Practice question
Question
How much heat is required to convert 0.35kg of ice at −28∘C to steam at 115∘C in a 0.1kg brass 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, latent heat of vaporization = 2.256×106J kg−1, brass = 386J kg−1K−1)
Explanation
Q1 = 0.35×2100×28 = 20580J (ice to 0°C). Q2 = 0.35×3.35×105 = 117250J (melting). Q3 = (0.35×4186+0.1×386)×100 = (1465.1+38.6)×100 = 1503.7×100 = 150370J (to 100°C). Q4 = 0.35×2.256×106 = 789600J (vaporization). Q5 = 0.35×4186×15 = 21976.5J (steam to 115°C). Calorimeter cools: 0.1×386×(25−0) = 965J. Total: Q = 20580+117250+150370+789600+21976.5−965 = 1097811.5J = 1097.81kJ.