Calculating Final Temperature and Ice Remaining in a Thermodynamics Problem

[physicsnnewbie]

You take a jug of water from the refrigerator, and pour 240mL of it into a glass. The temperature of the beverage is 10.5ºC. You then add one 45g ice cube at 0ºC.

Determine the final temperature and the amount of ice remaining if any.

I've tried to work it out below, but my I must've made an error because the temperature has to be between 0 and 10.5. Where have I gone wrong?

4.184*240(x-10.5) = -[45*333+4.184*45(x-0)]
1004.16x - 10543.68 = -[29970+376.56x]
1004.16x - 10543.68 = -29970-376.56x
1,380.72x - 10543.68 = -29970
1,380.72x = -19,426.32
x = -19,426.32/1,380.72
x = -14.06970276

thanks

 

[marcusl]

It is most convenient to work in calories rather than converting to joules.

1. First find the amount of heat available in the water.
240g * 10.5 C = 2520 cal

2. find the amoutn of ice this will melt.
Heat of fusion is 80 cal/g.
Amount of melt = 2520 cal / (80 cal/g) = 31.5 g

3. Find remaining ice
45g - 31.5g = 13.5g

4. Since we've pulled water temp to 0C, and there's ice remaining at 0C, final temp=0C

 

[physicsnnewbie]

thanks marcusl :)

 

[Stevedye56]

It is most convenient to work in calories rather than converting to joules.

I'm not all that sure about that. I haven't done a problem in calories before.

 

[marcusl]

It's a small difference, doesn't really matter either way. 4.18 Joules of heat energy raise the temperature of 1 gram of water by 1 degree C. By definition, one calorie of heat energy raises the temperature of 1 gram of water by 1 degree C. The calorie unit is a little simpler or more direct in this problem. End result is the same