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^oC (or by 1 K).

• The units of specific heat capacity are J ^oC^-1 g^-1 or J K^-1 g^-1

• 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 ^oC^-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^oC (or by 1 K).
  The units of molar heat capacity are J ^oC^-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

1. Calculate the amount of heat needed to increase the temperature of 250g of water from 20^oC to 56^oC.

   q = m x C_g x (T_f - T_i)
   m = 250g
   C_g = 4.18 J ^oC^-1 g^-1 (from table above)
   T_f = 56^oC
   T_i = 20^oC

   q = 250 x 4.18 x (56 - 20)
   q = 250 x 4.18 x 36
   q = 37 620 J = 38 kJ

2. 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^oC
   T_f = 60^oC

   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 ÷ 204.75 = 0.39 J^oC^-1 g^-1

3. 216 J of energy is required to raise the temperature of aluminium from 15^o to 35^oC. Calculate the mass of aluminium.
   (Specific Heat Capacity of aluminium is 0.90 J^oC^-1g^-1).

   q = m x C_g x (T_f - T_i)
   q = 216 J
   C_g = 0.90 J^oC^-1g^-1
   T_i = 15^oC
   T_f = 35^oC

   216 = m x 0.90 x (35 - 15)
   216 = m x 0.90 x 20
   216 = m x 18
   m = 216 ÷ 18 = 12g

4. The initial temperature of 150g of ethanol was 22^oC. 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^oC^-1g^-1).

   q = m x C_g x (T_f - T_i)
   q = 3240 J
   m = 150g
   C_g = 2.44 J^oC^-1g^-1
   T_i = 22^oC

   3240 = 150 x 2.44 x (T_f - 22)
   3240 = 366 (T_f - 22)
   8.85 = T_f - 22
   T_f = 30.9^oC

Search this Site

Become an AUS-e-TUTE Member

Find out more about AUS-e-TUTE membership.

Become an AUS-e-TUTE member.

Register for your Free Newsletter

Ask Chris, the Chemist, a Question

Chris can also send you to the relevant
AUS-e-TUTE tutorial topic page.

Practice

Enter mass of substance:	g
Enter specific heat capacity:	J oC-1 g-1
Enter initial Temperature:	oC
Enter final Temperature:	oC
Click Calculate:
Calculation:
q =	J
To start again click Reset:

Related AUS-e-TUTE Topics

Manipulating Enthalpy Change Terms

Calorimetry

Energy Profiles (diagrams)

Fuel Definitions

Bookmark AUS-e-TUTE