Heat Capacity Calculations 
Key Concepts
 The amount of heat energy (q) gained or lost by a substance is equal to the mass of the substance (m) multiplied by its specific heat capacity (C_{g}) multiplied by the change in temperature (final temperature  initial temperature)
q = m x C_{g} x (T_{f}  T_{i})
 Specific Heat Capacity (C_{g}) of a substance is the amount of heat required to raise the temperature of 1g of the substance by 1^{o}C (or by 1 K).
 The units of specific heat capacity are J ^{o}C^{1} g^{1} or J K^{1} g^{1}
Specific Heat Capacities of Some Substances [C_{g} (J K^{1} g^{1} or J ^{o}C^{1} g^{1})] 
aluminium  C_{g} = 0.90 
water  C_{g} = 4.18 
carbon  C_{g} = 0.72 
ethanol (ethyl alcohol)  C_{g} = 2.44 
copper  C_{g} = 0.39 
sulfuric acid (liquid)  C_{g} = 1.42 
lead  C_{g} = 0.13 
sodium chloride solid  C_{g} = 0.85 
mercury  C_{g} = 0.14 
potassium hydroxide solid  C_{g} = 1.18 
 The amount of heat energy gained or lost by a substance can also be calculated for moles of substance:
q = n x C_{n} x (T_{f}  T_{i})
q = amount of heat energy gained or lost by substance
n = moles of substance
C_{n} = molar heat capacity (J ^{o}C^{1} mol^{1} or J K^{1} mol^{1})
T_{f} = final temperature
T_{i} = initial temperature
 Molar Heat Capacity (C_{n}) of a substance is the amount of heat required to raise the temperature of 1 mole of the substance by 1^{o}C (or by 1 K).
The units of molar heat capacity are J ^{o}C^{1} mol^{1} or J K^{1} mol^{1}
 Heats of Reaction can be measured in the laboratory using a simple calorimeter made up of a polystyrene cup (these are appropriate because of their good insulating properties) that is fitted with a lid and a thermometer.
Examples
 Calculate the amount of heat needed to increase the temperature of 250g of water from 20^{o}C to 56^{o}C.
q = m x C_{g} x (T_{f}  T_{i})
m = 250g
C_{g} = 4.18 J ^{o}C^{1} g^{1} (from table above)
T_{f} = 56^{o}C
T_{i} = 20^{o}C
q = 250 x 4.18 x (56  20)
q = 250 x 4.18 x 36
q = 37 620 J = 38 kJ
 Calculate the specific heat capacity of copper given that 204.75 J of energy raises the temperature of 15g of copper from 25^{o} to 60^{o}.
q = m x C_{g} x (T_{f}  T_{i})
q = 204.75 J
m = 15g
T_{i} = 25^{o}C
T_{f} = 60^{o}C
204.75 = 15 x C_{g} x (60  25)
204.75 = 15 x C_{g} x 35
204.75 = 525 x C_{g}
C_{g} = 204.75 ÷ 525 = 0.39 J^{o}C^{1} g^{1}
 216 J of energy is required to raise the temperature of aluminium from 15^{o} to 35^{o}C. Calculate the mass of aluminium.
(Specific Heat Capacity of aluminium is 0.90 J^{o}C^{1}g^{1}).
q = m x C_{g} x (T_{f}  T_{i})
q = 216 J
C_{g} = 0.90 J^{o}C^{1}g^{1}
T_{i} = 15^{o}C
T_{f} = 35^{o}C
216 = m x 0.90 x (35  15)
216 = m x 0.90 x 20
216 = m x 18
m = 216 ÷ 18 = 12g
 The initial temperature of 150g of ethanol was 22^{o}C. What will be the final temperature of the ethanol if 3240 J was needed to raise the temperature of the ethanol?
(Specific heat capacity of ethanol is 2.44 J^{o}C^{1}g^{1}).
q = m x C_{g} x (T_{f}  T_{i})
q = 3240 J
m = 150g
C_{g} = 2.44 J^{o}C^{1}g^{1}
T_{i} = 22^{o}C
3240 = 150 x 2.44 x (T_{f}  22)
3240 = 366 (T_{f}  22)
8.85 = T_{f}  22
T_{f} = 30.9^{o}C

Practice Questions 
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