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Andrew Mason
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Aug8-11, 04:18 PM
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How can we achieve infinitesimal temperature difference during heat transfer

Quote Quote by weng cheong View Post
but in my textbook, in defining the change in entropy of surrounding,

d Ssurr= Qsurr/Tsurr
= Qreversible/Tsurr


so what i don't understand is that, why it equates Qsurr to Qreversible, in which the transfer of heat is impossible to be reversible? and i can't figure out why it relates this to the fact that the surrounding is huge.
The reversible heat flow, Qrev, must be used to calculate the change in entropy. This has to do with the way entropy is defined. It is defined that way for a good reason: if it is defined that way, the change in entropy of the system + surroundings is 0 if the process was reversible. It has nothing to do with the fact that the surroundings are huge. The size of the surroundings is immaterial and may not be huge at all (ie. a closed system).

I am not sure I understand your question. In your question you say that Qsurr = Qrev. This is not correct. To calculate the change in entropy of the surroundings, you must use the heat flow in the reversible process between the initial and final states of the surroundings, not the actual heat flow.

But this does not mean that in calculating the change in entropy for a non-reversible process that you will get a 0 change in entropy using the reversible heat flows. You have to calculate the reversible heat flows of the system and of the surroundings separately.

For example, to determine the change in entropy of the universe in an adiabatic free expansion of a gas into a vacuum, you do as follows:

1. calculate the change in entropy of the surroundings. No change in P,V, or T so dS = 0.

2. calculate change in entropy of the gas: change in P and V but no change in T. The reversible process between the initial and final states is a quasi-static isothermal expansion. Since such an expansion does work but results in no change of internal energy (constant T) there must be heat flow into the gas: dQ = 0 + dW = PdV. So in calculating the change in entropy in the reversible process, there is positive heat flow so there is an increase in entropy:

[tex]\Delta S = \int dQ_{rev}/T = \int PdV/T = \int nRdV/V = nR\ln\frac{V_f}{V_i}[/tex]

So, although in this process there is no heat flow at all, there is an increase in entropy since the reversible path between the beginning and end states involves a positive flow of heat.

AM