Carnot devices: A Carnot air conditioner operates between an indoor temperature of 20°C and an outdoor temperature of 39°C. How much energy does it need to remove 2000 J of heat from the interior of the house?
A) 105 J
B) 130 J
C) 780 J
D) 520 J
E) 340 J

Question

Quizplus · Accepted Answer

The efficiency of a Carnot air conditioner (or refrigerator) is given by the coefficient of performance (COP), which is the ratio of the heat removed from the cold reservoir (Qc) to the work input (W). The COP for a Carnot refrigerator is given by $COP = \frac{Q_c}{W} = \frac{T_c}{T_h - T_c}$, where $T_c$ and $T_h$ are the temperatures of the cold and hot reservoirs, respectively, in Kelvin. To find the work input required to remove 2000 J of heat from the interior, we first convert the temperatures from Celsius to Kelvin: $T_c = 20°C + 273.15 = 293.15 K$ and $T_h = 39°C + 273.15 = 312.15 K$. Then, $COP = \frac{293.15}{312.15 - 293.15} = \frac{293.15}{19}$. The work input required is $W = \frac{Q_c}{COP} = \frac{2000 J}{\frac{293.15}{19}} \approx 130 J$.

Carnot Devices: a Carnot Air Conditioner Operates Between an Indoor