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Chapter 19: Problem 34

Use appropriate energy-flow diagrams to analyze the situation in Got It? \(19.2 ;\) that is, show that using a refrigerator to cool the lowtemperature reservoir can't increase the overall efficiency of a Carnot engine when the work input to the refrigerator is included.

Short Answer

Expert verified
Using a refrigerator to cool the low-temperature reservoir of a Carnot engine cannot increase the overall efficiency. This is because the work input to the refrigerator, while contributing to the total energy input into the system, does not contribute to the work output of the Carnot engine, thereby reducing the overall efficiency.

Step by step solution

01

Understand the Carnot engine

A Carnot engine is a theoretical device that operates between two different temperature reservoirs and has the highest efficiency possible for a heat engine. It does so by first absorbing heat \(Q_H\) from a high-temperature reservoir, performing necessary work, \(W\), and then rejecting remaining heat, \(Q_C\), to a low-temperature reservoir.
02

Evaluate the effect of a refrigerator on the system

A refrigerator operates by taking heat \(Q_C'\) from a low temperature reservoir and dumping a higher amount of heat \(Q_H'\) into a high temperature reservoir. This is achieved by doing work \(W'\) on the system. Thus, if a refrigerator is used to cool the low-temperature reservoir of a Carnot engine, the work input \(W'\) to the refrigerator is increasing the total energy input into the system but not contributing to the work output of the original Carnot engine.
03

Analyze the net effect on efficiency

The efficiency of the Carnot engine is given by \( \eta = \frac{W}{Q_H} \). After adding in the work input \(W'\) to the refrigerator, the total input into the system becomes \(Q_H + W'\), but the work output remains the same. Therefore, the revised efficiency is \( \eta' = \frac{W}{Q_H + W'} \), which is lower than the original efficiency since \( \eta' \leq \eta \). This shows that using a refrigerator to cool the low-temperature reservoir cannot increase the overall efficiency of a Carnot engine when the work input to the refrigerator is included.

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Most popular questions from this chapter

A reversible engine contains 0.20 mol of ideal monatomic gas, initially at \(600 \mathrm{K}\) and confined to \(2.0 \mathrm{L} .\) The gas undergoes the following cycle: Isothermal expansion to \(4.0 \mathrm{L}\) \(\cdot\) Isovolumic cooling to \(300 \mathrm{K}\) Isothermal compression to \(2.0 \mathrm{L}\) \(\cdot\) Isovolumic heating to \(600 \mathrm{K}\) (a) Calculate the net heat added during the cycle and the net work done. (b) Determine the engine's efficiency, defined as the ratio of the work done to the heat absorbed during the cycle.

The molar specific heat at constant pressure for a certain gas is given by \(C_{p}=a+b T+c T^{2},\) where \(a=33.6 \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}\), \(b=2.93 \times 10^{-3} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{2},\) and \(c=2.13 \times 10^{-5} \mathrm{J} / \mathrm{mol} \cdot \mathrm{K}^{3} .\) Find the entropy change when 2 moles of this gas are heated from \(20^{\circ} \mathrm{C}\) to \(200^{\circ} \mathrm{C}\).

Could you cool the kitchen by leaving the refrigerator open? Explain.

You're engineering an energy-efficient house that will require an average of \(4.6 \mathrm{kW}\) to heat on cold winter days. You've designed a photovoltaic system for electric power, which will supply on average \(2.0 \mathrm{kW}\). You propose to heat the house with an electrically operated groundwater-based heat pump. What should you specify as the minimum acceptable COP for the pump if the photovoltaic system supplies its energy?

A Carnot engine absorbs \(900 \mathrm{J}\) of heat each cycle and provides \(350 \mathrm{J}\) of work. (a) What's its efficiency? (b) How much heat is rejected each cycle? (c) If the engine rejects heat at \(10^{\circ} \mathrm{C},\) what's its maximum temperature?
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