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Chapter 15: Problem 37

An inventor proposes a heat engine to propel a ship, using the temperature difference between the water at the surface and the water \(10 \mathrm{m}\) below the surface as the two reservoirs. If these temperatures are \(15.0^{\circ} \mathrm{C}\) and \(10.0^{\circ} \mathrm{C},\) respectively, what is the maximum possible efficiency of the engine?

Short Answer

Expert verified
Answer: The maximum possible efficiency of the engine is approximately 1.74%.

Step by step solution

01

Convert temperatures to Kelvin

To calculate the maximum efficiency of the engine, we must first convert the temperatures from Celsius to Kelvin. To do this, we simply add 273.15 to each Celsius temperature. \(T_H = 15.0 + 273.15 = 288.15 \, \mathrm{K}\) \(T_C = 10.0 + 273.15 = 283.15 \, \mathrm{K}\) Where \(T_H\) is the temperature of the hot reservoir (water surface) and \(T_C\) is the temperature of the cold reservoir (10m below the surface).
02

Calculate the maximum efficiency

Now that we have the temperatures in Kelvin, we can use the formula for the maximum efficiency of a heat engine, given by the Carnot efficiency: \(\eta_{max} = 1 - \frac{T_C}{T_H}\) Plug the values of \(T_H\) and \(T_C\) into this formula: \(\eta_{max} = 1 - \frac{283.15}{288.15}\)
03

Compute the result

Perform the calculation: \(\eta_{max} = 1 - 0.9826 \approx 0.0174\)
04

Express the efficiency as a percentage

To express the maximum efficiency as a percentage, multiply the result by 100: \(\eta_{max} \% = 0.0174 \times 100 \approx 1.74 \%\) The maximum possible efficiency of the engine is approximately 1.74%.

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Most popular questions from this chapter

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 \(2.0^{\circ} \mathrm{C} ?\)
A monatomic ideal gas at \(27^{\circ} \mathrm{C}\) undergoes a constant volume process from \(A\) to \(B\) and a constant pressure process from \(B\) to \(C\) Find the total work done during these two processes.
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. \(]\)
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.
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.
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