Ch. 9 Problems, 5

Moderators: Chem_Mod, Chem_Admin

Posts: 61
Joined: Sat Jul 22, 2017 3:01 am

Ch. 9 Problems, 5

Postby Maria1E » Fri Jan 26, 2018 4:31 pm

What is the total entropy change of a process in which 40 kJ of energy is transferred as heat from a large reservoir at 800 K to one at 200 K?

In the solution manual, the calculation for the entropy change at 800 K is shown as deltaS = -40000 J/800 K. For the entropy change at 200 K, the calculation is deltaS = +40000 J/200 K. Why is the 40000 J in the first equation negative? Thank you!

Angela G 2K
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Re: Ch. 9 Problems, 5

Postby Angela G 2K » Fri Jan 26, 2018 6:54 pm

For the first reservoir, heat is being transferred out and into the second reservoir, so the change of entropy is negative.

Varsha Sivaganesh 1A
Posts: 51
Joined: Thu Jul 13, 2017 3:00 am

Re: Ch. 9 Problems, 5

Postby Varsha Sivaganesh 1A » Sat Jan 27, 2018 1:30 pm

For these questions, does anyone know if we have to convert kJ to J to calculate entropy so that its units are J/K? Or can we leave it as kJ/K?

Farah Ahmad 2A
Posts: 30
Joined: Thu Jul 27, 2017 3:00 am
Been upvoted: 1 time

Re: Ch. 9 Problems, 5

Postby Farah Ahmad 2A » Sat Jan 27, 2018 10:45 pm

I think as long as you are consistent within the problem it shouldn't matter whether you put your answer in kJ/K or J/K.

Return to “Concepts & Calculations Using Second Law of Thermodynamics”

Who is online

Users browsing this forum: No registered users and 2 guests