Calorimetry

CALORIMETRY

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Calorimetry

The branch of physics that deals with the measurement of heat is called calorimetry. The word

calorimetry is coined from the word calorie (in abbreviation ‘cal’).

Calorie is defined as the amount of heat energy required to raise the temperature of one gram of
water by 1 ℃ (particularly from 14.5 ℃ to 15.5 ℃) The device which is used to measure the amount

of heat is called calorimeter.

Mechanical Equivalent of Heat

As the heat is produced from the mechanical work, so quantity of heat can be measured from its

mechanical equivalent.

We have amount of work done is directly proportional to quantity of heat produced

Mathematically,

W∝Q

where W = mechanical work done

Q = quantity of heat produced

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or, W = J Q
where J is proportionality constant. It is called Joule’s constant or Joule’s mechanical equivalent

of heat J = 4.18 Joules /cal

Q.1) What energy in calorie is absorbed by a hail stone of mass 20 gm when drops from a height

of 400 m on the earth’s surface?

Given:

Height (h) = 400 m
Mass (m) = 20 gm = 20 × 10-3 kg

If the potential energy is completely converted into the heat energy, then

W = mgh
= 20 × 10-3 × 9.8 × 400

= 78.4 J

Also, J = 4.2 J cal-1

Q = W = 78.4 = 18.67 cal
J 4.2

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Calorimeter:

The device which is used to measure the amount of heat is
known as calorimeter. It consists of a copper vessel
provided with a copper stirrer. The vessel is kept into a
wooden frame. The gap between the wood and the vessel
is filled with insulating material (like wool, wooden dust
etc). Thus, the calorimeter gets thermally insulated from
the surroundings.

The loss of heat due to radiation is further reduced by polishing the outer surface of the
calorimeter and the inner surface of the wooden frame. The lid is provided with holes for inserting
a thermometer and a stirrer in the calorimeter as shown in fig.

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Specific Heat Capacity

Let ∆Q be the amount of heat absorbed or rejected by a substance of mass m so that the
temperature change is ∆ .

Then, the amount of heat absorbed or rejected is,
∆Q ∝ m . . . . . . . . (i)

and ∆Q ∝ ∆ . . . . . . . (ii)
Combining both eqn (i) and (ii) we get

∆Q ∝ m ∆
or, ∆Q = S m ∆
or, ∆Q = mS ∆
Where S be the proportionality constant and is called specific heat capacity.

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Define Specific Heat Capacity

We have

∆Q = mS ∆

i.e S = ∆Q
m∆

If m =1 kg and ∆ = 1 ℃ or 1 K

Then, S = ∆Q

Hence, specific heat capacity of a substance is defined as the amount of heat required to raise

the temperature of unit mass of substance through 1 ℃ or 1 K.

Its SI unit is Jkg-1K-1 ( or cal g-1 ℃-1). It depends on nature of material.

Q.) The specific heat capacity of alcohol is 2400 Jkg-1 ℃-1. What does it mean?
It means that, 2400 J amount of heat is required to raise the temperature of 1 kg alcohol
through 1 ℃

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Q.) Explain why water is used as the cooling agent in automobile engines?

We know water has high specific heat capacity. That means, it absorbs more quantity of heat

from the engine than other substances for the same fall of temperature. That is why water is

used to cool the engines of the automobile.

Heat Capacity (Thermal Capacity)

Heat capacity of a substance is defined as the quantity of heat required to raise the temperature of

any mass of the substance through 1 ℃.
For a body of mass m whose temperature is raised by ∆ , heat capacity is written as

C = ∆Q = mS∆
∆ ∆

C = mS

where S is the specific heat capacity of a substance.

Its unit is J K-1 in SI and Cal ℃-1 in CGS.

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Principle of Calorimetry

When two substances at different temperatures are mixed together, then the hot substance loses

heat while the cold substance gains heat. If there is no loss of heat to the surroundings, the

amount of heat lost by the hot substance is equal to the heat gained by the cold substance.

i.e. Heat gained by the cold body = Heat lost by hot body
Q.2) During a bout with flu an 80 kg man ran a fever of 39 ℃ instead of the normal body
temperature of 37 ℃. Assuming that the human body has mostly water. How much heat is

required to raise his temperature by that amount.

