Question

Your 300 mL cup of coffee is too hot to drink when served at 95 ∘C

What is the mass of an ice cube, taken from a -15 ∘C freezer, that will cool your coffee to a pleasant 57 ∘C?

Specific Heat                                        Lf (J/kg)            Lv (J/kg)

Ice - 2090 J/kg * K                        

Water - 4190 J/kg * K                           333,000             2,260,000

Answer 1

here,

mass of coffee , m1 = 300 mL = 300 g = 0.3 kg

let the mass of ice cube required be m2

using conservation of heat energy

heat lost by ice cube = heat gained by coffee

m2 * (Ci * ( 0 - (-15)) + Lf + Cw * ( 57 - 0)) = m1 * Cw * (95 - 57 )

m2 * (2090 * ( 0 - (-15)) + 333000 + 4190 * ( 57 - 0)) = 0.3 * 4190 * (95 - 57 )

solving for m2

m2 = 0.0792 kg = 79.2 g

the mass of ice added is 79.2 g