Answer:
D. 130 J
Explanation:
The coefficient of performance for a machine that is being used to cool, is given by:
[tex]COP=\frac{Q_C}{W}=\frac{T_C}{T_H-T_C}[/tex]
Here [tex]Q_C[/tex] 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, [tex]T_C[/tex] is the cold temperature and [tex]T_H[/tex] is the hot temperature. Recall use absolutes temperatures([tex]273.15+^\circ C[/tex]). Replacing and solving for W:
[tex]W=Q_c\frac{T_H-T_C}{T_C}\\W=2000J\frac{312.15K-293.15K}{293.15K}\\W=129.63J[/tex]
