/*! 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} 4.37 A common (but imprecise) way of ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: 4.37 (page 148)

A common (but imprecise) way of stating the third law of thermodynamics is "You can't reach absolute zero." Discuss how the third law, as stated in Section 3.2, puts limits on how low a temperature can be attained by various refrigeration techniques.

Short Answer

Expert verified

The refrigeration techniques cannot attain absolute zero temperature

Step by step solution

01

Given Information

Given techniques: refrigeration techniques

How low a temperature can be attained by various refrigeration techniques

02

Explanation

As per the Third law of thermodynamics, entropy of the system tends to zero at absolute zero temperature.

As entropy approaches zero at absolute zero temperature therefore the heat capacity also goes to zero. This means that the heat capacity becomes negligibly low at very low temperature.

So the cooling process becomes ineffective for very low heat capacity. This is the reason why refrigeration requires temperature higher than absolute zero.

So we can say that the refrigeration techniques cannot attain absolute zero temperature.


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

Prove that if you had a heat engine whose efficiency was better than the ideal value (4.5), you could hook it up to an ordinary Carnot refrigerator to make a refrigerator that requires no work input.

An apparent limit on the temperature achievable by laser cooling is reached when an atom's recoil energy from absorbing or emitting a single photon is comparable to its total kinetic energy. Make a rough estimate of this limiting temperature for rubidium atoms that are cooled using laser light with a wavelength of 780 nm.

A heat pump is an electrical device that heats a building by pumping heat in from the cold outside. In other words, it's the same as a refrigerator, but its purpose is to warm the hot reservoir rather than to cool the cold reservoir (even though it does both). Let us define the following standard symbols, all taken to be positive by convention:
Th=temperature inside buildingTc=temperature outsideQh=heat pumped into building in1dayQc=heat taken from outdoors in1dayW=electrical energy used by heat pump in1day
(a) Explain why the "coefficient of performance" (COP) for a heat pump should be defined as Qh / W.
(b) What relation among Qh , Qc, and W is implied by energy conservation alone? Will energy conservation permit the COP to be greater than 1 ?
(c) Use the second law of thermodynamics to derive an upper limit on the COP, in terms of the temperatures Th and Tc alone.
(d) Explain why a heat pump is better than an electric furnace, which simply converts electrical work directly into heat. (Include some numerical estimates.)

Prove that if you had a refrigerator whose COP was better than the ideal value (4.9), you could hook it up to an ordinary Carnot engine to make an engine that produces no waste heat.

Consider an ideal Hampson-Linde cycle in which no heat is lost to the environment.

(a) Argue that the combination of the throttling valve and the heat exchanger is a constant-enthalpy device, so that the total enthalpy of the fluid coming out of this combination is the same as the enthalpy of the fluid going in.

(b) Let xbe the fraction of the fluid that liquefies on each pass through the cycle. Show that

x=Hout-HinHout-Hliq,

where Hinis the enthalpy of each mole of compressed gas that goes into the heat exchanger, Houtis the enthalpy of each mole of low-pressure gas that comes out of the heat exchanger, and Hliqis the enthalpy of each mole of liquid produced.

(c) Use the data in Table 4.5to calculate the fraction of nitrogen liquefied on each pass through a Hampson-Linde cycle operating between 1 bar and 100 bars, with an input temperature of 300K. Assume that the heat exchanger works perfectly, so the temperature of the low-pressure gas coming out of it is the same as the temperature of the high-pressure gas going in. Repeat the calculation for an input temperature of 200K.
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Classical Mechanics

Read Explanation

Waves Physics

Read Explanation

Measurements

Read Explanation

Particle Model of Matter

Read Explanation

Physical Quantities And Units

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.