14.7-14.10 - Specific heat capacity and latent heat

14.7-14.10 - Specific heat capacity and latent heat

Distinguish energy that warms a material from energy that changes its state. Select and use both thermal equations, then explain how insulation reduces unwanted transfer.

Two specific thermal quantities

The word specific means that the quantity is defined for 1 kg of a substance. This allows materials to be compared fairly without one sample winning simply because it has more mass.

Specific heat capacity, c, is the energy required to raise the temperature of 1 kg of a substance by 1 °C, without a change of state. Its unit is joules per kilogram degree Celsius, J/kg °C.

That definition describes warming within one state. The next quantity is used when the state changes instead.

Specific latent heat, L, is the energy required to change the state of 1 kg of a substance without changing its temperature. Its unit is joules per kilogram, J/kg.

Specific heat capacity concerns a temperature change. A material with a larger c needs more energy per kilogram for the same temperature rise. Specific latent heat concerns a state change. Its value depends both on the substance and on the change: the specific latent heat of fusion applies to melting or freezing, while the specific latent heat of vaporisation applies to liquid-to-gas changes such as boiling or evaporation, and to the reverse change of condensation.

The units expose the difference. J/kg °C includes “per degree Celsius” because temperature changes; J/kg does not, because the temperature does not change during the state change.

Calculating a temperature change

For a temperature change within one state, the Pearson equation list supplies:

change in thermal energy = mass × specific heat capacity × change in temperature

ΔQ=m×c×Δθ\Delta Q = m \times c \times \Delta \theta

  • ΔQ is the change in thermal energy in joules (J)
  • m is mass in kilograms (kg)
  • c is specific heat capacity in joules per kilogram degree Celsius (J/kg °C)
  • Δθ is the temperature change in degrees Celsius (°C), found from final temperature - initial temperature

Do not substitute the final temperature for Δθ. Also convert grams to kilograms before using a value of c in J/kg °C.

Worked example

A 750 g aluminium block warms from 18 °C to 65 °C. The specific heat capacity of aluminium is 900 J/kg °C. Calculate the block's change in thermal energy.

  1. The temperature changes and the state does not, so select ΔQ=m×c×Δθ\Delta Q = m \times c \times \Delta \theta .
  2. Convert the mass: 750g=0.750kg750 \mathrm{g} = 0.750 \mathrm{kg}.
  3. Find the temperature change: Δθ=65−18=47∘C\Delta \theta = 65 - 18 = 47 {}^{\circ}C.
  4. Substitute: ΔQ=0.750×900×47\Delta Q = 0.750 \times 900 \times 47.
  5. Calculate: ΔQ=31725J=3.2×104J\Delta Q = 31 725 \mathrm{J} = 3.2 \times 10^{4} \mathrm{J} to two significant figures.
  6. Sense-check: a positive energy transfer into the block should give a positive temperature rise, and a value of tens of kilojoules is plausible for this mass and temperature change.

The relationship can also be rearranged: c=ΔQ/(m×Δθ)c = \Delta Q / (m \times \Delta \theta ), m=ΔQ/(c×Δθ)m = \Delta Q / (c \times \Delta \theta ) or Δθ=ΔQ/(m×c)\Delta \theta = \Delta Q / (m \times c).

Calculating a change of state

For a change of state at constant temperature, the Pearson equation list supplies:

thermal energy for a change of state = mass × specific latent heat

Q=m×LQ = m \times L

  • Q is thermal energy in joules (J)
  • m is mass in kilograms (kg)
  • L is specific latent heat in joules per kilogram (J/kg)

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account