A Carnot air conditioner takes energy from the thermal energy of a room at $\mathbf{70}\mathbf{°}\mathbf{F}$and transfers it as heat to the outdoors, which is at $\mathbf{96}\mathbf{°}\mathbf{F}$. For each joule of electric energy required to operate the air conditioner, how many joules are removed from the room?

Temperatures,$T_L=70°F=294.26\hspace{0.33em}K$and

$T_H=96°F=308.71\hspace{0.33em}K$

It is given that $Q_H=T_H$and $Q_L=T_L$

We apply the concept of conservation of energy in the formula for the coefficient of performance of the Carnot engine to find the amount of heat in joules removed from the room.

Formula:

The formula for the coefficient of performance of a Carnot Engine,

$K=\frac{T_L}{T_H-T_L}$ (1)

The work done per cycle of a Carnot Engine,

$W_{in}=Q_H-Q_L$ (2)

Using equation (1), the coefficient of performance of the Carnot air conditioner is given as:

$$\begin{gathered} \mathrm{K}=\frac{294.26\mathrm{K}}{\left(308.71\mathrm{K}-294.26\mathrm{K}\right)} \\ =20.36 \end{gathered}$$

But, ${\mathrm{Q}}_{\mathrm{H}}={\mathrm{T}}_{\mathrm{H}}$and${\mathrm{Q}}_{\mathrm{L}}={\mathrm{T}}_{\mathrm{L}}$

Using equation (2) in equation (1) and the above-given values, we get that

$$\begin{gathered} K=\frac{Q_L}{Q_H-Q_L} \\ =\frac{Q_L}{W} \end{gathered}$$

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{L}}=\mathrm{K}\times \mathrm{W} \\ =\left(20.36\right)\left(1.0\mathrm{J}\right) \\ =20.36\mathrm{J} \end{gathered}$$

Therefore, the amount of heat in joules removed from the room is 20.36 J