The discussion centers on calculating the amount of ice that melts when added to 1 kg of lemonade, which cools from 20°C to 0°C. The heat extracted from the lemonade is calculated to be 83.38 kJ, using the formula Q = mcΔT, where the specific heat capacity of water is 4.169 kJ/kg°C. The latent heat of fusion for water is established as 333 kJ/kg, leading to the conclusion that 0.25 kg of ice melts as a result of this heat transfer.
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kingdomxiii said: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.
kingdomxiii said:A phase change occurs, but the problem does not give the mass of the ice.
kingdomxiii said: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.