## Homework Problem #19

Glenda Marshall DIS 3M
Posts: 68
Joined: Sat Sep 27, 2014 3:00 am

### Homework Problem #19

In this problem it says that a piece of copper at 100 degrees celsius is put into water at 22 degrees celsius. I am confused why the copper has a positive "q" value while the water has a negative "q" value. If the water is absorbing heat given off by the copper shouldn't it be the other way around? Thanks!

Sarah H Brown 1L
Posts: 56
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Homework Problem #19

You are right, if something is giving off heat, the sign is negative. If something is absorbing heat, the sign is positive. However, in this question, the most important concept is that qsystem + qsurroundings = 0. This means that one equals the negative version of the other. You can choose which one to subtract over to the right to reflect the conditions of the experiment. For example, in this case, you would want
qsurroundings = -qsystem because the surroundings are absorbing heat from the copper. If you would like me to do out the problem completely, just let me know. I hope this helped!

Glenda Marshall DIS 3M
Posts: 68
Joined: Sat Sep 27, 2014 3:00 am

### Re: Homework Problem #19

Thank you! That definitely helps! I am still a little confused though because in the solutions manual it says that "heat lost by metal = - heat gained by water. If the water is the surroundings, then why is it negative?

Neil DSilva 1L
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Homework Problem #19

It's not necessarily saying that the heat gained by the water is negative. It just means that the heat lost and heat gained are equal and opposite.

Justin Le 2I
Posts: 142
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Homework Problem #19

It might be easier to understand if you move the negative sign, so it would look like this:
-heat lost by metal = heat gained by water

Anuk Burli 2C
Posts: 29
Joined: Fri Sep 26, 2014 2:02 pm

### Re: Homework Problem #19

The way to remember any problem like these kind of calorimeter problems is to think of it in terms of the first law of thermodynamics where any heat released by the system has to be absorbed by the surroundings or aka q system + q surroundings=0. As long as you correctly identify the system and the surroundings the sign conventions of q will be easy to figure out.