## Heating water + cooling

$\Delta U=q+w$

sarahtang4B
Posts: 132
Joined: Fri Sep 28, 2018 12:27 am

### Heating water + cooling

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
Posts: 72
Joined: Fri Sep 28, 2018 12:26 am

### Re: Heating water + cooling

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
Posts: 60
Joined: Fri Sep 28, 2018 12:18 am

### Re: Heating water + cooling

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).