## 9.1

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

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

### 9.1

I guess this just proves how much I've forgotten about old material, but I'm having trouble on 9.1

The question asks the find at what rate does your body heat generate entropy in your surroundings, taken to be at 20. C? The body generates heat at a rate of about 100. W (J/s)

I can't seem to figure out why the equation is $\Delta S=\frac{-q}{T}$
rather than
$\Delta S=\frac{q}{T}$

And furthermore, in the solutions manual, the answer suddenly becomes positive.

Any help would be greatly appreciated! Thanks

Ishan Saha 1L
Posts: 60
Joined: Fri Sep 29, 2017 7:03 am

### Re: 9.1

Hi! Because it asks for the rate your body generates heat "in your surroundings" the formula has a -q. This because the system is releasing heat into the surroundings, or heat is flowing out of the system (with the system being the body).

Angela 1K
Posts: 80
Joined: Fri Sep 29, 2017 7:05 am

### Re: 9.1

Ishan Saha 1L wrote:Hi! Because it asks for the rate your body generates heat "in your surroundings" the formula has a -q. This because the system is releasing heat into the surroundings, or heat is flowing out of the system (with the system being the body).

That makes sense, but then why is the final answer positive if there's a negative component in the fraction?

Nisarg Shah 1C
Posts: 54
Joined: Sat Jul 22, 2017 3:00 am

### Re: 9.1

That is because the system is releasing heat, so it should be -q/t for the system, and positive q/t for the surroundings since they are absorbing heat.

Mitch Walters
Posts: 45
Joined: Fri Sep 29, 2017 7:07 am

### Re: 9.1

There is a negative because of the fact that the heat is leaving your body and going into the surroundings.

Rishi Khettry 1L
Posts: 29
Joined: Fri Sep 29, 2017 7:05 am

### Re: 9.1

Heat is leaving the body so you regard the body as the "system" and therefore q is negative