Practice question
Question
How much heat is required to convert 0.2kg of ice at −22∘C to steam at 110∘C in a 0.05kg aluminium calorimeter initially at 30∘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, aluminium = 900J kg−1K−1)
Explanation
Q1 = 0.2×2100×22 = 9240J (ice to 0°C). Q2 = 0.2×3.35×105 = 67000J (melting). Q3 = (0.2×4186+0.05×900)×100 = (837.2+45)×100 = 88220J (to 100°C). Q4 = 0.2×2.256×106 = 451200J (vaporization). Q5 = 0.2×4186×10 = 8372J (steam to 110°C). Calorimeter cools: Q6 = 0.05×900×(30−0) = 1350J (assume it cools to 0°C). Total: Q = 9240+67000+88220+451200+8372−1350 = 622682J = 622.68kJ.