Specific Heat Capacity
- Specific heat capacity is defined as:
the energy required to raise the temperature of one kilogram of a substance by one kelvin
- The increase in temperature of an object depends on:
- The amount of heat energy transferred
- The mass of the object
- The specific heat capacity of the material from which the object is made
Explanation of Heat Energy
- The energy input is in the form of heat energy
- The amount of heat energy needed is given by the equation:
ΔE = mcΔθ
- Where:
- ΔE = change in heat energy, in joules (J)
- m = mass, in kilograms (kg)
- c = specific heat capacity, in joules per kilogram per degree Kelvin (J/kg K or J/kg °C)
- Δθ = change in temperature, in Kelvin or Celsius
- (The symbol Δ in Maths is used to denote a change in value)
Examples of heat energy
- The thermodynamic Kelvin temperature scale is defined as:
An absolute temperature scale where each degree is the same size as those on the Celsius scale
- Different materials:
- Have different specific heat capacities because of their molecular structure
- Have different rises in temperature for the same change in heat energy
- Specific heat capacity is mainly used for liquids and solids
- Good electrical conductors, such as copper and lead, are excellent conductors of heat due to their low specific heat capacity
- On the other hand, water has a very high specific heat capacity, making it ideal for heating homes as the water remains hot in a radiator for a long time
- The specific heat capacity of different substances determines how useful they would be for a specific purpose eg. choosing the best material for kitchen appliances
Low v high specific heat capacity
- The specific heat capacity of some substances are given in the table below as examples:
Table of values of specific heat capacity for various substances
Worked example
Water of mass 0.48 kg is increased in temperature by 0.7 K. The specific heat capacity of water is 4200 J kg-1 K-1. Calculate the amount of energy transferred to the water.
Step 1: Write down the known quantities
-
- Mass, m = 0.48 kg
- Change in temperature, Δθ = 0.7 K
- Specific heat capacity, c = 4200 J kg-1 K-1
Step 2: Write down the relevant equation
ΔE = mcΔθ
Step 3: Calculate the energy transferred by substituting in the values
ΔE = (0.48) × (4200) × (0.7) = 1411.2
Step 4: Round the answer to 2 significant figures
ΔE = 1400 J
Examiner Tip
You will always be given the specific heat capacity of a substance, so you do not need to memorise any values. Make sure that Δθ is the change in temperature, therefore, it can be in K or °C.