Entropy Generation

Tags:
1. Apr 7, 2017

asteeves_

1. The problem statement, all variables and given/known data
5 kg of water at 60 degrees are put in contact with 1 kg of ice at 0 degrees and are thermally isolated from everything else. The latent heat of ice is 3.3x105 J/kg.

1. What is the change of entropy of the universe when 100J of energy are transferred from the water to the ice?
2. Relevant equations
Q=mcΔT
Q=ml
ΔS=Q/T

3. The attempt at a solution
Qi (ice -> water) + Qi (water -> Tf) -Qw (water -> Tf) = 100
ml + mcΔT - mcΔT = 100J

My plan was to use this set of equations to solve for the T value then plug that into the equation for entropy generation but once I solve for Tf I get a value of 95K which doesn't make sense since the Tf should e somewhere in between 273K and 333K. I think my approach to the problem might be flawed and would appreciate some guidance!

2. Apr 7, 2017

Staff: Mentor

When 100 J of heat are transferred to the ice, how much ice melts? When 100 J of heat are transferred from the original water, what is the resulting water temperature? If you mix the water formed from the ice melting with the initial water that is now at a lower temperature, what is the new temperature of the water mixture? What is the final state of the system in terms of the ice temperature, the remaining amount of ice, the water temperature, and the final amount of water?

3. Apr 7, 2017

asteeves_

Thanks so much I think I have made sense of it and the concepts make sense to me, however, working through this I am getting results that are almost trivial - i.e. only 0.0003 kg of ice melts, the temp of the initial water remains practically unchanged and as a result of the mass of the new water being so small the final temperature is also essentially unchanged. I don't have an answer key to this problem and am trying to justify it to myself but would love a second opinion!

4. Apr 8, 2017

Staff: Mentor

Well, I've thought about this some more, and I realized that, if the water formed from the ice is allowed to mix with the 60 degree water, then additional heat will be transferred, over and above the 100 J. So the only way of making sense out of the problem statement is to have a conductive barrier between the ice compartment and the 60 degree water compartment. This barrier allows heat to flow across it until 100 J have been transferred, and then shuts the heat flow off.

So the final state on the ice side of the barrier is ice and melted water at 0 C. The final state on the water side of the barrier is water at a temperature slightly lower than 60 C. So I am recommending that we try the problem this way. And we will be comparing the entropy change calculated for this situation (taking into account that the water temperature drops a little below 60 C) with the result we would get for entropy change if the water remained at 60 C. So, for the comparison case, $$\Delta S=\frac{100}{273}-\frac{100}{333}$$

Does this make sense, and do you consider it worthwhile?

P.S., the entropy generation in the case of the conductive barrier being present takes place mostly within the barrier.