Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 19: Problem 67

Refrigerators remain among the greatest consumers of electrical energy in most homes, although mandated efficiency standards have decreased their energy consumption by some \(80 \%\) in the past four decades. In the course of a day, one kitchen refrigerator removes \(30 \mathrm{MJ}\) of energy from its contents, in the process consuming \(10 \mathrm{MJ}\) of electrical energy. The electricity comes from a \(40 \%\) efficient coal-fired power plant. The fuel energy consumed at the power plant to run this refrigerator for the day is a. \(12 \mathrm{MJ}\). b. \(25 \mathrm{MJ}\). c. \(40 \mathrm{MJ}\). d. \(75 \mathrm{MJ}\).

Short Answer

Expert verified
The fuel energy consumed at the power plant to run this refrigerator for a day is \(25 \mathrm{MJ}\) (b)

Step by step solution

01

Understand energy consumption of the refrigerator

Firstly, it's noted that while the refrigerator removes \(30 \mathrm{MJ}\) of energy per day, it consumes \(10 \mathrm{MJ}\) of electrical energy to do this. So the overall efficiency of the refrigerator can be calculated.
02

Calculate the energy requirement at the power plant

Secondly, because power plants are only \(40 \%\) efficient, all energy produced is not utilized and is lost somewhere along the process. The overall energy required can be calculated by using the efficiency formula, which is \(\text{Efficiency} = \frac{\text{output energy}}{\text{input energy}}\) . Here, output energy refers to the energy used by the refrigerator and input energy refers to the energy produced at the power plant. Solving the equation for input energy gives \(\text{input energy}= \frac{\text{output energy}}{\text{Efficiency}}\) .
03

Get the fuel energy

Finally, substitute the known values, \(10 \mathrm{MJ}\) for output energy and \(0.40\) for efficiency to find the amount of fuel energy required to run the refrigerator for a day. It comes out to \(\frac{10 \mathrm{MJ}}{0.40} = 25 \mathrm{MJ}\). Hence, \(25 \mathrm{MJ}\) of energy is required to run the refrigerator for a day.

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

You're the environmental protection officer for a \(35 \%\) efficient nuclear power plant that produces 750 MW of electric power, situated on a river whose minimum flow rate is \(110 \mathrm{m}^{3} / \mathrm{s}\). State environmental regulations limit the rise in river temperature from your plant's cooling system to \(5^{\circ} \mathrm{C}\). Can you achieve this standard if you use river water for all your cooling, or will you need to install cooling towers that transfer some of your waste heat to the atmosphere?

A 5.0 -mol sample of an ideal diatomic gas is at 1.0 atm pressure and 300 K. Find the entropy change if the gas is heated to \(500 \mathrm{K}\) (a) at constant volume, (b) at constant pressure, and (c) adiabatically.

The temperature of \(n\) moles of ideal gas is changed from \(T_{1}\) to \(T_{2}\) with pressure held constant. Show that the corresponding entropy change is \(\Delta S=n C_{p} \ln \left(T_{2} / T_{1}\right)\).

A power plant extracts energy from steam at \(250^{\circ} \mathrm{C}\) and delivers 800 MW of electric power. It discharges waste heat to a river at \(30^{\circ} \mathrm{C} .\) The plant's overall efficiency is \(28 \% .\) (a) How does this efficiency compare with the maximum possible at these temperatures? (b) Find the rate of waste-heat discharge to the river. (c) How many houses, each requiring \(18 \mathrm{kW}\) of heating power, could be heated with the waste heat from this plant?

You operate a store that's heated by an oil furnace supplying 30 kWh of heat from each gallon of oil. You're considering switching to a heat-pump system. Oil costs \(\$ 1.75 /\) gallon, and electricity costs \(16.5 \notin / \mathrm{kWh} .\) What's the minimum heat-pump COP that will reduce your heating costs?
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Atoms and Radioactivity

Read Explanation

Fundamentals of Physics

Read Explanation

Engineering Physics

Read Explanation

Translational Dynamics

Read Explanation

Fluids

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