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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

  1. 400 J
  2. 1450 J
  3. 1600 J
  4. 2000 J
  5. 2760 J

Thermodynamics}Carnot Engine

Recall the common-sense definition of the efficiency e 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. (Q_{input} is heat put into the system to get stuff going, Wis 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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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
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?
2010-11-06 19:53:20
The temperatures are in Kelvin
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.
2013-09-30 18:27:41
i dont think you have to convert to kelvin. Tc/Th is a ratio...the units cancel out
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.
2007-09-19 02:47:04
Remenbering Q_1 / T_1 + Q_2 / T_2 = 0, you can get the answer easily.
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.
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.
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?
2012-10-06 12:08:55
should be

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