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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 heat engine consists of a heat source that causes a monatomic gas to expand, pushing against a piston, thereby doing work. The gas begins at a pressure of \(300 . \mathrm{kPa}\), a volume of \(150 . \mathrm{cm}^{3}\), and room temperature, \(20.0^{\circ} \mathrm{C}\). On reaching a volume of \(450 . \mathrm{cm}^{3}\), the piston is locked in place, and the heat source is removed. At this point, the gas cools back to room temperature. Finally, the piston is unlocked and used to isothermally compress the gas back to its initial state. a) Sketch the cycle on a \(p V\) -diagram. b) Determine the work done on the gas and the heat flow out of the gas in each part of the cycle. c) Using the results of part (b), determine the efficiency of the engine.

The change in entropy of a system can be calculated because a) it depends only on the c) entropy always increases. initial and final states. d) none of the above. b) any process is reversible.

An outboard motor for a boat is cooled by lake water at \(15.0^{\circ} \mathrm{C}\) and has a compression ratio of \(10.0 .\) Assume that the air is a diatomic gas. a) Calculate the efficiency of the engine's Otto cycle. b) Using your answer to part (a) and the fact that the efficiency of the Carnot cycle is greater than that of the Otto cycle, estimate the maximum temperature of the engine.

20.9a) The maximum efficiency of a Carnot engine is \(100 \%\) since the Carnot cycle is an ideal process. b) The Carnot cycle consists of two isothermal processes and two adiabatic processes. c) The Carnot cycle consists of two isothermal processes and two isentropic processes (constant entropy). d) The efficiency of the Carnot cycle depends solely on the temperatures of the two thermal reservoirs.

A certain refrigerator is rated as being \(32.0 \%\) as ef ficient as a Carnot refrigerator. To remove \(100 .\) J of heat from the interior at \(0^{\circ} \mathrm{C}\) and eject it to the outside at \(22^{\circ} \mathrm{C}\), how much work must the refrigerator motor do?
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