Chapter 22, Problem 77PQ

(a)

To determine

The change in entropy of the universe for each cycle.

Answer to Problem 77PQ

The change in entropy of the universe for each cycle is $48.3\text{J}/\text{K}$.

Explanation of Solution

Write the expression to calculate the change in entropy of the hot reservoir.

  $\Delta s_h=\frac{-Q_h}{T_h}$                                                                                                                 (I)

Here, $\Delta s_h$ is the change in entropy of the hot reservoir, $Q_h$ is the heat lost from the hot reservoir and $T_h$ is the temperature of hot reservoir.

Write the expression to calculate the change in entropy of the hot reservoir.

  $\Delta s_c=\frac{Q_c}{T_c}$                                                                                                               (II)

Here, $\Delta s_c$ is the change in entropy of the cold reservoir, $Q_c$ is the heat added to the cold reservoir and $T_c$ is the temperature of cold reservoir.

Write the expression to calculate the heat added to the cold reservoir.

  $Q_c=Q_h-W$

Here, W is the work done by the engine.

Substitute the above equation in (II) to rewrite.

  $\Delta s_c=\frac{Q_h-W}{T_c}$                                                                                                     (III)

Write the expression to calculate the entropy change in universe.

  $\Delta s=\Delta s_h+\Delta s_c$

Here, $\Delta s$ is the change in entropy of the universe.

Substitute the equation (I) and (III) in the above equation to rewrite.

  $\Delta s=\frac{-Q_h}{T_h}+\frac{Q_h-W}{T_c}$

Conclusion:

Substitute $3.25\times {10}^4\text{J}$ for $Q_h$, $673\text{K}$ $\left(400\text{°C}\right)$ for $T_h$, $295\text{K}$ $\left(22.0\text{°C}\right)$ for $\left(400\text{°C}\right)$ and $4.00\text{kJ}$ for W in the above equation to calculate $\Delta s$.

  $\begin{array}{c}\Delta s=\frac{-\left(3.25\times {10}^4\text{J}\right)}{673\text{K}}+\frac{3.25\times {10}^4\text{J}-\left(4.00\text{kJ}\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\right)}{295\text{K}}\\ =48.3\text{J}/\text{K}\end{array}$

Therefore, the change in entropy of the universe for each cycle is $48.3\text{J}/\text{K}$.

(b)

To determine

The required more work has to be done by the Carnot engine.

Answer to Problem 77PQ

The required more work has to be done by the Carnot engine is $1.43\times {10}^4\text{J}$.

Explanation of Solution

Write the expression to calculate the work done by the Carnot engine.

  $W^{\prime }=e{Q}_h$

Here, $W^{\prime }$ is the work done and e is the efficiency.

Write the expression to calculate the efficiency of the engine.

  $e=1-\frac{T_c}{T_h}$

Rewrite the equation for W using the above expression.

  $W^{\prime }=\left(1-\frac{T_c}{T_h}\right)Q_h$                                                                                                       (I)

Write the expression to calculate the more work required.

  $\Delta W=W^{\prime }-W$                                                                                                           (II)

Here, $\Delta W$ is the more work that has to be done by the engine.

Conclusion:

Substitute $3.25\times {10}^4\text{J}$ for $Q_h$, $673\text{K}$ $\left(400\text{°C}\right)$ for $T_h$ and $295\text{K}$ $\left(22.0\text{°C}\right)$ for $\left(400\text{°C}\right)$ in the above equation (I) to calculate $W^{\prime }$.

    $\begin{array}{c}{W}^{\prime }=\left(1-\frac{295\text{K}}{673\text{K}}\right)\left(3.25\times {10}^4\text{J}\right)\\ =1.83\times {10}^4\text{J}\end{array}$

Substitute $1.83\times {10}^4\text{J}$ for $W^{\prime }$ and $4.00\text{kJ}$ for W in the equation (II) to calculate $\Delta W$.

    $\begin{array}{c}\Delta W=1.83\times {10}^4\text{J}-4.00\text{kJ}\left(\frac{{10}^3\text{J}}{1\text{kJ}}\right)\\ =1.43\times {10}^4\text{J}\end{array}$

Therefore, the required more work has to be done by the Carnot engine is $1.43\times {10}^4\text{J}$.