## Problem 4I.5

Volume: $\Delta S = nR\ln \frac{V_{2}}{V_{1}}$
Temperature: $\Delta S = nC\ln \frac{T_{2}}{T_{1}}$

Savannah Mance 4G
Posts: 107
Joined: Fri Aug 30, 2019 12:17 am

### Problem 4I.5

When finding the equilibrium temperature of water I don't understand why you wouldn't multiply the grams given by the specific heat capacity for the how water and the cold water? Why do they just cancel?

DarrenKim_1H
Posts: 123
Joined: Fri Sep 20, 2019 12:17 am
Been upvoted: 1 time

### Re: Problem 4I.5

I think it's because since q = mCdeltaT and you set qhot equal to qcold, you can divide both sides by the specific heat capacity of water to cancel them out, since you're finding the equilibrium temperature.

Uisa_Manumaleuna_3E
Posts: 60
Joined: Wed Sep 21, 2016 2:56 pm

### Re: Problem 4I.5

the way I understand it, both the cold water and the hot water have the exact same specific heat capacity, so when we set them to one another (q of cold = - q of hot) it just makes it simpler knowing that we can divide both sides without having to do any extra calculations

I imagine that if you do end up multiplying both sides by the specific heat capacity (being sure that the units match up properly) you will still get the same answer. But for me I always run the risk of messing up with extra unnecessary calculations