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Carnot engine entropy

  • Thread starter tjackson3
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Homework Statement



You want to boil a pot of water at 20C by heating it to 100C. I suggest a way of heating the pot in a reversible manner: simply inserting a Carnot engine between the reservoir (at 100C) and the pot. The Carnot engine operates between two temperatures, absorbing heat [tex]dQ_1[/tex] from the 100C reservoir and rejecting heat [tex]dQ_2[/tex] to the pot at temperature T (heat capacity C). At the same time, work [tex]dW[/tex] is generated in this process (n.b.: I don't know if this is relevant; it was used in the second part of the question). Starting from 20C water temperature, receiving heat in a reversible manner, the pot of water reached 100C.

Show that the total entropy of the system (reservoir + pot) remains the same in the process. Ignore the heat capacity of the pot and the temperature dependence of the water heat capacity.

Homework Equations



For a Carnot engine:

[tex]\frac{Q_H}{Q_L} = \frac{T_H}{T_L}[/tex]

[tex]dS = \frac{dQ}{T}[/tex]

The Attempt at a Solution



Since all Carnot cycles imply zero change in entropy, rather than use numbers, I just use [tex]T_R,T_P[/tex] for the temperature of the reservoir and the pot, respectively. When I got stuck, I tried inserting the actual numbers but to no avail. For the reservoir, since its temperature remains constant, and we know how much work it does, its change in entropy is just [tex]\frac{-dQ_1}{T_R}[/tex]. Given the way that the problem is worded, I'm unsure if [tex]dQ_1[/tex] is already negative or not, so I put it here to be safe.

For the pot, its change in heat is [tex]dQ_2[/tex], so its change in entropy becomes:

[tex]\Delta S_P = dQ_2\int_{T_P}^{T_R} \frac{dT}{T} = dQ_2 \ln \frac{T_R}{T_P}[/tex]

We want [tex]\Delta S_P = -\Delta S_R[/tex], so we have

[tex]\frac{dQ_1}{T_R} = dQ_2 \ln \frac{T_R}{T_P}[/tex]

We know that

[tex]\frac{dQ_1}{dQ_2} = \frac{T_R}{T_P}[/tex], so dividing both sides by [tex]dQ_2[/tex], we get

[tex]\frac{1}{T_P} = \ln \frac{T_R}{T_P}[/tex]

But clearly this can't be true, and in fact, substituting in the given numbers show this. Where did I go wrong?

Thanks in advance! :)
 

Answers and Replies

  • #2
Andrew Mason
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For the pot, its change in heat is [tex]dQ_2[/tex], so its change in entropy becomes:

[tex]\Delta S_P = dQ_2\int_{T_P}^{T_R} \frac{dT}{T}[/tex]
Should this not be:

[tex]\Delta S_{P} = \int \frac{dQ_{P}}{T_{P}} = \int_{T_P}^{T_R} \frac{mCdT}{T} = mC\ln{\frac{T_R}{T_P}}[/tex]


AM
 
  • #3
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Andrew: Thanks for the help again! I tried it that way, however. If you do it like that, then your entropy expression for the reservoir is stuck with a [tex]dQ_R[/tex] term, which you can get rid of using the [tex]dQ_R/dQ_P = T_R/T_P[/tex] relation, but you ultimately still get the cancellation that I got in my final expression of the problem.
 
  • #4
Andrew Mason
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Andrew: Thanks for the help again! I tried it that way, however. If you do it like that, then your entropy expression for the reservoir is stuck with a [tex]dQ_R[/tex] term, which you can get rid of using the [tex]dQ_R/dQ_P = T_R/T_P[/tex] relation, but you ultimately still get the cancellation that I got in my final expression of the problem.
But the dimensions of entropy are Energy/Temperature (J/K). Your integral has dimensions of Energy.

You just have to determine the reversible path and then calculate the change in entropy using that path. For the reversible path (putting a Carnot engine between the reservoir and the pot) the flow of heat to the pot results in the pot gradually increasing its temperature until it reaches 100C. In such a path the change in entropy of the water in the pot is (temp. in K):

[tex]\Delta S_P = mc\ln{\frac{T_R}{T_{Pi}}[/tex]

where m is the mass of the water in the pot and c is the heat capacity of water.

Because work is being done by the heat engine, the magnitude of the heat flow out of the reservoir is GREATER than the magnitude of the heat flow into the pot. But the heat flow out of the reservoir is all at the same temperature T_R, so all you have to do to calculate entropy is find the heat flow and divide by T_R.

AM
 
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  • #5
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So do you mean the heat flow from the reservoir into the pot or just from the reservoir? Because it seems like either way, you just get dQ_R or dQ_P, respectively, which doesn't change anything, unless I'm missing something simple?
 
  • #6
Andrew Mason
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So do you mean the heat flow from the reservoir into the pot or just from the reservoir? Because it seems like either way, you just get dQ_R or dQ_P, respectively, which doesn't change anything, unless I'm missing something simple?
The heat flow to the pot is just [itex]mc\Delta T[/itex] where m is mass of the water and c is the specific heat. But this is not the heat flow FROM the reservoir. The heat flow from the 100C reservoir heats the water in the pot AND does work. The amount of work done PLUS the heat flow into the pot gives you the amount of heat flow out of the hot reservoir.

The object is to determine how much heat flows out of that reservoir. The hot reservoir remains at constant temperature, Th. The temperature of the water in the pot, T, increases.

To make the process completely reversible, you need an arbitrarily large number of cycles of the heat engine so that the temperature of the pot is effectively constant during each cycle. Let T be the temperature of the pot as it gradually increases from TP (TPi) to TR (= Th). Express [itex]dS = dQ_h/T_h[/itex] in terms of [itex]dQ_c[/itex] and the temperature of the water in the pot. Then integrate that. You will see that it is the same magnitude as the change (opposite sign) in entropy of the water in the pot.

AM
 
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