An air source heat pump transfers energy from the air outside a building to increase the temperature of the air inside the building - AQA - GCSE Physics Combined Science - Question 6 - 2021 - Paper 1

The energy needed to change the state of the liquid coolant is called the latent heat of vaporisation.

Stays the same

The increase in pressure in the pipe is due to the increase in density and temperature of the coolant vapour caused by the compressor. When the coolant is compressed, the molecules are forced closer together, resulting in more collisions per second and a greater force exerted on the walls of the pipe, leading to increased pressure.

To find the specific heat capacity (c), we can use the formula:

$ext{Efficiency} = \frac{\text{Useful Output Energy Transfer}}{\text{Total Input Energy Transfer}}$

The total energy input to the system is 1560 kJ, and the temperature change (∆T) is:

$22.1 °C - 11.6 °C = 10.5 °C$

From the efficiency, the useful output energy transfer can be calculated as:

$\text{Useful output energy transfer} = 0.875 \times 1560000 \text{ J} = 1365000 \text{ J}$

Using the formula for specific heat capacity:

$Q = mc\Delta T$

where (Q) is the useful output energy (1365000 J), (m) is the mass of the air (125 kg), and (\Delta T) is the temperature change (10.5 °C):

$1365000 = 125c(10.5)$

Solving for c gives:

$c = \frac{1365000}{125 \times 10.5} = 1040 J/kg °C$

In standard form: $c = 1.04 \times 10^3 \text{ J/kg °C}$

The advertisement is misleading because it suggests that the heat pump system can produce 400% more energy than the energy supplied from electricity. This is not possible as the total energy input must account for the energy drawn from the surrounding air as well. Furthermore, the efficiency must always be less than 100% since no system can create energy, it can only transfer it.