GREPhysics.NET: GR9277 #16

GR9277 #16

Problem

GREPhysics.NET Official Solution

\prob{16}
An engine absorbs heat at a temperature of 727 degrees Celsius and exhausts heat at a temperature of 527 degrees Celsius. If the engine operates at a maximum possible efficiency, for 2000 joules of heat input the amount of work the engine performs is most nearly

400 J
1450 J
1600 J
2000 J
2760 J

Thermodynamics $\Rightarrow$ }Carnot Engine

Recall the common-sense definition of the efficiency $e$ of an engine,

$e=\frac{W_{accomplished}}{Q_{input}},<br />$

where one can deduce from the requirements of a Carnot process (i.e., two adiabats and two isotherms), that it simplifies to

$e=1-\frac{T_{low}}{T_{high}}$

for Carnot engines, i.e., engines of maximum possible efficiency. ( $Q_{input}$ is heat put into the system to get stuff going, $W$ is work done by the system and $T_{low}$ ( $T_{high}$ ) is the isotherm of the Carnot cycle at lower (higher) temperature.)

The efficiency of the Carnot engine is thus $e=1-\frac{800}{1000}=0.2$ , where one needs to convert the given temperatures to Kelvin units. (As a general rule, most engines have efficiencies lower than this.) The heat input in the system is $Q_{input}=2000J$ , and thus $W_{accomplished}=400 J$ , as in choice (A).

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Comments

hooverbm
2012-10-05 08:58:41

I think you can get around not even calculating efficiency.

We know that the relationship between Qs and Ts:

Qc/Qh = Tc/Th (h = hot, c = cold)

The work:

W = Qh-Qc

Solve for Qc above, and plug into work. You get an answer for Qc ~ 1400,

So W ~ 600J, which is closest to 400J.

Thus A

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heypete
2010-11-05 14:56:52

Where, exactly, did the values $e=1-\frac{800}{1000}$ come from when explaining the Carnot efficiency? Those values are not found anywhere in the problem. Is that a typo?

When I solve for e, I find $e=1-\frac{527}{727}=0.27$ , so $e \times Q_{input} = W_{accomplished} = 0.27 \times 2000 = 550 J$

Certainly, 550 J is closer to 400 J (choice A) than any other value, but it still seems a bit rough. Am I doing something incorrectly, or is there an error in the posted solution?

z3phx
2010-11-06 19:53:20

The temperatures are in Kelvin

neon37
2010-11-10 11:22:00

yep all temperatures in Kelvin should be used for all thermo calculations in SI units. The numbers were suspiciously odd which made me realize this.

aziza
2013-09-30 18:27:41

i dont think you have to convert to kelvin. Tc/Th is a ratio...the units cancel out

justin_l
2013-10-15 11:29:55

okay, that's just silly. it's an affine transformation, so ratios aren't perserved.rnrncounterexample: 2 C / 1 C vs 279 K / 274 K.rnrnThe celsius ratio is 2, the kelvin ratio is very nearly 1.

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mm
2007-09-19 02:47:04

Remenbering Q_1 / T_1 + Q_2 / T_2 = 0, you can get the answer easily.

petr1243
2008-01-15 20:24:45

Since entropy is a state function, the net change in entropy must be zero for one complete cycle(this could be sought out, when one plots T as a function of S. So you should have a difference of the the high and low entropies, instead of a sum.

gt2009
2009-06-14 20:12:52

One of the Q's is implicitly negative. It would be a difference if you put absolute values around each Q.

Sagan_Man
2012-02-02 17:38:55

Wait, but this gives $\frac{T_{2}}{T_{1}}Q_{1}=Q_{2}$ , which is $\frac{800}{1000}2000=1500=Q_{2}$ ...which is wrong?

morenogabr
2012-10-06 12:08:55

should be
$(1-\frac{T_2}{T_1})Q_1=Q_2$

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I think you can get around not even calculating efficiency. We know that the relationship between Qs and Ts: Qc/Qh = Tc/Th (h = hot, c = cold) The work: W = Qh-Qc Solve for Qc above, and plug into work. You get an answer for Qc ~ 1400, So W ~ 600J, which is closest to 400J. Thus A

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