The document discusses heat capacity, which is the amount of heat required to change a substance's temperature by 1 degree Kelvin. There are three types of heat capacity: heat capacity, specific heat capacity (per gram), and molar heat capacity (per mole). Molar heat capacity at constant pressure (Cp) and constant volume (Cv) are also discussed. Calorimetry, the measurement of heat flow, can be used to calculate heat and enthalpy changes through temperature change measurements. Bomb calorimetry under constant volume is given as an example to determine the energy available from combusting foods like glucose.

1. Heat Capacity
XI FDC CHEMISTRY
CHAPTER 11
SIDRA JAVED

2. Heat Capacity
Amount of heat absorbed by a substance is directly proportional to the
temperature change
q ∝ ΔT
q = Heat Capacity x ΔT
Where heat capacity is constant of proportionality
Heat Capacity =
Heat
ΔT

3. Heat Capacity
The amount of Heat required to raise the temperature of given amount of
substance by 1 kelvin is called heat capacity.
It is expressed in Joules per Kelvin
Heat Capacity = ----------(i)
Heat
ΔT

4. Specific Heat Capacity
The amount of heat required to raise the temperature of 1 gram of a
substance by 1 Kelvin is called Specific Heat Capacity
It is expressed in joules per gram per kelvin
Specific Heat Capacity (C) = -----------(ii)
Heat
m x ΔT

5. Molar Heat Capacity
The amount of heat required to raise the temperature of one mole of a
substance by one kelvin is called its molar heat capacity.
SI unit of Molar heat Capacity is J.mol-1K-1
Molar Heat Capacity (C) = ------------(iii)
Heat
Moles x ΔT

6. Molar heat capacity at constant pressure
If the quantity of Heat q is absorbed by n moles of the substance and its
temperature raises from T1 to T2 , its molar heat capacity is given by:
C = =
Since heat absorbed by a substance at constant pressure is equal to ΔH
qp = ΔH
∴ CP = =
Or ΔH = nCPΔT
Where Cp is the Molar Heat Capacity of the substance at constant pressure.
q
n(T2-T1)
q
n(ΔT)
qP
n(ΔT)
ΔH
n(ΔT)

7. Molar heat capacity at constant volume
At constant volume:
qV = ΔH = ΔE
And Cv = =
ΔE = nCV ΔT
Where Cv is the molar heat capacity at constant volume
qV
n(ΔT)
ΔE
n(ΔT)

8. Using specific heat of a substance, the quantity of heat abosrbed or evolved
can be evaluated
C =
q = m C ΔT ------(iv)
Similarly by amount of heat can be calculare using molar heat capacity
C =
q = n C ΔT -------(v)
q
m x ΔT
q
n x ΔT

9. Calorimetry
Caorimetry is the science of
meausring heat of chemical
Reaction by measuring the
temperature change
A device that
measures heat flow
is called Calorimeter

10. Constant pressure Calorimetry
In constant pressure calorimetry pressure of the
system is kept constant
A thermally insulated container with a
thermometer and stirrer is used. Generally a
coffee cup calorimeter is used for this purpose.
q = m C ΔT
Where m = mass of reactants
C = specific heat of reaction mixture
ΔT = Change in temperature

11. Constant volume calorimetry
It is used for accurate determination of the
enthalpy of combustion for food fuel and
other compounds
A bomb calorimeter is used for this purpose.
In a bomb calorimeter chemical reaction
takes place under constant volume
conditions.

12. Estimation of energy available from food
Human being require three major classes of food:
1. Carbohydrates
2. Proteins
3. Fats
Carbohydrates and mainly glucose are the main source of energy.
The energy available in glucose can be measured by determining its heat of
combustion.

13. ● Weigh 1.8 g of glucose and place it in holder cup of bomb
calorimeter and seal it.
● Adjust the pressure of oxygen in the calorimeter to about 25 atm.
● Immerse calorimeter in an insulated water bath fixed with a
motorized stirrer and a thermometer.
● Record the temperature of water T1.
● Ignite the glucose electrically when it will burn energy will flow
from the chemicals to the calorimeter and water.
● Record the temperature of water again T2.

14. T1 = 25oC
T2 = 31.52oC
Mass of glucose = 1.8 g
Molar mass = 180 gmol-1
Heat capacity of calorimeter
= 4.321 KJK-1
Increase in temperature
= ΔT = T2 - T1
ΔT = 31.52 - 25 = 6.52
Heat evolved = ΔT x Total heat capacity of
calorimeter
Heat evolved = 6.52 x 4.321= 28.1729 KJ
No of moles of glucose = 1.8/180 = 0.01
moles
Heat evolved by 0.01 mol glucose =
28.1729KJ
Heat evolved by 1 mol glucose =
28.1729/0.01 = 2817.29 KJmol-1

15. End of Lesson