Melting Ice in Lemonade: Calculating Heat Transfer

[kingdomxiii]

Homework Statement

Ice at 0°C is added to a 1 kg of lemonade cooling it from 20°C to 0°C. How much ice melts? (The heat capacity of lemonade equals that of water.)

Homework Equations

Q = mcΔT

The Attempt at a Solution

Found the heat in of the lemonade to the ice... = 83.38 kJ and don't know where to go from there. c of water is 4.169.

 

[gneill]

Homework Statement

Ice at 0°C is added to a 1 kg of lemonade cooling it from 20°C to 0°C. How much ice melts? (The heat capacity of lemonade equals that of water.)

Homework Equations

Q = mcΔT

The Attempt at a Solution

Found the heat in of the lemonade to the ice... = 83.38 kJ and don't know where to go from there. c of water is 4.169.

What happens to the ice that melts? What's involved?

 

[kingdomxiii]

A phase change occurs, but the problem does not give the mass of the ice.

 

[gneill]

A phase change occurs, but the problem does not give the mass of the ice.

Right. Because it's the mass of ice that melts that you're looking for... Write out the formula that pertains to that phase change. Which of the variables do you you have values for?

 

[Chestermiller]

How much heat is removed from from the lemonade when it cooled from 20C to 0 C?

Chet

 

[kingdomxiii]

gneill: Ok, so the formula that pertains to that phase is ΔQ = Lm. I know that L is the latent heat of fusion for water so it is 333 kJ/kg. Would the change in heat be the 3.38 kJ?

Chestermiller: I found the change in heat for the lemonade form 20-0 °C to be 83.38 kJ.

 

[kingdomxiii]

Sorry, i meant the 83.38 kJ

 

[gneill]

gneill: Ok, so the formula that pertains to that phase is ΔQ = Lm. I know that L is the latent heat of fusion for water so it is 333 kJ/kg. Would the change in heat be the 3.38 kJ?

Chestermiller: I found the change in heat for the lemonade form 20-0 °C to be 83.38 kJ.

Yes. You calculated the heat that needed to be extracted from the lemonade in order to cool it to 0C, and that heat had to go somewhere. Clearly it went to melt some of the ice, providing the heat of fusion for the process.

 

[Chestermiller]

So if the heat removed from the lemonade was 83.38 kJ, and the amount of heat required to melt 1 kg of ice is 333 kJ/kg, how many kg of ice have to melt?

Chet

 

[kingdomxiii]

Right, so I got 83.38 kJ/(333 kJl/kg) = .25 kg. So .25 kg. Is this right?

 

[Chestermiller]

Yes.

 

[kingdomxiii]

Great! Thank you Chestermiller and gneill!