the temperature of the gas after process I, [2 marks] suhu gas selepas proses I, [2 markah] (ii) the volume of the gas after process II, and [2 marks] isi padu gas selepas proses II dan [2 markah] (iii) the work done by the gas in the complete cycle. [4 marks] kerja yang dilakukan oleh gas itu dalam satu kitar yang lengkap. [4 markah] (c) Draw a p-V graph to illustrate the processes I, II and III. [3 marks] Lukiskan graf p-V untuk menggambarkan proses-proses I, II dan III. [3 markah] ANSWERS PAPER 1 Section A 1. A 2. B 3. C 4. D 5. C 6. D 7. A 8. D 9. D 10. B 11. A 12. B 13. B 14. C 15. C Section B 16. (a) H = v2 sin2 45° 2g R = 2v2 sin45° cos45° g H R = 1 4 tan 45° = 1 4 (b) (i) 45o (ii) H = v2 sin2 45° 2g = 3.00 m Or maximum R = v2 g = 2(6.0) m v = 10.8 m s-1 17. (a) (i) Ideal gas: Gas that obeys the ideal gas equation pV = nRT for all temperatures and pressures. (ii) State of a gas is determine by its temperature, pressure and volume.

328 Physics Term 1 STPM STPM Model Paper (960/2) (b) p V0 V (i) (ii) (iii) 2V0 0 Section C 18. (a) Friction between the road and the tyres. Constant. Air resistance. Increases as speed increases. (b) (i) a = 20 – 0 5.0 m s–2 = 4.0 m s–2 (ii) F = ma = 3.84 × 103 N (iii) Forward force > friction + air resistance As the speed increases, air resistance increases. To maintain a constant acceleration the forward force must increase. (iv) Air resistance Forward force Friction (c) Power = Fv F = 25 × 103 20 N = 1.25 × 103 N 19. (a) (i) Young’s modulus = Stress Strain (ii) Y = F/A e/L = (ke)L Ae = kL A (b) (i) OP: Elastic deformation. When the wire is stretched, atoms are displaced from the mean positions. When force is removed, atoms return to the original positions. PQ: Plastic deformation. When the wire is stretched beyond the elastic limit, atomic planes slide over each other. When the force is removed, atoms do not return completely to their original positions. (ii) k = 1 gradient = YA L Y = ( 20 2.5 × 10–3 )( 20 π(0.60 × 10–3)2 )N m–2 = 2.1 × 1010 N m–2 (iii) 0 Force (N) Extension (mm) 1.0 8.0 16.0 24.0 2.0 3.0 0.3 O P Q Permanent extension = 0.3 mm (iv) Energy dissipated = area between loading and unloading graphs = 5 × 10–3 J 20. (a) (i) Q = ∆U + W Q : heat supplied, ∆U : change in internal energy W : work done by the gas (ii) Isothermal expansion, temperature constant. ∆U = 0 Q = 0 + W Heat supplied Q used to do work against external pressure when the gas expands. (b) (i) I: V = constant, p ∝ T p1 = 3p0 , hence T1 = 3T0 (ii) II: p = constant, V ∝ T When T1 = 3T0 , V = V0 Hence when T2 = T0 , V2 = V0 3 (iii) I : ∆V = 0, W1 = 0 II : W2 = (3p0 )(V0 3 – V0 ) = – 2p0 V0 III: W3 = nRT ln ( V2 V1 ) and (p0 V0 = nRT0 ) = p0 V0 ln ( V0 V0 /3 ) = p0 V0 (ln3) W = W1 + W2 + W3 = (ln3 – 2) p0 V0 = – 0.90p0 V0 II III I 3p0 p p0 V0 /3 V0 V 0

QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Mass Jisim m kg Length Panjang l m Time Masa t s Electric current Arus elektrik I A Thermodynamic temperature Suhu termodinamik T K Amount of substance Amaun bahan n mol Other Quantities Kuantiti Lain Distance Jarak d m Displacement Sesaran s, x m Area Luas A m2 Volume Isi padu V m3 Density Ketumpatan ρ kg m–3 Speed Laju u, v m s–1 Velocity Halaju u, v m s–1 Acceleration Pecutan a m s–2 Acceleration of free fall Pecutan jatuh bebas g m s–2 Force Daya F N Weight Berat W N Momentum Momentum p N s Work Kerja W J Energy Tenaga E, U J Potential energy Tenaga keupayaan U J Heat Haba Q J Power Kuasa P W Pressure Tekanan p Pa Torque Tork  N m Gravitational constant Pemalar graviti G N kg–2 m2 Gravitational field strength Kekuatan medan graviti g N kg–1 Gravitational potential Keupayaan graviti V J kg–1 Moment of inertia Momen inersia I kg m2 Angular displacement Sesaran sudut θ °, rad Angular speed Laju sudut ω, θ • rad s–1 Angular velocity Halaju sudut ω, θ • rad s–1 Angular acceleration Pecutan sudut α, θ •• rad s–2 Angular momentum Momentum sudut L kg m2 rad s–1 Period Tempoh T s Frequency Frekuensi f, v Hz Angular frequency Frekuensi sudut ω rad s–1 Wavelength Panjang gelombang λ m Electric charge Cas elektrik Q, q C Current density Ketumpatan arus J A m–2 Electric potential Keupayaan elektrik V V Electric potential difference Beza keupayaan elektrik V V Electromotive force Daya gerak elektrik ε, E V Summary of Key Quantities and Units 329

Physics Term 1 STPM Summary of Key Quantities and Units QUANTITY KUANTITI BASE QUANTITY KUANTITI ASAS SYMBOL SIMBOL UNIT UNIT Resistance Rintangan R Ω Resistivity Kerintangan ρ Ω m Conductance Konduktans G S = Ω–1 Conductivity Kekonduksian σ S m–1 = Ω–1 m–1 Electric field strength Kekuatan medan elektrik E N C–1 Permittivity Ketelusan ε F m–1 Permittivity of free space Ketelusan ruang bebas ε0 F m–1 Relative permittivity Ketelusan relatif εr – Capacitance Kapasitans C F Time constant Pemalar masa τ s Magnetic flux Fluks magnet φ Wb Magnetic flux density Ketumpatan fluks magnet B T Self inductance Swainduktans L H Mutual inductance Induktans saling M H Reactance Reaktans X Ω Impedance Impedans Z Ω Permeability Ketelapan μ H m–1 Permeability of free space Ketelapan ruang bebas μ0 H m–1 Relative permeability Ketelapan relatif μr – Force constant Pemalar daya k N m–1 Modulus Young Modulus Young E, Y Pa Tension Tegangan T N Stress Tegasan σ Pa Refractive index Indeks biasan n – Critical angle Sudut genting θc ° Temperature Suhu T, θ K, °C Heat capacity Muatan haba C J K–1 Specific heat capacity Muatan haba tentu c J K–1 kg–1 Latent heat Haba pendam L J Specific latent heat Haba pendam tentu l J kg–1 Molar heat capacity Muatan haba molar Cm J K–1 mol–1 Principle molar heat Muatan haba molar utama CV.m, Cp.m J K–1 mol–1 capacity Molar gas constant Pemalar gas molar R J K–1 mol–1 Boltzmann constant Pemalar Boltzmann k J K–1 Avogadro constant Pemalar Avogadro L, NA mol–1 Thermal conductivity Kekonduksian terma k, λ W m–1 K–1 Planck constant Pemalar Planck h J s Activity of radioactive source Keaktifan sumber radioaktif A s–1, Bq Decay constant Pemalar reputan λ s–1 Half-life Setengah hayat t 1/2 s Atomic mass Jisim atom ma kg Relative atomic mass Jisim atom relatif Ar – Mass of electron Jisim elektron me kg, u Mass of neutron Jisim neutron mn kg, u Mass of proton Jisim proton mp kg, u Molar mass Jisim molar M kg mol–1 Atomic mass unit Unit jisim atom u kg Relative molecular mass Jisim molekul relatif Mr – Proton number Nombor proton Z – Nucleon number Nombor nukleon A – Neutron number Nombor neutron N – 330

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