Given:

Mass (m) = 80 kg we have, = mS ( 2 - 1)
initial temperature ( 1) = 37 ℃ Q = 80 × 4200 × (39-37)
Final temperature ( 2) = 39 ℃ = 6.7 × 105 J
specific heat capacity of water (S) = 4200 J kg-1 ℃-1
amount of heat required (Q) = ?

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Water Equivalent of a Substance

It is the mass of water that will absorb or lose the same amount of heat as the substance for the

same change in temperature.

Suppose a substance has of mass m and specific heat capacity S. when its temperature rises
through ∆ , we have

Q = mS ∆ . . . . . . . . . . (i)

If w be the water equivalent of this substance, then Q is given by

Q = w Sw ∆ . . . . . . . . . (ii)

from eqn (i) and (ii)

w Sw ∆ = mS ∆

or, w Sw = mS

or, w = mS
Sw

In CGS system, specific heat capacity of water Sw = 1 cal gm-1 ℃-1

then, w = mS

So, water equivalent of a substance is numerically equal to its thermal capacity in CGS

system. But the unit of water equivalent is grams and that of thermal capacity is calories.

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Determination of Specific Heat Capacity by the method of Mixture

Regnault’s apparatus can be used to determine the specific heat capacity of a solid heavier than
water. The principle, procedure and calculation are described as follows.
Principle: This experiment is based on the principle of calorimetry. According to this principle, “ if

there is no exchange of heat with the surrounding, heat gained by the cold body is equal to
the heat lost by hot body ”.

Procedure:
First of all, a calorimeter with a stirrer is weighed

and about two third of calorimeter is filled with

water so that the experimental solid may be

completely immersed in water. Then the mass of

water is also determined by subtracting the mass

of empty calorimeter with stirrer from total mass

of calorimeter with stirrer containing water. Fig. Regnault’s apparatus to determine the specific heat
capacity of a solid

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The experimental solid whose specific heat capacity is to be measured is weighed and then suspended in

a steam chamber. The initial temperature of water is noted. Initial temperature of solid is also noted. Now,

the experimental solid is kept into the water in calorimeter. Finally, the steady (final) temperature of the

mixture is also noted.

Calculation
Let,

mass of solid = m

mass of calorimeter and stirrer = mc
mass of water = mw
initial temperature of solid = 1
initial temperature of calorimeter and water = 2
final temperature of the mixture =

specific heat capacity of water = Sw
specific heat capacity of calorimeter = Sc
specific heat capacity of solid = S = ?

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We have

Heat gained by calorimeter with stirrer = mc Sc ( - 2)
Also heat gained by water = mw Sw ( - 2)
Heat lost by solid = mS ( 1 - )
Now from the principle of calorimetry

heat lost by solid = heat gained by calorimeter and water

or, mS ( 1 - ) = mc Sc ( - 2) + mw Sw ( - 2)
or, mS ( 1 - ) = (mcSc + mwSw)( - 2)

or, S = mcSc + mwSw)( − 2)
m( 1 − )

Since the values of mw mc sw sc 1 2 and are known, the specific heat capacity of solid

can be calculated.

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Q.) A metal piece of mass 250 gm at 80 ℃ is dropped into 120 gm of water 20 ℃. Find the final

temperature reached, if the specific heat capacity of metal is 0.09 Cal gm-1 ℃-1

Given: Mass of metal piece (Mm) = 250 gm
Temperature of the metal piece (θ1) = 80 ℃
Specific heat capacity of metal (Sm) = 0.09 Cal gm-1 ℃-1
Mass of water (Mw) = 120 gm
Temperature of water (θ2) = 20 ℃
Specific heat capacity of water (Sw) = 1 Cal gm-1 ℃-1
Final temperature of the mixture (θ) = ?

Now, when the metal piece is dropped into water then

Heat lost by the metal piece

(Q1) = MmSm (θ1 – θ)
= 250 × 0.09 × (80 – θ)

= 22.5 (80 – θ) . . . . .. . . . . . .. .(i)

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Heat gained by the water (Q2) = MwSw (θ – θ2)
= 120 × 1 × ( – 20)
= 120 ( – 20) . . . . . . . . . (ii)

From the principle of calorimetry
heat gained = heat lost

or, 120 ( – 20) = 22.5 (80 – θ)
or, 120 – 2400 = 1800 -22.5
or, 97.5 = 4200
or = 29.5 ℃

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Q.) A copper pot with mass 0.5 kg contains 0.2 kg of water at a temperature of 30 ℃. A 0.25 kg of
iron at 85 ℃ is dropped into the pot. Find the final temperature assuming no heat loss to the

surrounding.

