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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 heat engine takes in \(125 \mathrm{kJ}\) of heat from a reservoir at $815 \mathrm{K}\( and exhausts \)82 \mathrm{kJ}\( to a reservoir at \)293 \mathrm{K}$ (a) What is the efficiency of the engine? (b) What is the efficiency of an ideal engine operating between the same two reservoirs?
If the pressure on a fish increases from 1.1 to 1.2 atm, its swim bladder decreases in volume from \(8.16 \mathrm{mL}\) to \(7.48 \mathrm{mL}\) while the temperature of the air inside remains constant. How much work is done on the air in the bladder?
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.
A container holding \(1.20 \mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) is placed in a freezer that is kept at \(-20.0^{\circ} \mathrm{C} .\) The water freezes and comes to thermal equilibrium with the interior of the freezer. What is the minimum amount of electrical energy required by the freezer to do this if it operates between reservoirs at temperatures of $20.0^{\circ} \mathrm{C}\( and \)-20.0^{\circ} \mathrm{C} ?$
What is the efficiency of an electric generator that produces $1.17 \mathrm{kW}\(.h per \)\mathrm{kg}$ of coal burned? The heat of combustion of coal is \(6.71 \times 10^{6} \mathrm{J} / \mathrm{kg}\).
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