## Question 9.5

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

Isabella Sanzi 2E
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

### Question 9.5

Question 5 asks: What is the total entropy change of a process in which 40.0 kJ of energy is transferred as heat from a large reservoir at 800. K to one at 200. K? I understand that you would carry out the deltaS = q/T two separate times, but the solution manual has 40.0 kJ as a negative value in the first calculation and positive in the second calculation. Why does the sign switch from the first step to the second? Thanks!

Sohini Halder 1G
Posts: 58
Joined: Thu Jul 13, 2017 3:00 am
Been upvoted: 1 time

### Re: Question 9.5

To calculate the total entropy change, you need to first find the entropy change for each reservoir: one is giving off heat, and one is absorbing heat. 400kj should therefore have two signs because q=(-) when the hotter reservoir is giving off heat and q=(+) when the cooler reservoir is absorbing heat. Then use those and the corresponding temperatures in Kelvin to find each entropy change and add to find the total entropy change.

Meredith Steinberg 2E
Posts: 50
Joined: Thu Jul 13, 2017 3:00 am

### Re: Question 9.5

Heat is being given off in the 800K reservoir, which makes the energy sign negative, and it's being absorbed in the 200K reservoir, which makes its energy sign positive.

Posts: 88
Joined: Fri Sep 29, 2017 7:03 am

### Re: Question 9.5

I believe the signs in the solutions manual are referring to the system itself

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