Given: Mass of Copper pot (Mc) = 0.5 kg
Mass of iron block (Mi) = 0.25 kg
Mass of water (Mw) = 0.2 kg
Specific heat capacity of copper (Sc) = 390 Jkg-1K-1
Specific heat capacity of iron (Si) = 470 Jkg-1K-1
Specific heat capacity of water (Sw) = 4200 J kg-1 K-1
Temperature of the iron (θ1) = 85 ℃
Temperature of water and copper pot (θ2) = 30 ℃
Final temperature of the mixture (θ) = ?

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Now, When hot iron block is dropped into the water From the principle of calorimetry
Heat gained = Heat lost
in a pot.
or, 1035 – 31050 = 9987.5 – 117.5 θ
Heat lost by the iron block or, 1152.5 = 41037.5
(Q1) = MiSi (θ1 – θ) or, 792.5 = 27600
or = 35.6 ℃
= 0.25 × 470 × (85 – θ)
= 117.5(85 – θ)
= 9987.5 – 117.5 . . . . .. . . . . . .. .(i)

Heat gained by the water and copper pot
(Q2) = MwSw (θ – θ2) + Mc Sc (θ – θ2)

= 0.2 × 4200 × ( – 30) + 0.5 × 390 × ( – 30)
= 840 ( – 30) + 195 ( – 30)
= 1035 – 31050 . . . . . . . . . (ii)

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Q.) A piece of iron of mass 0.1 kg is kept inside a furnace , till it attains the temperature of the

furnace. The hot iron piece is then dropped into a calorimeter containing 0.24 kg of water at 20 ℃.

The mixture attains an equilibrium temperature of 60 ℃. Find the temperature of the furnace.

Given that water equivalent of calorimeter = 0.01 kg and sp. Heat capacity of iron = 470 Jkg-1K-1..

Given:

Mass of iron (mi) = 0.1 kg

Mass of water (mw) = 0.24 kg

Specific heat capacity of iron (Si) = 470 Jkg-1K-1

Specific heat capacity of water (Sw) = 4200 J kg-1 K-1

Water equivalent of calorimeter w = 0.01 kg

Initial temperature water and calorimeter (θ2) = 20℃
Final temperature of the mixture (θ) = 60 ℃

Initial temperature hot iron = θ1 = ?

Heat capacity of calorimeter (mcSc) = w × sw = 0.01 × 4200 = 42 J K-1

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When iron piece is dropped into water then iron loses heat while water and calorimeter gains heat.
So, heat lost by iron is

Q1 = miSi ( 1 - ) = 0.1 × 470 × ( 1 - 60) = 47 1 - 2820 . . . . . . .(i)
Heat gained by water and calorimeter

Q2= (mwSw + mcSc ) ( - 2) = (0.24 × 4200 + 42)(60 – 20) = 42000 J . . . . . .(ii)
From the principle of calorimetry

Heat lost = Heat gained
Q1 = Q2

or, 47 1 - 2820 = 42000
or, 47 1 = 44820
∴ 1 = 953.62 ℃

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Newton’s Law of Cooling

The amount of heat lost by a liquid depends on the temperature of surroundings. For example, a

cup of tea cools faster in winter than in summer though the temperature of tea is same. This is due

the fact that loss of heat by tea depends on the temperature of surrounding as the surrounding

temperature in winter is smaller that in summer.
Therefore, Newton’s law of cooling states that, the rate of loss of heat of a liquid is directly

proportional to the difference in temperature between the liquid and the surroundings.

Let and o be the temperature of the liquid and its surroundings such that the liquid loses a

small amount of heat dQ in a small time dt. Then Newton’s law of cooling can be expressed as

dQ ∝ ( - o)
dt
dQ
or, dt = K ( - o)

where k is proportionality constant whose value depends on the area of the exposed surface and

its nature.

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Determination of Specific Heat Capacity of a Liquid by Newton’s Law of Cooling

The experimental arrangement to measure the specific heat capacity of liquid is shown in fig. The
principle, procedure and calculation is mentioned below.
Principle: This experiment is based on the principle of Newton’s law of cooling. According to this
principle, “when two liquids are cooled under identical condition, the rates of cooling are equal.”

