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Chapter 15: Problem 32

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?

Short Answer

Expert verified
Answer: The power output of the second engine is \(1.81 \times 10^4 \, \mathrm{W}\).

Step by step solution

01

Find the efficiency of the reversible engine

The efficiency of a reversible (Carnot) heat engine is given by the formula: $$\eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H}$$ where \(\eta_{\text{Carnot}}\) is the efficiency of the reversible engine, \(T_C\) is the temperature of the cold reservoir, and \(T_H\) is the temperature of the hot reservoir. In this problem, \(T_H = 750 \, \mathrm{K}\) and \(T_C = 350 \, \mathrm{K}\). Plug in these values to calculate the efficiency of the reversible engine: $$\eta_{\text{Carnot}} = 1 - \frac{350}{750} = 1 - \frac{7}{15} = \frac{8}{15}$$
02

Calculate the heat input of the reversible engine

The heat input \(Q_H\) to the reversible engine is related to its power output \(P_{\text{Carnot}}\) and efficiency \(\eta_{\text{Carnot}}\) by the formula: $$P_{\text{Carnot}} = \eta_{\text{Carnot}} Q_H$$ Plug in the known values for the reversible engine: \(\eta_{\text{Carnot}} = \frac{8}{15}\) and \(P_{\text{Carnot}} = 2.3 \times 10^4 \, \mathrm{W}\), and solve for the heat input \(Q_H\): $$Q_H = \frac{P_{\text{Carnot}}}{\eta_{\text{Carnot}}} = \frac{2.3 \times 10^4}{\frac{8}{15}} = 2.3 \times 10^4 \times \frac{15}{8} = 4.3125 \times 10^4 \, \mathrm{W}$$
03

Calculate the power output of the second engine

The power output \(P_2\) of the second engine is related to its heat input and efficiency \(\eta_2\) by the formula: $$P_2 = \eta_2 Q_H$$ The efficiency of the second engine is given as \(42 \%\), which is equal to \(\frac{42}{100}\). Plug in the values for the second engine: \(\eta_2 = \frac{42}{100}\) and \(Q_H = 4.3125 \times 10^4 \, \mathrm{W}\), and solve for the power output \(P_2\): $$P_2 = \frac{42}{100} \times 4.3125 \times 10^4 = 1.81 \times 10^4 \, \mathrm{W}$$
04

Final Answer

The power output of the second engine is \(1.81 \times 10^4 \, \mathrm{W}\).

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Most popular questions from this chapter

The efficiency of a muscle during weight lifting is equal to the work done in lifting the weight divided by the total energy output of the muscle (work done plus internal energy dissipated in the muscle). Determine the efficiency of a muscle that lifts a \(161-\mathrm{N}\) weight through a vertical displacement of \(0.577 \mathrm{m}\) and dissipates \(139 \mathrm{J}\) in the process.
Estimate the entropy change of \(850 \mathrm{g}\) of water when it is heated from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C} .\) [Hint: Assume that the heat flows into the water at an average temperature. \(]\)
A balloon contains \(160 \mathrm{L}\) of nitrogen gas at \(25^{\circ} \mathrm{C}\) and 1.0 atm. How much energy must be added to raise the temperature of the nitrogen to \(45^{\circ} \mathrm{C}\) while allowing the balloon to expand at atmospheric pressure?
Calculate the maximum possible efficiency of a heat engine that uses surface lake water at \(18.0^{\circ} \mathrm{C}\) as a source of heat and rejects waste heat to the water \(0.100 \mathrm{km}\) below the surface where the temperature is \(4.0^{\circ} \mathrm{C}\).
The United States generates about \(5.0 \times 10^{16} \mathrm{J}\) of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost \(100 \%\) efficiency by an electric motor. (a) If this energy is generated by power plants with an average efficiency of \(0.30,\) how much heat is dumped into the environment each day? (b) How much water would be required to absorb this heat if the water temperature is not to increase more than \(2.0^{\circ} \mathrm{C} ?\)
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