Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

An inventor claims to have developed a refrigerator that maintains the refrigerated space at \(40^{\circ} \mathrm{F}\) while operating in a room where the temperature is \(85^{\circ} \mathrm{F}\) and that has a COP of \(13.5 .\) Is this claim reasonable?

Short Answer

Expert verified
Answer: ___________ [Use the provided step-by-step solution to calculate the Carnot COP, compare it with the claimed COP, and conclude if the claim is reasonable or not.]

Step by step solution

01

Convert temperatures to Kelvin

Before we proceed with calculations, we need to convert the given temperatures from Fahrenheit to Kelvin. We use the following formula to convert Fahrenheit to Kelvin: $$ K = \frac{5}{9}(°F - 32) + 273.15 $$ Calculate the Kelvin temperatures for the refrigerated space and the room: $$ T_{cold} = \frac{5}{9}(40 - 32) + 273.15 $$ $$ T_{hot} = \frac{5}{9}(85 - 32) + 273.15 $$
02

Calculate Carnot COP

The Carnot COP is the theoretical maximum efficiency of any heat engine or refrigerator, and it is determined by the ratio of the temperatures of the cold and hot reservoirs. The formula for Carnot COP of a refrigerator is: $$ COP_{Carnot} = \frac{T_{cold}}{T_{hot} - T_{cold}} $$ Substitute the values of \(T_{cold}\) and \(T_{hot}\) obtained in Step 1 into the formula, and calculate the Carnot COP.
03

Compare Carnot COP with the claimed COP

Now that we have calculated the Carnot COP, we can compare it with the claimed COP of \(13.5\). If the claimed COP is indeed less than or equal to the Carnot COP, then the claim is reasonable. Otherwise, the claim is not reasonable. Compare the calculated Carnot COP with the claimed COP of \(13.5\).
04

Conclude if the inventor's claim is reasonable

Based on the comparison of the Carnot COP and the claimed COP in Step 3, we can conclude if the inventor's claim of developing a refrigerator that maintains the refrigerated space at \(40^{\circ} \mathrm{F}\) while operating in a room at \(85^{\circ} \mathrm{F}\) with a COP of \(13.5\) is reasonable or not.

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 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

A heat pump is used to maintain a house at \(25^{\circ} \mathrm{C}\) by extracting heat from the outside air on a day when the outside air temperature is \(4^{\circ} \mathrm{C}\). The house is estimated to lose heat at a rate of \(110,000 \mathrm{kJ} / \mathrm{h},\) and the heat pump consumes \(4.75 \mathrm{kW}\) of electric power when running. Is this heat pump powerful enough to do the job?

The drinking water needs of an office are met by cooling tab water in a refrigerated water fountain from 23 to \(6^{\circ} \mathrm{C}\) at an average rate of \(10 \mathrm{kg} / \mathrm{h}\). If the COP of this refrigerator is \(3.1,\) the required power input to this refrigerator is \((a) 197 \mathrm{W}\) (b) \(612 \mathrm{W}\) \((c) 64 \mathrm{W}\) \((d) 109 \mathrm{W}\) \((e) 403 \mathrm{W}\)

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at \(1400 \mathrm{K}\) and rejects the waste heat to another reservoir at temperature \(T\) The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at \(300 \mathrm{K}\). If the thermal efficiencies of both engines are the same, determine the temperature \(T .\)

Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one. If the source temperature of the first engine is \(1300 \mathrm{K}\) and the sink temperature of the \(\sec\) ond engine is \(300 \mathrm{K}\) and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is \((a) 625 \mathrm{K}\) (b) \(800 \mathrm{K}\) \((c) 860 \mathrm{K}\) \((d) 453 \mathrm{K}\) \((e) 758 \mathrm{K}\)

A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55 percent. The waste heat from this engine is rejected to a nearby lake at \(60^{\circ} \mathrm{F}\) at a rate of \(800 \mathrm{Btu} / \mathrm{min} .\) Determine \((a)\) the power output of the engine and \((b)\) the temperature of the source.

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.

Sign-up for free