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Chapter 15: Problem 45

Show that in a reversible engine the amount of heat \(Q_{C}\) exhausted to the cold reservoir is related to the net work done \(W_{\text {net }}\) by $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}} W_{\mathrm{net}} $$

Short Answer

Expert verified
Answer: The amount of heat exhausted to the cold reservoir \(Q_C\) is related to the net work done \(W_{\text{net}}\) by the equation: $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}W_{\mathrm{net}} $$

Step by step solution

01

Efficiency for Reversible Engines

A reversible engine operating between hot and cold reservoirs is the most efficient engine possible. Its efficiency is given by: $$ \eta_{\text{rev}} = 1 - \frac{T_C}{T_H} $$
02

Relating Heat and Work Transfer to Efficiency

Since the efficiency of an engine is defined as the ratio of the work done to the heat absorbed from the hot reservoir, we can write: $$ \eta_{\text{rev}} = \frac{W_{\text{net}}}{Q_H} $$ Then, we substitute the expression for the efficiency from Step 1: $$ 1 - \frac{T_C}{T_H} = \frac{W_{\text{net}}}{Q_H} $$
03

Heat Transfer to Cold Reservoir

In a reversible engine, the relationship between heat absorbed from the hot reservoir \(Q_H\) and heat exhausted to the cold reservoir \(Q_C\) is: $$ Q_H = Q_C + W_{\text{net}} $$
04

Substituting and Solving for \(Q_C\)

Now, we can substitute the expression for \(Q_H\) from Step 3 into the equation from Step 2: $$ 1 - \frac{T_C}{T_H} = \frac{W_{\text{net}}}{Q_C + W_{\text{net}}} $$ Then, we can solve for \(Q_C\): $$ Q_C\left(1 - \frac{T_C}{T_H}\right) = W_{\text{net}} - \frac{T_C}{T_H}W_{\text{net}} $$ $$ Q_C =\frac{T_C}{T_H - T_C} W_{\text{net}} $$
05

Conclusion

We have shown that, for a reversible engine, the amount of heat exhausted to the cold reservoir \(Q_C\) is related to the net work done \(W_{\text{net}}\) by the given equation: $$ Q_{\mathrm{C}}=\frac{T_{\mathrm{C}}}{T_{\mathrm{H}}-T_{\mathrm{C}}}W_{\mathrm{net}} $$

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Most popular questions from this chapter

A balloon contains \(200.0 \mathrm{L}\) of nitrogen gas at $20.0^{\circ} \mathrm{C}$ and at atmospheric pressure. How much energy must be added to raise the temperature of the nitrogen to \(40.0^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure?
A reversible engine with an efficiency of \(30.0 \%\) has $T_{\mathrm{C}}=310.0 \mathrm{K} .\( (a) What is \)T_{\mathrm{H}} ?$ (b) How much heat is exhausted for every \(0.100 \mathrm{kJ}\) of work done?
A town is considering using its lake as a source of power. The average temperature difference from the top to the bottom is \(15^{\circ} \mathrm{C},\) and the average surface temperature is \(22^{\circ} \mathrm{C} .\) (a) Assuming that the town can set up a reversible engine using the surface and bottom of the lake as heat reservoirs, what would be its efficiency? (b) If the town needs about \(1.0 \times 10^{8} \mathrm{W}\) of power to be supplied by the lake, how many \(\mathrm{m}^{3}\) of water does the heat engine use per second? (c) The surface area of the lake is $8.0 \times 10^{7} \mathrm{m}^{2}$ and the average incident intensity (over \(24 \mathrm{h})\) of the sunlight is \(200 \mathrm{W} / \mathrm{m}^{2} .\) Can the lake supply enough heat to meet the town's energy needs with this method?

On a cold winter day, the outside temperature is \(-15.0^{\circ} \mathrm{C} .\) Inside the house the temperature is \(+20.0^{\circ} \mathrm{C}\) Heat flows out of the house through a window at a rate of \(220.0 \mathrm{W}\). At what rate is the entropy of the universe changing due to this heat conduction through the window?
Suppose 1.00 mol of oxygen is heated at constant pressure of 1.00 atm from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} .\) (a) How much heat is absorbed by the gas? (b) Using the ideal gas law, calculate the change of volume of the gas in this process. (c) What is the work done by the gas during this expansion? (d) From the first law, calculate the change of internal energy of the gas in this process.
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