Thermal energy transferAQA A-Level Physics: Revision notes

Internal energy is the sum of the randomly distributed kinetic energies and potential energies of all the particles in a body. It is not the motion or height of the body as a whole.

The internal energy of a system increases when energy is transferred to it by heating or when work is done on it, and decreases when it loses energy by heating or does work on its surroundings. Qualitatively this is the first law of thermodynamics:

$\Delta U = Q + W$

where $Q$ is energy transferred to the system by heating and $W$ is work done on the system.

Worked example: 450 J of work is done on a gas while it loses 120 J by heating: $\Delta U = -120 + 450 = +330$ J.

For a change of temperature with no change of state,

$Q = mc\Delta\theta$

where $c$ is the specific heat capacity, the energy needed to raise the temperature of 1 kg of a substance by 1 K (units J kg⁻¹ K⁻¹). During a temperature rise, the mean kinetic energy of the particles increases.

If a heater of power $P$ is on for time $t$ with no losses, $Pt = mc\Delta\theta$.

Worked example: heating 2.0 kg of water from 20 °C to 100 °C needs $Q = 2.0 \times 4180 \times 80 = 6.7 \times 10^5$ J. A 2.2 kW heater at 100 % efficiency takes $6.69 \times 10^5 / 2200 = 304$ s.

For a change of state at constant temperature,

$Q = ml$

where $l$ is the specific latent heat, the energy needed to change the state of 1 kg of a substance without a change of temperature. There is one value for fusion (solid to liquid) and a larger one for vaporisation (liquid to gas).

During a change of state the potential energies of the particles change, as they are moved apart against the attractive forces, but the mean kinetic energy does not change, so the temperature stays constant.

Worked example: melting 0.250 kg of ice with $l = 3.34 \times 10^5$ J kg⁻¹ needs $8.35 \times 10^4$ J. A 150 W heater takes 557 s.

On a graph of temperature against energy supplied, the sloping sections show temperature rising, where the gradient is $1/(mc)$, and the flat sections show changes of state. A steeper slope means a smaller specific heat capacity for the same mass.

The flat section for boiling is longer than for melting because the specific latent heat of vaporisation is much larger than that of fusion: the molecules have to be separated completely, not just loosened.

When heating a mixture of two states, use $Q = ml$ for the changing part and $Q = mc\Delta\theta$ for any part whose temperature changes, and add them.

In continuous flow a fluid flows steadily past a heater. In each second the heater supplies energy $P$, and the fluid of mass $\dot{m}$ (mass per second) gains $\dot{m}c\Delta\theta$:

$P = \dot{m}c\Delta\theta$

If there is a loss rate $L$, then $P = \dot{m}c\Delta\theta + L$.

Worked example: an 8.5 kW heater with a 30 K rise: $\dot{m} = 8500/(4180 \times 30) = 0.0678$ kg s⁻¹.

To find $c$ without measuring $L$, use two flow rates $\dot{m}_1$ and $\dot{m}_2$ with powers $P_1$ and $P_2$ adjusted to give the same $\Delta\theta$. Then $L$ is the same, so

$c = \frac{P_1 - P_2}{(\dot{m}_1 - \dot{m}_2)\Delta\theta}$