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Chapter 4: Q. 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.

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Most popular questions from this chapter

Imagine that your dog has eaten the portion of Table 4.1 that gives entropy data; only the enthalpy data remains. Explain how you could reconstruct the missing portion of the table. Use your method to explicitly check a few of the entries for consistency. How much of Table 4.2 could you reconstruct if it were missing? Explain.

To get more than an infinitesimal amount of work out of a Carnot engine, we would have to keep the temperature of its working substance below that of the hot reservoir and above that of the cold reservoir by non-infinitesimal amounts. Consider, then, a Carnot cycle in which the working substance is at temperatureThwas it absorbs heat from the hot reservoir, and at temperatureTcwas it expels heat to the cold reservoir. Under most circumstances the rates of heat transfer will be directly proportional to the temperature differences:

QhΔt=KTh-ThwandQcΔt=KTcw-Tc

I've assumed here for simplicity that the constants of proportionality Kare the same for both of these processes. Let us also assume that both processes take the

same amount of time, so theΔt''s are the same in both of these equations.*

aAssuming that no new entropy is created during the cycle except during the two heat transfer processes, derive an equation that relates the four temperaturesTh,Tc,Thwand Tcw

bAssuming that the time required for the two adiabatic steps is negligible, write down an expression for the power (work per unit time) output of this engine. Use the first and second laws to write the power entirely in terms of the four temperatures (and the constant K), then eliminateTcwusing the result of part a.

cWhen the cost of building an engine is much greater than the cost of fuel (as is often the case), it is desirable to optimize the engine for maximum power output, not maximum efficiency. Show that, for fixed Thand Tc, the expression you found in part bhas a maximum value at Thw=12Th+ThTc. (Hint: You'll have to solve a quadratic equation.) Find the correspondingTcw.

dShow that the efficiency of this engine is 1-Tc/ThEvaluate this efficiency numerically for a typical coal-fired steam turbine with Th=600°CandTc=25°C, and compare to the ideal Carnot efficiency for this temperature range. Which value is closer to the actual efficiency, about 40%, of a real coal-burning power plant?

Suppose that heat leaks into your kitchen refrigerator at an average rate of 300 watts. Assuming ideal operation, how much power must it draw from the wall?

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.)

Table 4.5 gives experimental values of the molar enthalpy of nitrogen at 1 bar and 100 bars. Use this data to answer the following questions about a nitrogen throttling process operating between these two pressures.

(a) If the initial temperature is 300K, what is the final temperature? (Hint: You'll have to do an interpolation between the tabulated values.)

(b) If the initial temperature is 200K, what is the final temperature?

(c) If the initial temperature is 100K, what is the final temperature? What fraction of the nitrogen ends up as a liquid in this case?

(d) What is the highest initial temperature at which some liquefaction takes place?

(e) What would happen if the initial temperature were 600K? Explain.
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