Practice question
Question
How much heat is required to convert 0.25kg of ice at −30∘C to steam at 120∘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 = 2.256×106Jkg−1)
Explanation
Q1 = 0.25×2100×30 = 15750J (ice to 0°C). Q2 = 0.25×3.35×105 = 83750J (melting). Q3 = 0.25×4186×100 = 104650J (water to 100°C). Q4 = 0.25×2.256×106 = 564000J (vaporization). Q5 = 0.25×4186×20 = 20930J (steam to 120°C). Total: Q = 15750+83750+104650+564000+20930 = 789080J = 789.08kJ.