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 latent heat (or specifically, the latent heat of vaporization).

The mass of the coolant stays the same as it evaporates and becomes a vapor.

The pressure in the pipe increases because the compressor reduces the volume of the coolant vapor, leading to more collisions between the particles per second. This results in a greater force per collision, thereby increasing the pressure inside the pipe.

To calculate the specific heat capacity (c):

Using the efficiency equation:

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

Let Useful Output Energy Transfer = 1560 kJ x 0.875 = 1365 kJ

Convert to Joules: 1365 kJ = 1,365,000 J

Calculation for c:

$c = \frac{\text{Useful Output Energy}}{\text{Mass} \times \Delta T}$

Where:

• Mass = 125 kg
• \Delta T = 22.1 °C - 11.6 °C = 10.5 °C

So,

$c = \frac{1,365,000}{125 \times 10.5} = 1040 \text{ J/kg °C}$

In standard form:

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

The advertisement is incorrect because it ignores the energy input from the surrounding air. The total energy input must account for both the energy supplied from electricity and the energy absorbed from the air outside the building. Additionally, it is impossible for the efficiency to exceed 100%, meaning that the claim of 400% efficiency is unrealistic.