91Ó°ÊÓ

Temperatures,${\mathrm{T}}_{\mathrm{L}}=5.0°\mathrm{C}=268.15\mathrm{K}$and

$T_H=17°C=290.15K$

We use the concept of conservation of energy. The ratio of heat taken from outside to the corresponding temperature is equal to the ratio of the heat absorbed by the room to the corresponding room temperature. We use this relation to find the amount of energy.

Formula:

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

$$\begin{gathered} \mathrm{K}=\frac{{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{Q}}_{\mathrm{H}}-{\mathrm{Q}}_{\mathrm{L}}} \\ \mathrm{K}=\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}-{\mathrm{T}}_{\mathrm{L}}} \end{gathered}$$ (1)

The work done per cycle of a Carnot Engine,

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

From equation (1), we can get the relation of heat transfer and temperature as given:

$\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}=\frac{{\mathrm{Q}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{L}}}$

Now, using equation (2) in the above equation, we can get the relation as follows:

$$\begin{gathered} \frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}=\frac{{\mathrm{Q}}_{\mathrm{H}}-\mathrm{W}}{{\mathrm{T}}_{\mathrm{L}}} \\ \frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{L}}}=-\frac{\mathrm{W}}{{\mathrm{T}}_{\mathrm{L}}} \\ \frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{H}}}=\frac{{\mathrm{Q}}_{\mathrm{H}}}{{\mathrm{T}}_{\mathrm{L}}}=\frac{\mathrm{W}}{{\mathrm{T}}_{\mathrm{L}}} \\ {\mathrm{Q}}_{\mathrm{H}}=\left(\frac{1}{{\mathrm{T}}_{\mathrm{L}}}-\frac{1}{{\mathrm{T}}_{\mathrm{H}}}\right)=\frac{\mathrm{W}}{{\mathrm{T}}_{\mathrm{L}}} \end{gathered}$$

$$\begin{gathered} {\mathrm{Q}}_{\mathrm{H}}=\frac{{\displaystyle \frac{\mathrm{W}}{{\mathrm{T}}_{\mathrm{L}}}}}{\left({\displaystyle \frac{1}{{\mathrm{T}}_{\mathrm{L}}}}-{\displaystyle \frac{1}{{\mathrm{T}}_{\mathrm{H}}}}\right)} \\ =\frac{1}{{\mathrm{T}}_{\mathrm{L}}\left({\displaystyle \frac{1}{{\mathrm{T}}_{\mathrm{L}}}}-{\displaystyle \frac{1}{{\mathrm{T}}_{\mathrm{H}}}}\right)} \\ =\frac{\mathrm{W}}{\left({\displaystyle 1-\frac{{\mathrm{T}}_{\mathrm{L}}}{{\mathrm{T}}_{\mathrm{H}}}}\right)} \\ =\frac{1.0\mathrm{J}}{\left(1-{\displaystyle \frac{268.15\mathrm{K}}{290-15\mathrm{K}}}\right)} \\ =13.2\mathrm{J} \end{gathered}$$

Therefore, the amount of energy that would be transferred as heat to the room for each joule of electric energy consumed is 13.2J