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[Thermodynamics] Two bodies exchanging heat through a Carnot engine 
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#1
Oct2212, 03:17 PM

P: 13

1. The problem statement, all variables and given/known data
It's basically the classic problem I've seen here a lot. There are two bodies, both with equal heat capacity ([itex]C_P[/itex]), one at temperature [itex]T_1[/itex] and the other at [itex]T_2[/itex]. They exchange heat using an infinitesimalreversible Carnot Engine, which will work until thermal equilibrium is reached. I must find, first, the final temperature of the bodies ([itex]T_F[/itex]) and then the entropy variation of the entire system. 2. Relevant equations [itex] dU = \partial Q + \partial W [/itex] (I know Q and W are nonexact differentials but I can't find the [itex]\LaTeX[/itex] code for it. [itex] dS = \frac {\partial Q}{T} [/itex] [itex] \eta = 1  \frac {T_2}{T_1} [/itex] For a Carnot Engine. 3. The attempt at a solution I've read at least 10 threads in the site about this and still not convinced. Some of the threads affirm the entropy variation of the universe is 0, and that way is pretty easy, but in my exercise I should get the results in the opposed order (moreover, I don't see trivially that it's 0). Then the other way, which I liked more work it from the first law of thermodynamics and the Carnot efficiency (subtly implying the 2nd Law). But they are made in a total quantities way and, although the solution is fine, I think it's more appropriate to make it in differential form as the quantities of heat vary after each cycle. Applying the 1st law to the engine should result in [itex] \partial Q_H + \partial Q_C + W = 0 [/itex] And the heat quantities are opposite from the one I get by evaluating the bodies (if Q is flowing out of the hotter body, Q is negative, but enters to the engine as a positive amount of heat). [itex]  C_P dT_c  C_P dT_h =  \partial W [/itex] But here the work would be the one done ON the engine which sounds weird, so if I consider the work done by the engine [itex] C_P dT_c + C_P dT_h = \partial W [/itex] But I get some problems while writing [itex] \partial W [/itex] as for a Carnot Engine. The answer is [itex]T_F = \sqrt {T_1 T_2} [/itex] Thanks for your time. 


#2
Oct2212, 04:21 PM

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P: 4,761

If ##dW## represents the work done by the engine, then the first law implies ##dW = dQ_h  dQ_c##
Keeping in mind that ##dT_h## is going to be negative, think about whether ##dQ_h = C_P\,dT_h## or ##dQ_h = C_P\,dT_h##. Similarly, think about ##dQ_c##. 


#3
Oct2212, 06:43 PM

P: 13

Thanks for your reply.
Which information am I adding that way? The first law is already implied. I think I have to incorporate the fact that it's not an ordinary engine, but a Carnot one, which won't come up that way. I thought something like [itex] \frac {\delta W}{\delta Q_H} = \frac {Q_H  Q_C}{Q_H} = \frac {T'_H  T'_C}{T'_H} [/itex] Where in the last equality I've used the fact that it's a reversible engine (I think), where the [itex]'[/itex] symbolizes the "instant" temperature, I mean, the appropriate temperature for each cycle. But mathematically that hasn't led anywhere. What are your opinions on basic Thermodynamics books? (Or should I open a new thread for this subject?). I'm reading:
As always, thank you for your time and patience guys. 


#4
Oct2212, 07:07 PM

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P: 4,761

[Thermodynamics] Two bodies exchanging heat through a Carnot engine
Then substitue for ##Q_H## in terms of ##C_P## and ##dT'_H## and similarly for ##Q_C##. 


#5
Oct2212, 07:32 PM

P: 13

Thanks for such quick answers man. I'm quitting study now, it's beer time :P, but tomorrow I'll look into it and write what I get.



#6
Oct2312, 01:29 PM

P: 13

I'll start again to be more clear on the notation
First law applied to the engine: [itex] \delta Q + \delta W_{on} = 0 [/itex] [itex] \delta Q_H + \delta Q_C =  \delta W_{on} [/itex], where the work here is the on done ON the engine, so, changing for the work done BY it and considering absolute values of heat... [itex] \delta Q_H   \delta Q_C = \delta W \ \ \ \ (1)[/itex] Then for a Carnot Engine I know the relation [itex]\frac {\delta W}{ \delta Q_H} = \frac {\delta Q_H  \delta Q_C} {\delta Q_H} = 1  \frac {\delta Q_C}{\delta Q_H} = 1  \frac {T_C}{T_H} [/itex] Where the last step is possible due to the information that is a Carnot Engine. Conclusion from that: [itex]\delta W = \left ( 1  \frac {T_C}{T_H} \right ) \delta Q_H [/itex] Using this information in [itex](1)[/itex] I get: [itex] \delta Q_H  \delta Q_C = \left ( 1  \frac {T_C}{T_H}) \right )  \delta Q_H [/itex] Which means [itex] 1  \frac {\delta Q_C}{\delta Q_H} = 1  \frac {T_C}{T_H} [/itex] Then [itex] \frac {\delta Q_C}{\delta Q_H} = \frac {T_C}{T_H} [/itex] At this point I thought I really had it going, as it's kind of an instantbyinstant Carnot HeatTemperatures relation. But then... [itex] \delta Q_C = C_P dT_C [/itex] As the cold temperature will be getting higher, then [itex]dT_C[/itex] will be positive, making the whole expression greater than 0, as it should be. [itex] \delta Q_H =  C_P dT_H [/itex] As, contrary to the above case, the differential here will be negative. [itex] T_C = T^{0}_C + dT_C [/itex] Here [itex]T^{0}_C[/itex] is the initial temperature of the cold body. [itex] T_H = T^{0}_H + dT_H [/itex] Same as above Then putting all together I get nowhere. [itex] \frac {C_P dT_C}{C_P dT_H} = \frac {T^0_C + dT_C}{T^0_H + dT_H} \Longrightarrow  \frac{T^0_H}{dT_H}  1 = \frac {T^0_C}{dT_C} + 1 [/itex] Which doesn't seems very useful. What am I missing? Thanks for your time. EDIT: Now I think of it, the expressions for the temperatures are bad, but can really see how to get the correct one. 


#7
Oct2312, 05:59 PM

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P: 4,761

Then try to rearrange so that you can integrate. 


#8
Oct2312, 08:51 PM

P: 13

You are right, I don't know why I try to make it so complicated.
Just for the exercise to be complete if someone looks for this: [itex] \frac {C_P dT_C}{C_P dT_H} = \frac {T_C}{T_C} \Longrightarrow \frac {dT_C}{T_C} =  \frac {dT_H}{T_H} [/itex] Then integrating that condition through the whole process [itex] \int_{T^0_C}^{T_f} \frac {dT_C}{T_C} =  \int_{T^0_H}^{T_f} \frac {dT_H}{T_H} \Longrightarrow \log \left ( {\frac{T_f}{T^0_C}} \right ) =  \log \left ( \frac {T_f}{T^0_H} \right ) \Longrightarrow \frac {T_f}{T^0_C}= \frac {T_H}{T_f} \Longrightarrow T_f = \sqrt {T^0_C T^0_H}[/itex] Thanks a lot for your help!!!! 


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