Question

Your 300mL cup of coffee is too hot to drink when served at 85.0 ∘C.What is the mass of an ice cube, taken from a -17.0 ∘C freezer, that will cool your coffee to a pleasant 55.0 ∘?

Answer 1

density of water = 1 g/mL
mass of cofee = 300 mL*1g/mL = 300 g
specific heat capacity of water, C1 = 4.186 J/goC
specific heat capacity of ice, C2 = 2.03 J/goC
Latent heat of fusion of ice, Lf = 334 J/g

Let mass of ice be m g

Heat given out by coffee =300*4.186*(85-55) = 37674 J

this heat must be supplied by ice

Heat supplied= m*Lf + m*C2*delta T + m* C1*delta T1
37674 = m* [Lf + C2*delta T2 + C1*delta T1]
37674 = m* [334 + 2.03*(17) + 4.186*55]
37674 = m* 264.74
m= 142.3 g
Answer: 142.3 g