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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
Question 6
An air source heat pump transfers energy from the air outside a building to increase the temperature of the air inside the building. Figure 13 shows an air source h... show full transcript
Worked Solution & Example Answer: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
Step 1
1. Explain what happens to the internal energy of the coolant as its temperature increases.
Answer
As the temperature of the coolant increases, its internal energy also increases. This is because the internal energy comprises the kinetic energy of the particles, which is related to their movement. The rise in temperature causes the speed of the particles to increase, which in turn elevates the kinetic energy, thereby raising the internal energy.
Step 2
Step 3
Step 4
4. Explain why the pressure in the pipe increases.
Answer
The pressure in the pipe increases because the compressor compresses the coolant vapour, leading to a reduction in its volume. According to the ideal gas law, if the volume decreases while the amount of gas remains constant, the pressure must increase. Additionally, as the temperature of the coolant vapour rises, this further contributes to increased pressure.
Step 5
5. Calculate the specific heat capacity of the air in the building.
Answer
First, determine the useful output energy transferred to the air:
Total input energy (from heat pump) = 1560 kJ Efficiency = 87.5%
Useful output energy = 1560 kJ × 0.875 = 1365 kJ
The temperature change (ΔT) is:
ΔT = 22.1 °C - 11.6 °C = 10.5 °C
Using the formula:
Specific heat capacity (c) = ( \frac{E}{m \cdot \Delta T} )
Where E is the useful output energy, m is the mass of the air, and ΔT is the temperature change.
Substituting the values:
( c = \frac{1365000 J}{125 kg \cdot 10.5 °C} = 1040 J/kg °C )
Therefore, the specific heat capacity of the air in standard form is ( 1.04 \times 10^3 J/kg °C ).
Step 6
6. Explain why the advertisement is not correct.
Answer
The advertisement is incorrect because it has ignored the energy input from the surrounding air. The total energy input must include the energy drawn from both the mains electricity supply and the surrounding air. Additionally, the claim of 400% efficiency implies that the system produces more energy than it consumes, which is not possible according to the laws of thermodynamics. The efficiency of any system must be less than 100%.