Practice question
Question
How much heat is required to convert 0.3kg of ice at −18∘C to water at 40∘C in a 0.1kg copper calorimeter initially at 25∘C? (Specific heat of ice = 2100J kg−1K−1, latent heat of fusion = 3.35×105J kg−1, specific heat of water = 4186J kg−1K−1, copper = 386J kg−1K−1)
Explanation
Q1 = 0.3×2100×18 = 11340J (ice to 0°C). Q2 = 0.3×3.35×105 = 100500J (melting). Heat gained by water and calorimeter: (0.3×4186+0.1×386)×(40−0) = (1255.8+38.6)×40 = 1294.4×40 = 51776J. Heat lost by calorimeter: 0.1×386×(25−0) = 965J (assume it cools to 0°C first). Total heat supplied: Q = 11340+100500+51776−965 = 162651J = 162.65kJ.