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 55 ∘C?

Answer 1

Heat lost by coffee= heat gained by ice cubes

                   mcdt = m'cdt' + m'L

Where

m = mass of coffee = volume x density = 300 mL x 1.0g/mL = 300 g

c = specific heat capacity of water = specific heat capacity of water = 4.184 J/g-K

dt = change in temperature of coffee = 95 - 55 = 40 oC

m' = mass of ice = ?

dt' = change in temperature of ice from -15oC to ice at 0oC = 0(-15) = 15 oC

L = Heat of fusion of ice = 334.9 J/g

Plug the values we get m' = 126.2 g