Question

Marianne really likes coffee, but on summer days she doesn't want to drink a hot beverage. If she is served 220 mL of coffee at 83 ∘C in a well-insulated container, how much ice at 0 ∘C should she add to obtain a final temperature of 40 ∘C?

Answer 1

Assumptions: Pressure is constant at 1atm and the coffee can be treated as water. Well insulated means that any heat lost to the surrounding can be treated as being nil.

Some Data:
specific heat capacity of water = 4.19kJ/kg
Density of water = 1g/mL

Q out of coffee = m*c*deltaT = 250g*4.190J/gC*(83-40) = 45042J. This amount of heat will be absorbed by the ice when it first melts and then increases it temperature to 40C. So

m*Lf * m*c*deltaT = 45042J. So m*(Lf +c*(40-0) = 45042 Lf for water is 3.34x10^2J/g So

m = 45042/(3.34x10^2+4.190*40) = 90g