Specific heat capacity (c)
Thermal Physics - OCR A-Level Physics
Key Definition
Specific heat capacity (c)
The energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 \degree C). Units: $J kg^{-1}$ $K^{-1}$.
The energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 \degree C). Units: $J kg^{-1}$ $K^{-1}$.
Key Definition
Specific latent heat (L)
The energy required to change the state of 1 kg of a substance without changing its temperature. Units: $J kg^{-1}$.
The energy required to change the state of 1 kg of a substance without changing its temperature. Units: $J kg^{-1}$.
$$Q = mc\Delta\theta$$
$$Q = mL$$
- Specific latent heat of fusionThe energy required to change 1 kg of a substance from solid to liquid at its melting point, without changing temperature. (L_f): energyThe capacity to do work. Measured in joules (J). to change 1 kg from solid to liquid at the melting point.
- Specific latent heat of vaporisationThe energy required to change 1 kg of a substance from liquid to gas at its boiling point, without changing temperature. (L_v): energyThe capacity to do work. Measured in joules (J). to change 1 kg from liquid to gas at the boiling point.
- L_v is always greater than L_f because molecules must completely overcome intermolecular forces to become a gas.
- In experiments, energyThe capacity to do work. Measured in joules (J). losses to the surroundings mean the measured value of c or L is often an overestimate.
Worked Example
A 0.50 kg block of aluminium at 20 \degree C is heated to 80 \degree C. Calculate the energy transferred. ($c_{Al}$ = 900 $J kg^{-1}$ $K^{-1}$)
Show Solution
1
Identify
$m = 0.50 kg, c = 900 J kg^{-1} K^{-1}, \Delta\theta = 80 - 20 = 60 K.$
2
$Apply Q = mc\Delta\theta.$
3
$Q = 0.50 \times 900 \times 60 = 27000 J.$
4
$Q = 27 kJ.$
Answer
$Q = 27 kJ$
Common Mistake
MEDIUM
Wrong: Using \Delta\theta in \degree C for gas law equations but K for specific heat capacityThe energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C)..
Right: For Q = mc\Delta\theta, a change of 1 K equals a change of 1 \degree C, so either unit works for \Delta\theta. But gas law equations ($pV = nRT) always require absolute temperature$ in K.
Right: For Q = mc\Delta\theta, a change of 1 K equals a change of 1 \degree C, so either unit works for \Delta\theta. But gas law equations ($pV = nRT) always require absolute temperature$ in K.