Chapter 15, Problem 74P

(a)

To determine

The efficiency of reversible engine.

Answer to Problem 74P

Efficiency is $0.051$.

Explanation of Solution

Average temperature difference between top and bottom is $15°\text{C}$ and the average surface temperature is $22°\text{C}$.

Write the equation to calculate the efficiency.

$\epsilon =1-\left(\frac{T_{sur}-T_{diff}}{T_{sur}}\right)$

Here, $\epsilon$ is the efficiency, $T_{sur}$ is the average surface temperature and $T_{diff}$ is the average temperature difference between top and bottom.

Conclusion:

Substitute $22°\text{C}$ for $T_{sur}$ and $15°\text{C}$ or $T_{diff}$ in the above equation to find $\epsilon$.

$\begin{array}{c}\epsilon =1-\frac{\left(\frac{22°\text{C}-15°\text{C}}{22°\text{C}-15°\text{C}}\right)\left(\text{273}\text{.15}\text{K+22}\text{K}-\text{15}\text{K}\right)}{\left(\frac{22°\text{C}}{22°\text{C}}\right)\left(\text{273}\text{.15}\text{K+22}\text{K}\right)}\\ =1-0.949\\ =0.051\end{array}$

Therefore, the efficiency is $0.051$.

(b)

To determine

The cubic meters of water used by heat engine per second.

Answer to Problem 74P

Water used by heat engine is $31{\text{m}}^{\text{3}}$.

Explanation of Solution

Average temperature difference between top and bottom is $15°\text{C}$ and the average surface temperature is $22°\text{C}$, power supplied by the lake is $1.0\times {10}^8\text{W}$.

Write the equation to find the power supplied to the town.

$P=\epsilon \frac{\Delta Q}{\Delta t}$

Here, $P$ is the power supplied to the town, $\Delta Q$ is the heat energy supplied, and $\Delta t$ is the time duration.

Write the equation to find $\Delta Q$.

$\Delta Q=\left(\rho V\right)c\Delta T$

Here, $\rho$ is the density of water, $V$ is the volume of water, $c$ is the specific heat capacity of water, and $\Delta T$ is the temperature change.

Rewrite the equation for $P$ by substituting the above relation for $\Delta Q$.

$\begin{array}{c}P=\epsilon \frac{\left(\rho V\right)c\Delta T}{\Delta t}\\ V=\frac{P\Delta t}{e\rho c\Delta T}\end{array}$

Conclusion:

Substitute $1.0\times {10}^8\text{W}$ for $P$, $1.0\text{s}$ for $\Delta t$, $0.051$ for $\epsilon$, $1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $4186\text{J}/\text{kg}\cdot \text{K}$ for $c$, and $15\text{K}$ for $\Delta T$ in the above equation to find $V$.

$\begin{array}{c}V=\frac{\left(1.0\times {10}^8\text{W}\right)\left(1.0\text{s}\right)}{\left(0.051\right)\left(1.00\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left(4186\text{J}/\text{kg}\cdot \text{K}\right)\left(15\text{K}\right)}\\ =31{\text{m}}^{\text{3}}\end{array}$

Therefore, the water used by heat engine is $31{\text{m}}^{\text{3}}$.

(c)

To determine

Check whether the lake can supply enough heat energy to meet the town’s energy need.

Answer to Problem 74P

Yes. Lake can supply enough heat energy to meet the town’s energy need.

Explanation of Solution

Average temperature difference between top and bottom is $15°\text{C}$ and the average surface temperature is $22°\text{C}$, power supplied by the lake is $1.0\times {10}^8\text{W}$, surface area of lake is $8.0\times {10}^7{\text{m}}^{\text{2}}$, and the average incident intensity is $200\text{W}/{\text{m}}^{\text{2}}$.

Write the equation to find the power delivered by sun.

$P_{sun}=IA$

Here, $P_{sun}$ is the average power of sunlight, $I$ is the intensity, and $A$ is the area of cross section.

Write the equation to calculate the power of engine.

$P_{engine}=\frac{P_{town}}{\epsilon }$

Here, $P_{engine}$ is the power of engine, and $P_{town}$ is the power required by the town.

Conclusion:

Substitute $200\text{W}/{\text{m}}^{\text{2}}$ for $I$ and $8.0\times {10}^7{\text{m}}^{\text{2}}$ for $A$ in the equation for $P_{sun}$.

$\begin{array}{c}{P}_{sun}=\left(200\text{W}/{\text{m}}^{\text{2}}\right)\left(8.0\times {10}^7{\text{m}}^{\text{2}}\right)\\ =1.6\times {10}^{10}\text{W}\end{array}$

Substitute $1.0\times {10}^8\text{W}$ for $P_{town}$ and $0.051$ for $\epsilon$ in the equation for $P_{engine}$.

$\begin{array}{c}{P}_{engine}=\frac{1.0\times {10}^8\text{W}}{0.051}\\ =2.0\times {10}^9\text{W}\end{array}$

It can be seen that $P_{sun}>P_{engine}$.This means that the lake can supply enough heat energy to meet the town’s energy need.

Therefore, Yes. Lake can supply enough heat energy to meet the town’s energy need.