/*! 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 56 The motor that drives a reversib... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 56

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 \(8.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
Answer: The maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.

Step by step solution

01

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin to work with them in thermodynamic calculations. To do this, we will add 273.15 to each temperature. \(T_H = 20.0^\circ C + 273.15 K = 293.15 K\) \(T_C = -5.0^\circ C + 273.15 K = 268.15 K\) \(T_W = 8.0^\circ C + 273.15 K = 281.15 K\)
02

Calculate the Coefficient of performance (COP) for the refrigerator

For a reversible refrigerator, the Coefficient of Performance (COP) is given by the following formula: \(\text{COP} = \frac{T_C}{T_H-T_C}\) Plug in the values of \(T_H\) and \(T_C\) to get the COP. \(\text{COP} = \frac{268.15}{293.15 - 268.15} = \frac{268.15}{25} = 10.726\)
03

Determine the heat removed from the cold reservoir

Since we know the power output of the refrigerator (148 W), we can calculate the heat removed from the cold reservoir using the formula: \(Q_C = \text{COP} \times W = 10.726 \times 148 W = 1587.448 W\) Now we need to find the heat removed in 2 hours, so we multiply this by the time: \(Q_C = 1587.448 W \times 2 h \times 3600 s/h = 11445504.64 J\)
04

Calculate the heat removed from the water to produce ice

To produce ice from the water, firstly, we need to cool down the water from \(281.15 K\) to \(273.15 K\), and then freeze it. The specific heat capacity of water (\(c_w\)) is \(4.18 \times 10^3 \mathrm{J/kg K}\) and the heat of fusion for water (\(L\)) is \(3.34 \times 10^5 \mathrm{J/kg}\). Let's denote the mass of water to be produced into ice by \(m\). Then, we have: \(\text{Heat removed to cool down water} = m \cdot c_w \cdot (T_W - 273.15 K )\) \(\text{Heat removed to freeze water} = m \cdot L \) The total heat removed can be found by summing these two terms, thus: \(Q_C = m \cdot (c_w \cdot 8 K + L)\)
05

Calculate the mass of ice produced

Now we can find the mass of ice produced by solving for \(m\): \(m = \frac{Q_C}{c_w \cdot 8 K + L} = \frac{11445504.64 J}{(4.18 \times 10^3 J/kgK) \cdot 8 K + 3.34 \times 10^5 J/kg} = 30.4 kg\) So, the maximum amount of ice that can be produced in 2 hours is approximately 30.4 kg.

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

A town is considering using its lake as a source of power. The average temperature difference from the top to the bottom is \(15^{\circ} \mathrm{C},\) and the average surface temperature is \(22^{\circ} \mathrm{C} .\) (a) Assuming that the town can set up a reversible engine using the surface and bottom of the lake as heat reservoirs, what would be its efficiency? (b) If the town needs about \(1.0 \times 10^{8} \mathrm{W}\) of power to be supplied by the lake, how many \(\mathrm{m}^{3}\) of water does the heat engine use per second? (c) The surface area of the lake is $8.0 \times 10^{7} \mathrm{m}^{2}$ and the average incident intensity (over \(24 \mathrm{h})\) of the sunlight is \(200 \mathrm{W} / \mathrm{m}^{2} .\) Can the lake supply enough heat to meet the town's energy needs with this method?
How much heat does a heat pump with a coefficient of performance of 3.0 deliver when supplied with \(1.00 \mathrm{kJ}\) of electricity?

The intensity (power per unit area) of the sunlight incident on Earth's surface, averaged over a 24 -h period, is about \(0.20 \mathrm{kW} / \mathrm{m}^{2} .\) If a solar power plant is to be built with an output capacity of \(1.0 \times 10^{9} \mathrm{W},\) how big must the area of the solar energy collectors be for photocells operating at \(20.0 \%\) efficiency?
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?
Estimate the entropy change of \(850 \mathrm{g}\) of water when it is heated from \(20.0^{\circ} \mathrm{C}\) to \(50.0^{\circ} \mathrm{C} .\) [Hint: Assume that the heat flows into the water at an average temperature. \(]\)
See all solutions

Recommended explanations on Physics Textbooks

Astrophysics

Read Explanation

Physics of Motion

Read Explanation

Linear Momentum

Read Explanation

Wave Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Atoms and Radioactivity

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.