HW #41 vs #21

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Michelle Steinberg2J
Posts: 73
Joined: Fri Sep 29, 2017 7:04 am

HW #41 vs #21

Postby Michelle Steinberg2J » Tue Jan 23, 2018 11:53 am

I am really confused by the negative sign that the solution manual puts on either the system or the surroundings. In the HW problem with the copper put into water, heat flow would go from the copper to the water... so why is the negative sign on the water? Isn't the copper the one that's losing heat? However, in the HW problem with the ice put into water, heat flow will go from the water to the ice. I understand why the negative sign is on the water in this situation, but I don't understand how it could be the same in the other HW problem when the direction of heat flow is opposite.

Someone mentioned that it doesn't matter where the negative sign goes... is this true?

Sirajbir Sodhi 2K
Posts: 47
Joined: Sat Jul 22, 2017 3:00 am

Re: HW #41 vs #21

Postby Sirajbir Sodhi 2K » Tue Jan 23, 2018 12:00 pm

The negative sign comes from one entity gaining the heat that is lost by the other. In my opinion, it makes the most sense if the negative sign goes on the entity that is gaining heat. But, it really does not matter where you put it, as long as it is in the equation. Switching the side of the negative sign is just like dividing both sides of the equation by -1.
Last edited by Sirajbir Sodhi 2K on Tue Jan 23, 2018 12:00 pm, edited 1 time in total.

Nehal Banik
Posts: 64
Joined: Thu Jul 13, 2017 3:00 am

Re: HW #41 vs #21

Postby Nehal Banik » Tue Jan 23, 2018 12:00 pm

It doesn't matter where you place the negative sign, because it will lead you to the same final answer. As long as you remember to put the negative somewhere during your calculation you should be fine. Just remember:
qsystem+qsurroundings=0

Therefore when the system releases heat it is absorbed by the surroundings it doesn't necessarily matter where you put it.

qsystem=-qsurroundings

Janine Chan 2K
Posts: 71
Joined: Fri Sep 29, 2017 7:04 am

Re: HW #41 vs #21

Postby Janine Chan 2K » Tue Jan 23, 2018 12:04 pm

In lecture, we discussed how in a perfect system, qsystem + qsurroundings = 0. If this is the case, qsystem = - qsurroundings. In this sense, you could say it's true that the negative sign can go on either one, because qsystem = - qsurr is the same as qsurroundings = - qsystem. If you assign copper to be the "system" and the water around it to be the "surrounding", then you could say q lost by copper = -q absorbed by water.


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