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GR9277 #16
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Problem
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\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
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Thermodynamics }Carnot Engine
Recall the common-sense definition of the efficiency  of an engine,
where one can deduce from the requirements of a Carnot process (i.e., two adiabats and two isotherms), that it simplifies to
for Carnot engines, i.e., engines of maximum possible efficiency. ( is heat put into the system to get stuff going,  is work done by the system and  ( ) is the isotherm of the Carnot cycle at lower (higher) temperature.)
The efficiency of the Carnot engine is thus  , 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  , and thus  , as in choice (A).
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Alternate Solutions |
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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 |  | heypete
2010-11-05 14:56:52 | Where, exactly, did the values  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  , so 
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
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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.
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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
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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.
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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.
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Sagan_Man
2012-02-02 17:38:55 |
Wait, but this gives  , which is  ...which is wrong?
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morenogabr
2012-10-06 12:08:55 |
should be
Q_1=Q_2)
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You are replying to:
Where, exactly, did the values 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 , so
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? |
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