Question

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. 340 J
B. 780 J
C. 105 J
D. 130 J
E. 520 J

Answer 1

Answer:

D. 130 J

Explanation:

The coefficient of performance for a machine that is being used to cool, is given by:

$COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}$

Here $Q_C$  is the heat removed from the cold reservoir, W is the work required, that is, the energy required to remove the heat from the interior of the house, $T_C$ is the cold temperature and $T_H$ is the hot temperature. Recall use absolutes temperatures($273.15+^\circ C$). Replacing and solving for W:

$W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J$