Ice cube melting in lemonade at different temperatures

[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

 

[ehild]

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)(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

 

[mikep]

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

 

[ehild]

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

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

 

[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

 

[ehild]

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

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

ehild