Procedure: first of all, two identical calorimeters A
and B are taken and the weight of each is noted.
Calorimeter A is filled with water and calorimeter B
is filled with liquid whose specific heat capacity is
to be measured. The mass of water and mass of
liquid are measured. Both calorimeters are kept
inside a box such that they are in identical
condition in every respect.

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Now, both the calorimeters are heated upto the equal temperature and are left to cool upto same

temperature. Time interval for cooling of water and liquid are noted.

Calculation: Let

Mass of calorimeter A = m1
Mass of calorimeter B = m2
Mass of water in A = mw
Mass of liquid in B = mL
Specific heat capacity of calorimeters = S

Specific heat capacity of water = Sw

Specific heat capacity of liquid = SL = ?

Initial temperature of both water and liquid = 1

Final temperature of both water and liquid = 2

Time required for water to cool from 1 to 2 is = t1

Time required for liquid to cool from 1 to 2 is = t2

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Now, heat lost by water and calorimeter A = mwSw ( 1- 2) + m1S ( 1- 2)
= (mwSw + m1S)( 1- 2)

Then rate of cooling of water and calorimeter A is

dQ = (mwSw + m1S)( 1− 2) . . . . . . . . . .(i)
dt t1


Similarly, rate of coolig of liquid and calorimeter B is

dQ = (mLSL + m2S)( 1− 2) . . . . . . . . . .(ii)
dt t2


From the Newton’s law of cooling,

dQ dQ
dt = dt



or, (mwSw + m1S)( 1− 2) = (mLSL + m2S)( 1− 2)
t1 t2
or,
(mLSL + m2S) = t2 (mwSw + m1S)
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or, (mLSL + m2S) = t2 (mwSw + m1S)
t1

or, mLSL = t2 (mwSw + m1S) - m2S
t1

or, SL = t2 mwSwm+L m1S - mmL2 S
t1

knowing the values of specific heat capacities of water and calorimeters, the specific heat

capacity of unknown liquid can be calculated.

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1) A copper calorimeter weighing 106 gm is filled with 100 g of water at 70 ℃ and the time taken

for the temperature to fall to 60 ℃ is 5 minutes. When filled with another liquid to the same

level, the weight of the liquid is 80 gm and time for the same fall of temperature is 2.7

minutes. Find the specific heat capacity of the liquid. (sp. Heat capacity of copper = 400 Jkg-

1K-1)

Mass of calorimeter (mc) = 106 gm = 0.106 kg

Mass of water (mw) = 100 gm = 0.1 kg

Mass of liquid (mL) = 80 gm = 0.08 kg

Sp. Heat capacity of calorimeter (Sc) = 400 Jkg-1K-1

Sp. Heat capacity of water (Sw) = 4200 Jkg-1K-1

Time taken by water to cool (tw) = 5 min = 300 sec

Time taken by liquid to cool (tw) = 2.7min = 162 sec

Temperature fall from 1 = 70 ℃ to 2 = 60 ℃

Sp. Heat capacity of liquid (SL) = ?

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Now rate of heat loss by water and calorimeter

dQ = mwSw ( 1 − 2) + mc Sc ( 1 − 2) = (0.1 × 4200 + 0.106 × 400)(70−60) = 15.41 J
dt tw 300
water

Rate of heat loss by liquid and calorimeter

dQ = (mL SL + mcSc )( 1 − 2)= (0.08 ×SL+ 0.106 × 400)(70−60) = 0.8 SL+ 424J
dt tL 162 162
liquid

From Newton’s law of cooling

dQ = dQ
dt water dt liquid

or, 15.41 =0.8 SL+ 424
162

or, 0.8 SL+ 424 = 2496.42

or, 0.8 SL = 2072.42

or, SL = 2590.52 J kg-1K-1

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2) Two identical calorimeters are each of heat capacity 12 JK-1, one contains 8 × 10-5 m3 of water

and takes 150 sec to cool from 325 K to 320K and the other contain an equal volume of

unknown liquid which takes 50 sec to cool over the same rangeof temperature if the density of

liquid is 800 kgm-3 . What the average sp. Heat capacity of liquid over the temperature range?

Take density of water = 1000 kgm-3 sp. Heat capacity of water = 4200 J kg-1K-1 .

