0
$\begingroup$

I recently solved a problem where I had to study the change of entropy for a lake that is freezing. It started off as being plain water at $0$ degrees, that then freezes to some negative temperature.

So the process consists of the formation of ice, and thereupon freezing. In order to calculate the change in entropy, I firstly had to calculate the heat dissipated by the lake for the two subprocesses, which we can call $Q_{ice}$ and $Q_{freezing}$ respectively. Both of these were calculated to be positive.

When calculating the entropy, I had made a mistake in my use of sign. At first I calculated the entropy change as

$$ \Delta S = Q_{ice}/T_{ice} + \int_{T_{ice}}^{T_{freeze}} C/T dT$$

when I apparently should have used

$$ \Delta S = -Q_{ice}/T_{ice} + \int_{T_{ice}}^{T_{freeze}} C/T dT$$

to arrive at the correct answer. I can't really tell why the negative sign was there and what convention is used here. Is the convention that heat dissipates should be negative? I'd be glad if anyone could give any tips on how these sign conventions work in general, since I often get sign errors in my problems.

Thanks.

$\endgroup$

3 Answers 3

1
$\begingroup$

Heat leaving the system transfers entropy out of the system, thus the change in entropy of the system (the ice-water mixture undergoing constant temperature freezing), is negative. Likewise, the reversible cooling of the ice is also a negative entropy change.

Hope this helps.

$\endgroup$
0
$\begingroup$

The lake is losing energy in the form of heat. So, the change of entropy ($\frac{dQ_{rev}}{T})$ is negative all the time. Note that it also happens for the integral, because $T_{freeze} < T_{ice}$.

$\endgroup$
0
$\begingroup$

Heat flow dQ from the surroundings to the system is considered positive. If heat flows from the system to the surroundings, dQ is negative.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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