SUMMARY CHAPTER 2 CALORIMETRY (2)

MIND MAP
CALORIMETRY

CALORIMETRY

Heat released from reaction = Heat absorbed by ‘surrounding’ (water, calorimeter, water + calorimeter)

q = mcΔT @
CΔT

∆H = q
∆H( enthalpy) ≠ q (heat).. unless n=1

BASIC PRINCIPLE IN CALORIMETER

NO SITUATION THERMOCHEMICAL PRINCIPLE EXAMPLE

1 Calorimeter with water Heat released from reaction = Heat absorbed by Example 2:
Water + calorimeter Calculate the amount of heat released in a
reaction in an aluminium calorimeter with a
q = mwcwΔT + mcccΔT mass of 3087.00g and contains 1700.00mL of
water. The initial temperature of the
@ calorimeter is 25.00 °C and it increased to
q = mwcwΔT + CcΔT 27.80 °C.
Given:
Specific heat capacity of Al = 0.553 Jg -1 °C-1
Specific heat capacity of water = 4.18 Jg -1 °C-1
Water density = 1.0 gmL-1

Example 3 also using this concept.

2 Calorimeter without Heat released from reaction = Heat absorbed by calorimeter Example 1:
water
q = mcccΔT In an experiment, 0.10g of H2 and excess of O2

@ were compressed and placed into a calorimeter
q = CcΔT with heat capacity of 9.08 x 104 J˚C−1 . The

initial temperature of the calorimeter was
25.00˚C and finally it increased to 25.155˚C.

Calculate the amount of heat released in the

reaction to form H2O, expressed in kJ per mole

3 Simple calorimeter @ Heat released from reaction = Heat absorbed by solution + Example 4:
constant-pressure A quantity of 100mL of 0.500M HCl is mixed
calorimeter calorimeter with 100mL of 0.500M NaOH in a constant-
pressure calorimeter that has a heat capacity of
q = mscsΔT + mcccΔT @ 335 J ˚C-1 . The initial temperature of the HCl
and NaOH solution is the same, 22.50 °C and
q = mscsΔT + CcΔT the final temperature of the mixed solution is
24.90°C. Calculate the heat change for the
OR neutralization reaction.
Assume that the density and specific heat of
Eg: Neutralisation reaction the solution are the same as water (1.00gmL-1
and 4.18 Jg -1 °C-1 respectively.)
Heat released from reaction = Heat absorbed by solution
Example 5 also using this concept.
q = mscsΔT
Another example:
4 Heating the water Heat released from combustion of gas = Heat absorbed by
When 100.0 mL of 0.50 M HCl(aq) and 100.0
water mL of 0.60 M NaOH(aq), both at 22.0 °C, are
added to a coffee cup calorimeter, the
q = mwcwΔT temperature of the mixture reaches a maximum
of 28.9 °C. Calculate enthalpy of neutralization
reaction. Assume that the density and specific
heat of the solution are the same as water

HCl(aq)+NaOH(aq)⟶NaCl(aq)+H2O(l)

Practice Question 4 :

A student wanted to boil 2.0kg of water at
20°C in a kettle. Calculate the mass of natural
gas, CH4, must be burned to raise this quantity
of water to 100°C? Assume there is no heat
loss to the surrounding.
[specific heat capacity of water = 4.18 Jg-1°C-1
and ∆Hc of CH4 = -890 kJmol-1 ]

Another example:
A flask containing 8.0 × 102 g of water is
heated, and the temperature of the water
increases from 21 °C to 85 °C. How much heat
did the water absorb?
Ans: 210kJ

5 Putting hot metal in Heat released from hot substance = Heat absorbed by water A 248-g piece of copper is dropped into 390
water mL of water at 22.6 °C. The final temperature
mhch∆Th = mwcw∆Tw of the water was measured as 39.9 °C.
Calculate the initial temperature of the piece of
@ copper. Assume that all heat transfer occurs
Ch∆Th = mwcw∆Tw between the copper and the water.

[specific heat capacity copper = 0.385 J°C-1
specific heat capacity water = 4.18 Jg-1°C-1 ]

Ans : 335.3 °C

6 Making lukewarm Heat released from hot water = Heat absorbed by cold water Example 6:
water
m1c1∆T1 = m2c2∆T2 A calorimeter contains 400mL of water at
25˚C. If 600mL of water at 60˚C is added to it,

determine the final temperature. Assume that
the heat absorbed by the calorimeter is
negligible.