/*! 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 34 A reversible engine with an effi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 34

A reversible engine with an efficiency of \(30.0 \%\) has $T_{\mathrm{C}}=310.0 \mathrm{K} .\( (a) What is \)T_{\mathrm{H}} ?$ (b) How much heat is exhausted for every \(0.100 \mathrm{kJ}\) of work done?

Short Answer

Expert verified
Answer: The hot temperature reservoir is approximately 442.9 K, and the amount of heat exhausted for every 0.100 kJ of work done is approximately 0.233 kJ.

Step by step solution

01

Use the formula for efficiency of a reversible engine

The efficiency of a reversible engine, \(\eta\), is related to the hot and cold temperature reservoirs (\(T_H\) and \(T_C\)) by the equation: $$ \eta = 1 - \frac{T_C}{T_H} $$ We are given the efficiency, \(\eta\), as \(30.0\%\) or \(0.30\), and we're given the cold temperature reservoir, \(T_C = 310.0 K\). We need to solve for the hot temperature reservoir, \(T_H\). Rearrange the equation above to solve for \(T_H\): $$ T_H = \frac{T_C}{1 - \eta} $$
02

Calculate \(T_H\) using the efficiency and \(T_C\) values

Now, plug in the given values of efficiency, \(\eta = 0.30\), and the cold temperature reservoir, \(T_C = 310.0 K\), into the above equation: $$ T_H = \frac{310.0}{1 - 0.30} = \frac{310.0}{0.70} $$ Upon evaluating the expression, we get the hot temperature reservoir, \(T_H \approx 442.9 K\). So, \(T_H = 442.9 K\).
03

Find the heat exhausted for every \(0.100 kJ\) of work done

To find the heat exhausted for every \(0.100 kJ\) of work done, first, we need to find how much heat is supplied to the engine, represented by \(Q_H\). We use the formula, $$ W = \eta Q_H $$ Rearrange the equation above to solve for \(Q_H\): $$ Q_H = \frac{W}{\eta} $$ Given the work done, \(W = 0.100 kJ\) and efficiency, \(\eta = 0.30\). $$ Q_H = \frac{0.100}{0.30} = 0.333... kJ $$
04

Find the heat exhausted, \(Q_C\), using the energy conservation principle

Now, we'll use the energy conservation principle, which states that the total heat supplied to the engine, \(Q_H\), is equal to the sum of the heat exhausted, \(Q_C\), and the work done, \(W\). Rearrange the equation as: $$ Q_C = Q_H - W $$ Substitute the value of \(Q_H = 0.333... kJ\) and \(W = 0.100 kJ\) into the equation: $$ Q_C = 0.333... - 0.100 = 0.233... kJ $$ So, the heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\). In summary, the hot temperature reservoir, \(T_H\), is approximately \(442.9 K\) and the amount of heat exhausted for every \(0.100 kJ\) of work done is approximately \(0.233 kJ\).

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 91影视!

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

For a more realistic estimate of the maximum coefficient of performance of a heat pump, assume that a heat pump takes in heat from outdoors at $10^{\circ} \mathrm{C}$ below the ambient outdoor temperature, to account for the temperature difference across its heat exchanger. Similarly, assume that the output must be \(10^{\circ} \mathrm{C}\) hotter than the house (which itself might be kept at \(20^{\circ} \mathrm{C}\) ) to make the heat flow into the house. Make a graph of the coefficient of performance of a reversible heat pump under these conditions as a function of outdoor temperature (from $\left.-15^{\circ} \mathrm{C} \text { to }+15^{\circ} \mathrm{C} \text { in } 5^{\circ} \mathrm{C} \text { increments }\right)$
A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.
Consider a heat engine that is not reversible. The engine uses $1.000 \mathrm{mol}\( of a diatomic ideal gas. In the first step \)(\mathrm{A})$ there is a constant temperature expansion while in contact with a warm reservoir at \(373 \mathrm{K}\) from \(P_{1}=1.55 \times 10^{5} \mathrm{Pa}\) and $V_{1}=2.00 \times 10^{-2} \mathrm{m}^{3}$ to \(P_{2}=1.24 \times 10^{5} \mathrm{Pa}\) and $V_{2}=2.50 \times 10^{-2} \mathrm{m}^{3} .$ Then (B) a heat reservoir at the cooler temperature of \(273 \mathrm{K}\) is used to cool the gas at constant volume to \(273 \mathrm{K}\) from \(P_{2}\) to $P_{3}=0.91 \times 10^{5} \mathrm{Pa} .$ This is followed by (C) a constant temperature compression while still in contact with the cold reservoir at \(273 \mathrm{K}\) from \(P_{3}, V_{2}\) to \(P_{4}=1.01 \times 10^{5} \mathrm{Pa}, V_{1} .\) The final step (D) is heating the gas at constant volume from \(273 \mathrm{K}\) to \(373 \mathrm{K}\) by being in contact with the warm reservoir again, to return from \(P_{4}, V_{1}\) to \(P_{1}, V_{1} .\) Find the change in entropy of the cold reservoir in step \(\mathrm{B}\). Remember that the gas is always in contact with the cold reservoir. (b) What is the change in entropy of the hot reservoir in step D? (c) Using this information, find the change in entropy of the total system of gas plus reservoirs during the whole cycle.
A reversible refrigerator has a coefficient of performance of \(3.0 .\) How much work must be done to freeze \(1.0 \mathrm{kg}\) of liquid water initially at \(0^{\circ} \mathrm{C} ?\)
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Fundamentals of Physics

Read Explanation

Modern Physics

Read Explanation

Waves Physics

Read Explanation

Work Energy and Power

Read Explanation

Astrophysics

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.