Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 25

An ideal engine has an efficiency of 0.725 and uses gas from a hot reservoir at a temperature of \(622 \mathrm{K}\). What is the temperature of the cold reservoir to which it exhausts heat?

Short Answer

Expert verified
Based on the given efficiency of 0.725 and a hot reservoir temperature of 622 K, the temperature of the cold reservoir for an ideal engine is approximately 275 K.

Step by step solution

01

Write down the given information

We are given: - Efficiency (\(\eta\)): 0.725 - Temperature of hot reservoir (\(T_h\)): 622 K
02

Find the temperature of the cold reservoir using the Carnot efficiency formula

We can use the equation \(\eta = 1 - \frac{T_c}{T_h}\) and solve for the temperature of the cold reservoir, \(T_c\). First, plug in the given values: \(0.725 = 1 - \frac{T_c}{622}\)
03

Solve for \(T_c\)

To solve for \(T_c\), first subtract 1 from both sides of the equation: \(0.725 - 1 = - \frac{T_c}{622}\) Now, multiply both sides of the equation by 622: \((0.725 - 1) \times 622 = -T_c\) Next, simplify the equation: \(-275 \approx -T_c\) Finally, multiply both sides by -1: \(T_c \approx 275 \mathrm{K}\)
04

State the final answer

The temperature of the cold reservoir to which the ideal engine exhausts heat is approximately \(275 \mathrm{K}\).

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

If the pressure on a fish increases from 1.1 to 1.2 atm, its swim bladder decreases in volume from \(8.16 \mathrm{mL}\) to \(7.48 \mathrm{mL}\) while the temperature of the air inside remains constant. How much work is done on the air in the bladder?
(a) What is the entropy change of 1.00 mol of \(\mathrm{H}_{2} \mathrm{O}\) when it changes from ice to water at \(0.0^{\circ} \mathrm{C} ?\) (b) If the ice is in contact with an environment at a temperature of \(10.0^{\circ} \mathrm{C}\) what is the entropy change of the universe when the ice melts?
A town is planning on using the water flowing through a river at a rate of \(5.0 \times 10^{6} \mathrm{kg} / \mathrm{s}\) to carry away the heat from a new power plant. Environmental studies indicate that the temperature of the river should only increase by \(0.50^{\circ} \mathrm{C} .\) The maximum design efficiency for this plant is \(30.0 \% .\) What is the maximum possible power this plant can produce?
Suppose 1.00 mol of oxygen is heated at constant pressure of 1.00 atm from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C} .\) (a) How much heat is absorbed by the gas? (b) Using the ideal gas law, calculate the change of volume of the gas in this process. (c) What is the work done by the gas during this expansion? (d) From the first law, calculate the change of internal energy of the gas in this process.
In a certain steam engine, the boiler temperature is \(127^{\circ} \mathrm{C}\) and the cold reservoir temperature is \(27^{\circ} \mathrm{C} .\) While this engine does \(8.34 \mathrm{kJ}\) of work, what minimum amount of heat must be discharged into the cold reservoir?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Energy Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Torque and Rotational Motion

Read Explanation

Fundamentals of Physics

Read Explanation
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.

Sign-up for free