Question

to maintain a room at a comfortable temperature of 21.0 °c, a carnot heat pump performs 345 j of work and delivers 3240 j of heat to the room. what is the temperature of the outside air?

a. - 10 °c
b. - 78 °c
c. - 38 °c
d. - 64 °c
e. - 25 °c

Explanation:

Step1: Calculate heat absorbed from outside

The work done by the heat - pump $W = 345\ J$ and the heat delivered to the room $Q_H=3240\ J$. According to the energy conservation $Q_H = Q_C+W$, so the heat absorbed from the outside $Q_C=Q_H - W$.
$Q_C=3240 - 345=2895\ J$

Step2: Use the Carnot heat - pump efficiency formula

The efficiency of a Carnot heat - pump $\eta_{HP}=\frac{Q_H}{Q_H - Q_C}=\frac{T_H}{T_H - T_C}$, where $T_H$ is the temperature of the hot reservoir (room temperature) and $T_C$ is the temperature of the cold reservoir (outside air temperature). First, convert the room temperature to Kelvin: $T_H=(21.0 + 273.15)\ K=294.15\ K$.
Since $\frac{Q_H}{Q_H - Q_C}=\frac{T_H}{T_H - T_C}$, we can cross - multiply: $Q_H(T_H - T_C)=T_H(Q_H - Q_C)$.
Substitute $Q_H = 3240\ J$, $Q_C = 2895\ J$, and $T_H=294.15\ K$ into the equation:
$3240\times(294.15 - T_C)=294.15\times(3240 - 2895)$
$3240\times(294.15 - T_C)=294.15\times345$
$294.15 - T_C=\frac{294.15\times345}{3240}$
$294.15 - T_C=\frac{101481.75}{3240}\approx31.32$
$T_C=294.15 - 31.32 = 262.83\ K$

Step3: Convert the temperature back to Celsius

$t_C=T_C - 273.15=(262.83 - 273.15)^{\circ}C=- 10.32^{\circ}C\approx - 10^{\circ}C$

Answer:

a. $- 10^{\circ}C$