Question

What is the mass of an ice cube, taken from a -15.0 C freezer, that will cool your coffee to a pleasant 65.0 C?

Answer 1

energy used to cool the coffee tp 62º equals the energy needed to warm the ice to 0º, melt it, and warm the resulting water to 62º

specific heat of water is 4.186 kJ/kgC
specific heat of ice is 2.06 kJ/kgC
heat of fusion of ice is 334 kJ/kg
density of fresh water at 20C = 0.998 g/cm³
= 998 kg/m³ = 8.33 lb/gal = 62.1 lb/ft³
300 mL = 300 cm³
300 cm³ x 0.998 g/cm³ = 299 g = 0.299 kg

E = m ( (2.06 kJ/kgC x 16) + 334 kJ/kg + (4.186 kJ/kgC x (62–0) )
E = 0.299 kg x 4.186 kJ/kgC x (88–62)

m ( (2.06 kJ/kgC x 16) + 334 kJ/kg + (4.186 kJ/kgC x (62–0) ) = 0.299 kg x 4.186 kJ/kgC x (88–62)

m(32.96 + 334 + 260) = 32.54
m = 32.54 / 627 = 0.0519 kg or about 52 g