Question 9.5

[Isabella Sanzi 2E]

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]

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]

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.

[Adrian Lim 1G]

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