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

AI Thread Summary
The discussion revolves around calculating the maximum work output of a heat engine using a metal block as a heat reservoir and the ocean as a cold reservoir. The user presents a formula derived from the Carnot efficiency and integrates it to express work done in terms of initial temperature, cold reservoir temperature, and heat capacity. There is a request for verification of the correctness of this formula, which involves integrating the efficiency over the temperature range. Participants express uncertainty and confusion regarding the solution, indicating that the problem is complex and may require further clarification. The conversation highlights challenges in applying thermodynamic principles to finite heat capacity systems.
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1. A heat engine is run with a large block of metal as a reservoir with initial temperature T_i and constant heat capacitance C. The ocean is used as a cold reservoir, with constant temperature T_0. What is the maximum work that could be done by the engine in terms of T_i T_0 and C
2. C = \frac{dQ}{dT} For one cycle of the engine: Efficiency E = \frac{dW}{dQ}Here's My attempt: (Can someone please verify if my answer is correct?)
If E = \frac{dW}{dQ} for one cycle
Then:
E = \frac{dW}{CdT}
dW = CE(T)dT where E(T) = formula for Carnot Efficiency = \frac{T - T_0}{T}

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

W = C\int_{T_0}^{T_i} E(T)dT

W = C\int_{T_0}^{T_i} \frac{T - T_0}{T}dT

W = C\int_{T_0}^{T_i} 1 - \frac{T_0}{T} dT

W = C((T_i- T_0) - T_0\ln(\frac{T_i}{T_0}))


Does this formula look correct?
 
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bumping the thread...
 
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...
 
hi, nope.. it aint made right..will post for, if requested for..
 
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