Question

The electric motor of a heat pump transfers energy as heat from the outdoors, which is at $\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{°}\mathbf{C}$, to a room that is at$\mathbf{17}\mathbf{°}\mathbf{C}$. If the heat pump were a Carnot heat pump (a Carnot engine working in reverse), how much energy would be transferred as heat to the room for each joule of electric energy consumed?

Short answer

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

Step-by-step solution

The given data

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

$T_H=17°C=290.15K$

Understanding the concept of the Carnot engine

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)

Calculation of the amount of energy to be transferred

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