A Carnot power cycle operates between fixed temperature reservoirs with a thermal efficiency of 20%. When the cycle is reversed to function as a refrigerator between the same fixed temperature reservoirs the coefficient of performance must be
The correct answer and explanation is:
The coefficient of performance (COP) of a Carnot refrigerator is related to its thermal efficiency when it is operating as a power cycle. The thermal efficiency (η) of a Carnot engine is defined by the equation: η=1−TcTh\eta = 1 – \frac{T_c}{T_h}
where:
- η\eta is the thermal efficiency,
- TcT_c is the temperature of the cold reservoir, and
- ThT_h is the temperature of the hot reservoir (both in absolute temperatures, Kelvin).
Given that the thermal efficiency of the Carnot power cycle is 20%, we can use this information to find the ratio of the temperatures of the two reservoirs: 0.20=1−TcTh0.20 = 1 – \frac{T_c}{T_h}
Rearranging the equation, we get: TcTh=0.80\frac{T_c}{T_h} = 0.80
This means the temperature of the cold reservoir is 80% of the temperature of the hot reservoir.
Now, for the reversed Carnot cycle (as a refrigerator), the coefficient of performance (COP) is given by the equation: COP=TcTh−TcCOP = \frac{T_c}{T_h – T_c}
Substituting TcTh=0.80\frac{T_c}{T_h} = 0.80 into the equation: COP=0.80ThTh−0.80Th=0.80Th0.20Th=4COP = \frac{0.80 T_h}{T_h – 0.80 T_h} = \frac{0.80 T_h}{0.20 T_h} = 4
Thus, the coefficient of performance of the Carnot refrigerator is 4.
In summary, when the Carnot cycle operates as a refrigerator between the same fixed temperature reservoirs, its coefficient of performance must be 4. This indicates that for every unit of work input into the refrigerator, it can transfer four units of heat from the cold reservoir to the hot reservoir.