## 9.1

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

Virpal Gill 1B
Posts: 32
Joined: Thu Jul 27, 2017 3:00 am

### 9.1

The problem states that the human body generated heat at the rate of about 100. W (1 W= 1 J/s). And the question asks at what rate does your body heat generate entropy in your surroundings, taken to be at 20. degrees Celsius? So would the entropy be that of the system or of the surroundings? The solutions manual starts with a negative sign for the entropy calculation but then the answer appears without a negative sign.

Qining Jin 1F
Posts: 53
Joined: Fri Sep 29, 2017 7:07 am

### Re: 9.1

I think it is talking about the surroundings

Minie 1G
Posts: 63
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.1

100/(20+273.15)=0.341 Joules/(Kelvin second)

Payton Schwesinger 1J
Posts: 45
Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.1

Your body is generating the 100 J/s, which means that the surroundings is absorbing 100 J/s (positive) which is why the entropy of the surroundings is a positive value.

Charles Ang 1E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: 9.1

The solutions manual uses the entropy generation formula which has a negative answer because the system will be releasing entropy. But the question asks for the rate of entropy your body generates. So we are answering the question, at what rate is the entropy in our surrounding changing, which would result in a positive entropy of the surroundings.