## 4f.1

$\Delta S = \frac{q_{rev}}{T}$

Elizabeth Bowen 1J
Posts: 53
Joined: Wed Nov 14, 2018 12:20 am

### 4f.1

question: A human body generates heat at the rate of about 100W. (1W 1J s-1)
a. what rate does your body heat generate entropy in your surroundings, taken to be at 20 C

Can someone explain why we're able to use delta S = q(rev)/T for this problem? How do we know that the q we get (q=100J) is reversible, is there any indication that the process is reversible? (I thought reversible processes basically mean the pathway is isothermic, but it seems like generating heat wouldn't be an isothermic thing, since it's generating heat which would change the temperature?)

Thanks

Isha_Maniyar_Dis2E
Posts: 110
Joined: Thu Jul 11, 2019 12:16 am

### Re: 4f.1

Because entropy is a state function, it doesn't matter which pathway we take, because we will end up getting to the same end result. Reversible pathways are more convenient to calculate for, so we tend to use that.

In reality, irreversible pathways will have a slightly lower entropy than reversible. However, for the purposes of this class, we just use reversible because we are looking for the maximum possible entropy.

Hope this helped!

Noe BM 1J
Posts: 50
Joined: Sat Sep 07, 2019 12:17 am

### Re: 4f.1

Under which conditions would you use delta S = nRln(V2/V1) and delta S = qrev/T?