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Chapter 15: Problem 34

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?

Short Answer

Expert verified
Answer: The hot temperature reservoir is approximately 442.9 K, and the amount of heat exhausted for every 0.100 kJ of work done is approximately 0.233 kJ.

Step by step solution

01

Use the formula for efficiency of a reversible engine

The efficiency of a reversible engine, \(\eta\), is related to the hot and cold temperature reservoirs (\(T_H\) and \(T_C\)) by the equation: $$ \eta = 1 - \frac{T_C}{T_H} $$ We are given the efficiency, \(\eta\), as \(30.0\%\) or \(0.30\), and we're given the cold temperature reservoir, \(T_C = 310.0 K\). We need to solve for the hot temperature reservoir, \(T_H\). Rearrange the equation above to solve for \(T_H\): $$ T_H = \frac{T_C}{1 - \eta} $$
02

Calculate \(T_H\) using the efficiency and \(T_C\) values

Now, plug in the given values of efficiency, \(\eta = 0.30\), and the cold temperature reservoir, \(T_C = 310.0 K\), into the above equation: $$ T_H = \frac{310.0}{1 - 0.30} = \frac{310.0}{0.70} $$ Upon evaluating the expression, we get the hot temperature reservoir, \(T_H \approx 442.9 K\). So, \(T_H = 442.9 K\).
03

Find the heat exhausted for every \(0.100 kJ\) of work done

To find the heat exhausted for every \(0.100 kJ\) of work done, first, we need to find how much heat is supplied to the engine, represented by \(Q_H\). We use the formula, $$ W = \eta Q_H $$ Rearrange the equation above to solve for \(Q_H\): $$ Q_H = \frac{W}{\eta} $$ Given the work done, \(W = 0.100 kJ\) and efficiency, \(\eta = 0.30\). $$ Q_H = \frac{0.100}{0.30} = 0.333... kJ $$
04

Find the heat exhausted, \(Q_C\), using the energy conservation principle

Now, we'll use the energy conservation principle, which states that the total heat supplied to the engine, \(Q_H\), is equal to the sum of the heat exhausted, \(Q_C\), and the work done, \(W\). Rearrange the equation as: $$ Q_C = Q_H - W $$ Substitute the value of \(Q_H = 0.333... kJ\) and \(W = 0.100 kJ\) into the equation: $$ Q_C = 0.333... - 0.100 = 0.233... kJ $$ So, the heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\). In summary, the hot temperature reservoir, \(T_H\), is approximately \(442.9 K\) and the amount of heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\).

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Most popular questions from this chapter

Two engines operate between the same two temperatures of \(750 \mathrm{K}\) and \(350 \mathrm{K},\) and have the same rate of heat input. One of the engines is a reversible engine with a power output of \(2.3 \times 10^{4} \mathrm{W} .\) The second engine has an efficiency of \(42 \% .\) What is the power output of the second engine?
List these in order of increasing entropy: (a) 1 mol of water at $20^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)20^{\circ} \mathrm{C}$ in separate containers; (b) a mixture of 1 mol of water at \(20^{\circ} \mathrm{C}\) and 1 mol of ethanol at \(20^{\circ} \mathrm{C} ;\) (c) 0.5 mol of water at \(20^{\circ} \mathrm{C}\) and 0.5 mol of ethanol at \(20^{\circ} \mathrm{C}\) in separate containers; (d) a mixture of 1 mol of water at $30^{\circ} \mathrm{C}\( and 1 mol of ethanol at \)30^{\circ} \mathrm{C}$.
Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
An electric power station generates steam at \(500.0^{\circ} \mathrm{C}\) and condenses it with river water at \(27^{\circ} \mathrm{C} .\) By how much would its theoretical maximum efficiency decrease if it had to switch to cooling towers that condense the steam at \(47^{\circ} \mathrm{C} ?\)
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?
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