TB #4I.5

[Christine Nguyen 3D]

The textbook solution for this said the first step was setting the heat transfer, q, of the colder water to the heat transfer of the warmer water, which is negative. Why would it be negative is what I'm confused about. The question is:

Suppose that 50g of water at 20˚C is mixed with 65g of water at 50˚C at constant atmospheric pressure in a thermally isolated vessel. Calculate ∆S and ∆S(tot) for the process.

Thanks!

[Anna Li 3B]

The heat transfer of the warmer water is negative because heat is leaving the water, and heat transfer of the colder water is positive because it is gaining heat. We're saying all the heat leaving the warmer water will be taken by the colder water, thus q(cold) = -q(hot).

[Christine Nguyen 3D]

The heat transfer of the warmer water is negative because heat is leaving the water, and heat transfer of the colder water is positive because it is gaining heat. We're saying all the heat leaving the warmer water will be taken by the colder water, thus q(cold) = -q(hot).

That's super helpful, thanks!