Solving Carnot Heat Engine: Tf = SQRT(TcTh)

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SUMMARY

The discussion centers on deriving the final temperature of two equal mass heat reservoirs in a Carnot heat engine, concluding that Tf = SQRT(TcTh) when maximal work is achieved. The relevant equations include Qin = Wout + Qout and mc(Th-Tf) = Wout + mc(Tf - Tc). Participants emphasized the importance of differentiating the efficiency equation to find maximum work output, clarifying that only one differentiation is necessary to determine if a maximum or minimum is present.

PREREQUISITES
  • Understanding of thermodynamics, specifically heat engines and heat pumps.
  • Familiarity with the Carnot efficiency equation and its implications.
  • Basic calculus skills, particularly differentiation and optimization techniques.
  • Knowledge of specific heat capacity and its role in thermal systems.
NEXT STEPS
  • Study the derivation of the Carnot efficiency formula in detail.
  • Learn about the implications of equal mass and specific heat capacity in thermal systems.
  • Explore optimization techniques in calculus, focusing on finding maxima and minima.
  • Investigate the relationship between work output and heat transfer in thermodynamic cycles.
USEFUL FOR

Students studying thermodynamics, engineers working with heat engines, and anyone interested in optimizing thermal systems and understanding heat transfer principles.

Carlo09
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Homework Statement


A heat pump takes heat from a hot resevoir and dissipates heat to a cold one. Both resevoirs are equal mass and specific heat capacity. Show that as the heat engine does maximal work the final temp of the resevoirs = Tf = SQRT(TcTh)


Homework Equations


Qin = Wout + Qout
mc(Th-Tf) = Wout + mc(Tf -Tc)

Efficiency = Wout/Win = 1- (Qin/Qout)


The Attempt at a Solution


Well I know i somehow need to get TcTh^2 in order to get the solution so I used efficiecy as 1 for maximal work out but I also assumed no work in which means I'm dividing by 0! Or if I say 1= 1-Qin/Qout then Qout = 1-Qin and now I am just confused. Help please :)
 
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Remember from calculus how to find the maximum (or minimim) of a function?
 
hotvette said:
Remember from calculus how to find the maximum (or minimim) of a function?

Are you meaning differentiate and set to 0 and then solve? Differentiate again and is D^2f(x)/Dx^2 < 0 then it's a maximum?
 
Carlo09 said:
Are you meaning differentiate and set to 0 and then solve?

Yep, but you should only have to differentiate once. It should be pretty clear whether you have a max or min.
 
hotvette said:
Yep, but you should only have to differentiate once. It should be pretty clear whether you have a max or min.

Thank you, Which equation do I differentiate?
 
Carlo09 said:
Thank you, Which equation do I differentiate?

Ooo right ok so I have dw/dQ_h = (T1-T2)/T1 = 0

therefore T1-T2 = 0 so T1=T2=T and so

dw/dQ_h = (T-T)/T which is 0 which doesn't help me :S
 

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