/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A Carnot engine operates between... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Is it a violation of the Second Law of Thermodynamics to capture all the exhaust heat from a steam engine and funnel it back into the system to do work? Why or why not?

1 .00 mole of a monatomic ideal gas at a pressure of 4.00 atm and a volume of \(30.0 \mathrm{~L}\) is isothermically expanded to a pressure of 1.00 atm and a volume of \(120.0 \mathrm{~L}\). Next, it is compressed at a constant pressure until its volume is \(30.0 \mathrm{~L}\), and then its pressure is increased at the constant volume of \(30.0 \mathrm{~L}\). What is the efficiency of this heat engine cycle?

Why might a heat pump have an advantage over a space heater that converts electrical energy directly into thermal energy?

A key feature of thermodynamics is the fact that the internal energy, \(E_{\text {int }}\) of a system and its entropy, \(S\), are state variables; that is, they depend only on the thermodynamic state of the system and not on the processes by which it reached that state (unlike, for example, the heat content, \(Q\) ). This means that the differentials \(d E_{\text {int }}=T d S-p d V\) and \(d S=\) \(T^{-1} d E_{\text {int }}+p T^{-1} d V,\) where \(T\) is temperature (in kelvins), \(p\) is pressure, and \(V\) is volume, are exact differentials as defined in calculus. What relationships follow from this fact?

A volume of \(6.00 \mathrm{~L}\) of a monatomic ideal gas, originally at \(400 . \mathrm{K}\) and a pressure of \(3.00 \mathrm{~atm}\) (called state 1 ), undergoes the following processes, all done reversibly: \(1 \rightarrow 2\) isothermal expansion to \(V_{2}=4 V_{1}\) \(2 \rightarrow 3\) isobaric compression \(3 \rightarrow 1\) adiabatic compression to its original state Find the entropy change for each process.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.