## Question 4B.3

$\Delta U=q+w$

Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

### Question 4B.3

Hello,
I got the first part of this question right, but not the second, I was curious if someone could help me figure out what I'm missing. Thank you!

The internal energy of a system increased by 982 J when it was supplied with 492 J of energy as heat. (a) Was work done by or on the system? (b) How much work was done?

b) Since change in internal energy = q + w => 982 = 492J + w => w=490
The answer key said 900J. I am a bit lost. Any help would be super appreciated!

Thank you,
Prasanna

Michael Nguyen 1E
Posts: 120
Joined: Sat Aug 17, 2019 12:17 am

### Re: Question 4B.3

In the solution manual that I have, 490 J is the correct answer. It is written as 4.90*102 J, but it is still the correct answer.

Osvaldo SanchezF -1H
Posts: 122
Joined: Wed Sep 18, 2019 12:21 am

### Re: Question 4B.3

Yea the answer cannot be 900J because that would not make sense. if the change in internal energy is a + 982 and the supplied energy energy is also a + 492 then the equation would be +982= +492+ w where if you do math the work would have to be +490. My book also showed this as the answer(solution Manuel) it would also be why the since work is positive, work was done "on" the system.

Posts: 111
Joined: Thu Jul 25, 2019 12:17 am

### Re: Question 4B.3

Thank you guys for your replies! I used the answer key at the back of the book (I don't have the solution manual) and it said 90 * 10^2 J...I think that the answer key in the book is wrong since the solution manual also says 490.

Leila_4G
Posts: 114
Joined: Sat Sep 14, 2019 12:17 am

### Re: Question 4B.3

I feel you, it seems like there are so many typos in the textbook I don't really trust the answer key that much.

Pablo 1K
Posts: 118
Joined: Sat Feb 02, 2019 12:15 am

### Re: Question 4B.3

It was a type I also stressed over it but yea it seems you were right.