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Heat Engine with Finite Heat Capacity- Is my answer correct?

  • Thread starter G01
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G01
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1. A heat engine is run with a large block of metal as a reservoir with initial temperature [tex]T_i[/tex] and constant heat capacitance C. The ocean is used as a cold reservoir, with constant temperature [tex]T_0[/tex]. What is the maximum work that could be done by the engine in terms of [tex]T_i T_0 [/tex] and C



2. [tex]C = \frac{dQ}{dT}[/tex] For one cycle of the engine: Efficiency [tex] E = \frac{dW}{dQ}[/tex]


Here's My attempt: (Can someone please verify if my answer is correct?)
If [tex] E = \frac{dW}{dQ}[/tex] for one cycle
Then:
[tex] E = \frac{dW}{CdT}[/tex]
[tex] dW = CE(T)dT[/tex] where E(T) = formula for Carnot Efficiency = [tex]\frac{T - T_0}{T}[/tex]

Adding up the work done in every cycle for every infinitessimal change in the metal block's temp gives:

[tex] W = C\int_{T_0}^{T_i} E(T)dT [/tex]

[tex] W = C\int_{T_0}^{T_i} \frac{T - T_0}{T}dT [/tex]

[tex] W = C\int_{T_0}^{T_i} 1 - \frac{T_0}{T} dT [/tex]

[tex] W = C((T_i- T_0) - T_0\ln(\frac{T_i}{T_0})) [/tex]


Does this formula look correct?
 

Answers and Replies

  • #2
G01
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bumping the thread...
 
  • #3
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Wow..

I can't find anything wrong. I have been in trouble with this problem. (also with two-finite heat reservoir problem). I think this is correct...
 
  • #4
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hi, nope.. it aint made right..will post for, if requested for..
 

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