Answer :
Final answer:
The Carnot air conditioner's coefficient of performance (COP) for the given temperatures can be calculated as 14.55, indicating that for every watt of electrical power used, it removes 14.55 watts of heat energy from the room. So, none of the given options is correct.
Explanation:
To determine the rate at which energy is removed from the room by a Carnot air conditioner, we use the coefficient of performance (COP) for a Carnot refrigerator. The COP is the ratio of the heat extracted from the cold reservoir (QL, the room in this case) to the work input (W). For a Carnot cycle, the COP is given by TL / (TH - TL), where TL and TH are the absolute temperatures of the cold and hot reservoirs, respectively. Given the temperatures are in degrees Fahrenheit, we must convert them to Kelvin. The conversion formula is K = (F - 32) * 5/9 + 273.15.
Converting the room temperature (65 ⁰F) and the outside temperature (101 ⁰F) to Kelvin, we have:
TL = (65 - 32) * 5/9 + 273.15 ≈ 291 K
TH = (101 - 32) * 5/9 + 273.15 ≈ 311 K
Substituting these values into the COP equation, we get:
COP = 291 / (311 - 291) = 291 / 20 = 14.55.
Therefore, for every watt of electrical power, the Carnot air conditioner removes 14.55 watts of heat energy from the room. Thus, the correct answer explaining the rate at which energy is removed would be expressed in terms of energy per time, such as watt or joules per second, which is not provided in the options (a) to (d).