 Heat capacity of calorimeter (mcSc) = 12 JK-1 Now, Mass of water
Volume of water (Vw) = 8 × 10-5 m3 (mw) = Vw × w = 8 × 10-5 × 1000 = 8 × 10-2 Kg
Volume of liquid (VL) = 8 × 10-5 m3
Density of water ( w) = 1000 kg m-3 Mass of liquid
Density of liquid ( L) = 800 kgm-3 (mL) = VL × L = 8 × 10-5 × 800 = 6.4 × 10-2 Kg
Time taken by water to cool (tw) = 150 sec

Time taken by liquid to cool (tl) = 50 sec

Temperature fall from 1 = 325 K to 2 = 320 K

Sp. Heat capacity of liquid (SL) = ?

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From Newton’s law of cooling

dQ water = dQ
dt dt liquid

or, mwSw ( 1 − 2) + mc Sc ( 1 − 2) = mL SL ( 1 − 2) + mcSc ( 1 − 2)
tw tL

or, mwSw + mc Sc = mL SL + mcSc
tw tL

or, 8 × 10−2 × 4200 + 12 = 6.4×10−2 × SL+ 12
150 50

or, 348 × 50 = 6.4 × 10-2 × SL + 12
150

or, 116 – 12 = 6.4 × 10-2 × SL

or, SL = 6.4 104
× 10−2

∴ SL = 1625 Jkg-1K-1

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3) A bucket full of water is kept in a room at constant temperature. The water in the bucket

cools from 55 ℃ to 50 ℃ in 5 minutes and from 50 ℃ to 45 ℃ in 7 minutes. Estimate the room

temperature. What time does it take to cool from 45 ℃ to 40 ℃ . (Neglect heat loss through the

bucket.)

Solution:

Time taken to cool from 55 ℃ to 50 ℃ is dt1 = 5 min

Time taken to cool from 50 ℃ to 45 ℃ is dt2 = 7 min

Time taken to cool from 45 ℃ to 40 ℃ is dt3 = ?

Average temperature between 55 ℃ and 50 ( 1) = 55 + 50 = 52.5 ℃
2

Average temperature between 50 ℃ and 45 ( 2) = 50 + 45 = 47.5 ℃
2

Average temperature between 45 ℃ and 40 ( 3) = 45 + 40 = 42.5 ℃
2

Let o be the temperature of surroundings.

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From Newton’s law of cooling Or, 262.5 - 5 o = 332.5 - o

For first case, dQ1 = - k ( 1 - o) Or, 2 o = 70
dt1
d 1 Or, o = 35 ℃
dt1
or, ms = - k (52.5- o) . . . .(i) For third case, dQ3 = - k ( 3 - o)
dt3
dQ2 d 3
For second case, dt2 = - k ( 2 - o) Or, ms dt3 = - k (42.5- 35) . . . . . . (iii)

or, ms d 2 = - k (47.5- o) . . . . (ii) Dividing eqn (ii) by (iii) we get
dt2

Dividing eqn (i) by (ii), we get dddd tt 3223

dddd tt 2112 = 47.5− o = 47.5− 35 = 12.5
42.5− 35 42.5− 35 7.5
52.5− o
Or, = 47.5− o

50 – 45 12.5
7 7.5
55 – 50 Or, 45 − 40 =

Or, 5 = 52.5− o dt3
50 − 45 47.5− o
dt3 =172.5.5
7 Or, 7

Or, 7 = 52.5− o ∴ dt3 = 12.5 × 7 = 11.67 minutes
5 47.5− o 7.5

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1) Groundnuts are fried along with sand, why?
 This is because sand has low specific heat capacity. Due to low specific heat capacity, it gets

quickly hot and it provides heat to the groundnuts so that the groundnuts fried very fast.

2) The greater the mass of a body, the greater is its specific heat capacity. Is it right? Explain.

 No, this is not right. This is because the specific heat capacity is the characteristic of the

material of the body and is independent of mass of the body.

3) Why do animals eat more in winter than in summer?

 In winter there is more temperature difference between animals and their surroundings, and

therefore according to Newton’s law of cooling the rate of loss of heat from animals is more

and to compensate this energy loss they should require more food and hence they eat more.

4) Milk boils faster than water. Why?

 The specific heat capacity of milk is smaller than that of water. So, water needs a larger

amount of heat to boil whereas the milk needs a less amount of heat to boil. Due to this reason

milk boils faster than water. Department Of Physics 30

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