# Homework Help: Phase Changes and Energy Conservation

1. Nov 24, 2004

### mikep

A large punch bowl holds 4.75 kg of lemonade (which is essentially water) at 20.0°C. A 2.00 kg ice cube at -10.2°C is placed in the lemonade. What is the final temperature of the system, and the amount of ice (if any) remaining? Ignore any heat exchange with the bowl or the surroundings.

can someone please tell me how to solve this problem?
so far i have:
(4.75kg)(4186J/kg°C)(20°C) = 397670J
(2kg)(2090J/kg°C)(10.2°C) = 426360J
(2kg)(335000J/kg) = 670000J

2. Nov 24, 2004

### ehild

(2kg)(2090J/kg°C)(10.2°C) = 42636J

You see the heat energy of water is enough to rise the temperature of all the ice up to 0.0 °C. The remaining 355034 J is enough to melt 335034/335000=1.06 kg ice only. I hope you can proceed from here.

ehild

3. Nov 25, 2004

### mikep

wouldn't it be 335034/670000 = 0.53kg ? because Q = m L = (2kg)(335000) = 670000

4. Nov 25, 2004

### ehild

L= 335000 J/kg. It is the heat needed to melt 1 kg ice at 0 °C. You have Q = 355034 J available heat energy from the warm lemonade.
m=Q/L=355034 J/335000 (J/kg)=1.06 kg.

That 670000 J you suggested is the heat needed to melt all the 2kg of ice. The heat available is enough to melt 0.53 times the original amount , that is 2*0.53=1.06 kg.

ehild

5. Nov 26, 2004

### mikep

oh i get it. so te amount of heat left is 0.94kg and the final temperature would be 0°C because there is still some ice left

6. Nov 26, 2004

### ehild

This is all right if you meant 0.94 kg ice instead of heat

ehild