Chapter 11, Problem 3SP

A Carnot engine operating in reverse as a heat pump moves heat from a cold reservoir at 7°C to a warmer one at 22°C.

a.    What is the efficiency of a Carnot engine operating between these two temperatures?

b.    If the Carnot heat pump releases 250 J of heat into the higher-temperature reservoir in each cycle, how much work must be provided in each cycle ?

c.    How much heat is removed from the 7°C reservoir in each cycle?

d.    The performance of a refrigerator or heat pump is described by a “coefficient of performance” defined as K = Q_c/W. What is the coefficient of performance for our Carnot heat pump?

e.    Are the temperatures used in this example appropriate to the application of a heat pump for home heating? Explain.

To determine

The efficiency of the Carnot engine.

Answer to Problem 3SP

The efficiency of the Carnot engine is $5.1\%$.

Explanation of Solution

Given info:

Temperature of hot reservoir is $22°\text{C}$ and temperature of cold reservoir is $7°\text{C}$.

Write an expression to calculate the efficiency.

$\eta =1-\frac{T_C}{T_H}$

Here,

$\eta$ is the efficiency of the Carnot engine

$T_C$ is the temperature of cold reservoir

$T_H$ is the temperature of hot reservoir

Substitute $7°\text{C}$ for $T_C$ and $22°\text{C}$ for $T_H$ to find $\eta$.

$\begin{array}{c}\eta =1-\frac{\left(7+273\right)\text{K}}{\left(22+273\right)\text{K}}\\ =0.051\\ =\left(0.051\right)\left(100\%\right)\\ =5.1\%\end{array}$

Thus, the efficiency of the Carnot engine is $5.1\%$.

Conclusion:

The efficiency of the Carnot engine is $5.1\%$.

To determine

The work provided in each cycle.

Answer to Problem 3SP

The work provided in each cycle is $12.75\text{J}$.

Explanation of Solution

Given info:

Heat released to hot reservoir is $250\text{J}$.

Write an expression for work provided in each cycle.

$W=\eta Q_H$

Here,

$Q_H$ is the heat released to hot reservoir.

$W$ is the work done.

Substitute $0.051$ for $\eta$ and $250\text{J}$ for $Q_H$ to find $W$.

$\begin{array}{c}W=\left(0.051\right)\left(250\text{J}\right)\\ =12.75\text{J}\end{array}$

Thus, work provided in each cycle is $12.75\text{J}$.

Conclusion:

The work provided in each cycle is $12.75\text{J}$.

To determine

The heat released from cold reservoir.

Answer to Problem 3SP

The heat released from cold reservoir is $237.25\text{J}$.

Explanation of Solution

Write an expression for heat released from cold reservoir.

$Q_C=Q_H-W$

Here,

$Q_C$ is the heat released from cold reservoir

Substitute $250\text{J}$ for $Q_H$ and $12.75\text{J}$ for $W$ to find $Q_C$.

$\begin{array}{c}{Q}_C=250\text{J}-12.75\text{J}\\ =237.25\text{J}\end{array}$

Thus, the heat released from cold reservoir is $237.25\text{J}$.

Conclusion:

The heat released from cold reservoir is $237.25\text{J}$.

To determine

The coefficient of performance of the heat pump.

Answer to Problem 3SP

The coefficient of performance of the heat pump is $18.6$.

Explanation of Solution

Write an expression for coefficient of performance of the heat pump.

$K=\frac{Q_C}{W}$

Here,

$K$ is the coefficient of performance of the heat pump

Substitute $237.25\text{J}$ for $Q_C$ and $12.75\text{J}$ for $W$ to find $K$.

$\begin{array}{c}K=\frac{237.25\text{J}}{12.75\text{J}}\\ =18.6\end{array}$

Thus, the coefficient of performance of the heat pump is $18.6$.

Conclusion:

The coefficient of performance of the heat pump is $18.6$.

To determine

The possibility of application of the heat pump for home heating.

Answer to Problem 3SP

Yes, the heat pump can be used for home heating.

Explanation of Solution

For home heating only moderate energy range is required. That will be sufficient to increase the temperature of home slightly. Here for working of the pump, only moderate range of energy is required.

Since the energy required is moderate, the energy provided by the heat pump will be adequate. The energy release will be less compared to the energy required to run the heat pump.

Conclusion:

Yes, the heat pump can be used for home heating.