Heating water + cooling

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Heating water + cooling

Postby sarahtang4B » Wed Feb 13, 2019 12:06 pm

When we have a block of ice added into water is it always q(ice)= -q(water)

then when we have water at 25 degrees added to water at 90 degrees is it still q(cooler water)= -q(hot water)

I noticed my TA was just doing q(ice)=q(water) and some homework problems do it with the above equation.

MaanasO 1A
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Re: Heating water + cooling

Postby MaanasO 1A » Wed Feb 13, 2019 12:32 pm

Hi Sarah!

The heat gained by the ice [q(ice)] is equal to the heat LOST by the water [-q(water)]. If you add the negative sign to the heat of water, the q value comes out as positive (because of the double negative). Their magnitudes are the same, so it's not technically incorrect for your TA to write it the way they do. Just don't forget to add the negative sign for the heat LOST by water.

Hope that helps!

Vicky Lu 1L
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Joined: Fri Sep 28, 2018 12:18 am

Re: Heating water + cooling

Postby Vicky Lu 1L » Wed Feb 13, 2019 1:29 pm

If we start out with qtotal = qsystem + qsurrounding, you can get qsystem=-qsurrounding if no energy can be lost to the surrounding.
Therefore, if you pick ice to be your system and water to the be surrounding, qice=-qwater. which means that heat gained by ice will be from the heat of water, (making water losing heat).

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