Final answer:
The design condition imposed on the refrigerator to extract 2100 J of heat per cycle with 300 J of work and a freezer temperature of -10°C leads to an unrealistic COP and maximum air temperature that is not achievable. This indicates that the condition is not reasonable and cannot be met with any of the given temperature options.
Step-by-step explanation:
To determine the maximum air temperature for which a refrigerator can do 300 J of work per cycle to extract 2100 J of heat per cycle from a freezer at -10°C, we need to apply concepts from thermodynamics, specifically the second law of thermodynamics and refrigerator coefficients of performance (COP).
The COP for a refrigerator is defined as the amount of heat removed from the cold reservoir (Qc) divided by the work done by the refrigerator (W). The formula for COP is:
COP = Qc / W
For an ideal refrigerator operating on a Carnot cycle, the COP is also related to the temperatures of the hot (Th) and cold (Tc) reservoirs in Kelvin:
COP = Tc / (Th - Tc)
Where,
- Tc = Temperature of the cold reservoir (-10°C or 263 K)
- Th = Maximum air temperature
- Qc = 2100 J
- W = 300 J
Plugging Qc and W into the COP formula gives us COP = 7. Rearranging the second formula to solve for Th, we get:
Th = Tc * COP + Tc
Now, substituting the values we have:
Th = 263 K * 7 + 263 K = 2107 K
Since this temperature is unrealistically high, we can say that the maximum air temperature for the condition is not achievable. Thus, none of the options (a, b, c, d) provided are valid because they all fall within normal ambient temperature ranges, which are far lower than 2107 K.
The design condition imposed on the refrigerator is not reasonable as it suggests a COP far greater than what real refrigerators can achieve, indicating a significant misunderstanding or misapplication of thermodynamics.