Final answer:
To determine the hot reservoir temperature for an ideal refrigerator with a cold temperature of -10.0°C and a coefficient of performance of 7.00, you can use the formula COP = Tc / (Th - Tc). After converting temperatures to Kelvin and rearranging the equation, you find that the hot reservoir temperature should be approximately 26.0°C.
Step-by-step explanation:
If you want to operate an ideal refrigerator with a cold temperature of -10.0°C and a coefficient of performance (COP) of 7.00, you can determine the hot reservoir temperature using the relationship between the COP and the temperatures of the cold and hot reservoirs.
The COP for a refrigerator is defined as the ratio of the heat removed from the cold reservoir (Qc) to the work W done by the refrigerator:
COP = Qc / W
For an ideal refrigerator operating on the Carnot cycle, COP is also related to the temperatures of the hot reservoir (Th) and the cold reservoir (Tc) in degrees Kelvin, by the following equation:
COP = Tc / (Th - Tc)
First, convert the cold reservoir temperature from Celsius to Kelvin:
Tc = -10.0°C + 273.15 = 263.15 K
Then, rearrange the equation to solve for Th:
Th = Tc / COP + Tc
Th = 263.15 K / 7 + 263.15 K = 37.593 K + 263.15 K = 300.743 K
Convert Th back to Celsius:
Th = 300.743 K - 273.15 = 27.593°C
The closest answer to this calculated temperature is 26.0°C, which is option (a).