Question

Your 300 ml cup of coffee is too hot to drink when served at 90 C. What is the mass of an ice cube, taken from a -10 C freezer, that will cook your coffee to a pleasant 60 C? You can take coffee’s physical properties to be the same as those of water l. Cice = 2090 J/(kgK), cwater = 4190 J/(kgK) and Lf= 3.33*10^5 J/kg

Answer 1

heat you have to remove is:

specific heat of water is 4.186 kJ/kgC

300 mL is 0.3(0.998 kg) or 0.299 kg

E = 4.190 kJ/kgC x 0.299 kg x (90-60)C = 37.58 kJ

how much ice do you melt to put out that energy?

It removes energy/heat to the water in three ways, first by warming up from –10 to zero, second by melting, third by the melted ice warming up to 60C

heat of fusion of ice is 333 kJ/kg

specific heat of ice is 2.09 kJ/kgC

37.58 kJ = 333 kJ/kg x M + 4.19 kJ/kgC x M x 60C + 2.09 x M x 10C

M = 0.062 kg of ice

answered by: anonymous