0
$\begingroup$

I get the same answers for 2 parts of a question that I think should be different. The question is... If 0 $^{\circ}$ C water is heated by a 100 $^{\circ}$ C reservoir to thermal equilibrium, what is the total change of entropy of the universe? Instead, if it is heated by a 50 $^{\circ}$ C reservoir to thermal equilibrium, then a 100 $^{\circ}$ C reservoir, how much does the entropy of the universe change? This is what I did: (heat capacity of the mass of water is C)

$$ \Delta S = \int_{273}^{373} \frac{CdT}{T} $$

$$ \Delta S = C\ln\frac{373}{273} $$

Then for the second part:

$$ \Delta S = \int_{273}^{323} \frac{CdT}{T} + \int_{323}^{373} \frac{CdT}{T} $$

$$ \Delta S = C\ln\frac{323}{273} + C\ln\frac{373}{323} $$

$$ \Delta S = C\ln\frac{373}{273} $$

I am assuming that I need to take the entropy change of the reservoir into account, but how would I do this as the temperature of the reservoir doesn't change?

$\endgroup$
  • $\begingroup$ Note that the entropy change of the supersystem can be made arbitrarily close to zero, if we heat the cold water in a more complicated way, using reversible heat engine that extracts heat from the reservoir, provides useful work which is extracted away from the supersystem, and dumps the waste heat into the cold water. This can achieve equilibrium at 100 Celsius as well, but due to the reversibility, entropy of the supersystem won't increase. However, if no such engine is used and there is direct heat exchange, then the calculation you and Chet Miller describe is correct. $\endgroup$ – Ján Lalinský May 2 at 0:01
1
$\begingroup$

In thermodynamics, a reservoir is virtually always treated as ideal, so, in the first case, $$\Delta S_{res}=\frac{-Q}{T_{res}}=\frac{-C(373-273)}{373}$$By analogy, how do you think it would be handled for the 2nd case?

$\endgroup$
  • $\begingroup$ Thanks I think so. So the second case would be S = - C(50)/323 - C(50)/373 for the reservoir? As the increments get smaller does this tend to a certain value for an infinite number of reservoirs? because it seems to converge. $\endgroup$ – TSpoon May 1 at 19:43
  • $\begingroup$ Yes. That is correct. So the overall increase in entropy of the system plus surroundings is greater in the (more irreversible) 1st case than in the (less irreversible) 2nd case. $\endgroup$ – Chet Miller May 1 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.