Figure 6 shows a low-voltage solid-state thermoelectric cooling element - AQA - A-Level Physics - Question 4 - 2022 - Paper 6

To determine the validity of the claim regarding the coefficient of performance (COP) of the thermoelectric refrigerator, we first need to calculate the COP using the formula:

$COP = \frac{Q_c}{P}$

where:

• $Q_c$ is the heat absorbed from the cold side (in watts)
• $P$ is the electrical power supplied (in watts)

In this case, we have:

• The heat dissipated from the hot side (which equals the heat absorbed on the cold side) is 65 W.
• The electrical power supplied is 28 W.

Substituting the values into the COP equation:

$COP = \frac{65 \, \text{W}}{28 \, \text{W}} \approx 2.32$

Now, for an ideal refrigerator, the COP can be derived as:

$COP_{ideal} = \frac{T_c}{T_h - T_c}$

where:

• $T_c = 278 \, K$ (0°C in Kelvin)
• $T_h = 308 \, K$ (35°C in Kelvin)

Calculating:

$COP_{ideal} = \frac{278}{308 - 278} = \frac{278}{30} \approx 9.27$

Comparing the two COP values:

• Actual COP: 2.32
• Ideal COP: 9.27

Clearly, the actual COP is significantly lower than the ideal COP, validating the claim that the COP of a thermoelectric refrigerator is much less than that of an ideal refrigerator.