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Chapter 20: Problem 33

A Carnot engine operates between a warmer reservoir at a temperature \(T_{1}\) and a cooler reservoir at a temperature \(T_{2}\). It is found that increasing the temperature of the warmer reservoir by a factor of 2 while keeping the same temperature for the cooler reservoir increases the efficiency of the Carnot engine by a factor of 2 as well. Find the efficiency of the engine and the ratio of the temperatures of the two reservoirs in their original form.

Short Answer

Expert verified
Answer: The efficiency of the engine in its original form is 1/3 or 33.33%, and the ratio between the temperatures of the two reservoirs in their original form is 2/3.

Step by step solution

01

Write down the efficiency equation for a Carnot Engine

A Carnot engine's efficiency is given by the equation: \[e =1 - \frac{T_{2}}{T_{1}}\]
02

Setup initial and final efficiency equations

We are given that the temperature of the warmer reservoir T1 is increased by a factor of 2, so the new temperature will be 2T1. Let's denote the initial efficiency as e and the final efficiency as e'. We now have: \[e' = 2e\] Now, using the efficiency equation for the engine with the original temperature and the increased temperature, we get: \[e = 1 - \frac{T_{2}}{T_{1}}\] \[e' = 1 - \frac{T_{2}}{2T_{1}}\]
03

Substitute expressions and solve for the desired variables

We have two equations with two unknowns, e and T2/T1. We can substitute e from one equation to the other, so we get: \[2(1-\frac{T_2}{T_1})=1-\frac{T_2}{2T_1}\] Now, solve the equation for the ratio \(\frac{T_{2}}{T_{1}}\): Expanding and simplifying the equation: \[2-2\frac{T_{2}}{T_{1}} = 1 - \frac{T_{2}}{2T_{1}}\] \[2\frac{T_{2}}{T_{1}}-\frac{T_{2}}{2T_{1}} = 1\] \[(4-1)\frac{T_{2}}{T_{1}} = 2\] \[\frac{T_{2}}{T_{1}} = \frac{2}{3}\] We found that the ratio between the temperatures of the two reservoirs in their original form is 2/3. Next, we will plug this value into the efficiency equation to find the efficiency e: \[e = 1 - \frac{2}{3}\] \[e = \frac{1}{3}\]
04

Present the final answer

The efficiency of the engine in its original form is 1/3 or 33.33% and the ratio between the temperatures of the two reservoirs in their original form is 2/3.

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Most popular questions from this chapter

A water-cooled engine produces \(1000 .\) W of power. Water enters the engine block at \(15.0^{\circ} \mathrm{C}\) and exits at \(30.0^{\circ} \mathrm{C}\). The rate of water flow is \(100 . \mathrm{L} / \mathrm{h}\). What is the engine's efficiency?

What is the magnitude of the change in entropy when \(6.00 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is condensed to water at \(100{ }^{\circ} \mathrm{C} ?\) a) \(46.6 \mathrm{~J} / \mathrm{K}\) c) \(36.3 \mathrm{~J} / \mathrm{K}\) b) \(52.4 \mathrm{~J} / \mathrm{K}\) d) \(34.2 \mathrm{~J} / \mathrm{K}\)

Which of the following processes always results in an increase in the energy of a system? a) The system loses heat and does work on the surroundings. b) The system gains heat and does work on the surroundings. c) The system loses heat and has work done on it by the surroundings. d) The system gains heat and has work done on it by the surroundings. e) None of the above.

A Carnot refrigerator is operating between thermal reservoirs with temperatures of \(27.0^{\circ} \mathrm{C}\) and \(0.00^{\circ} \mathrm{C}\) a) How much work will need to be input to extract \(10.0 \mathrm{~J}\) of heat from the colder reservoir? b) How much work will be needed if the colder reservoir is at \(-20.0^{\circ} \mathrm{C}^{2}\)

Consider a Carnot engine that works between thermal reservoirs with temperatures of \(1000.0 \mathrm{~K}\) and \(300.0 \mathrm{~K}\). The average power of the engine is \(1.00 \mathrm{~kJ}\) per cycle. a) What is the efficiency of this engine? b) How much energy is extracted from the warmer reservoir per cycle? c) How much energy is delivered to the cooler reservoir?
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