Practice question
Question
How much heat is required to convert 0.2kg of ice at −15∘C to steam at 100∘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 = 0.2×2100×15 = 6300J (ice to 0°C). Q2 = 0.2×3.35×105 = 67000J (melting). Q3 = 0.2×4186×100 = 83720J (water to 100°C). Q4 = 0.2×2.256×106 = 451200J (vaporization). Total: Q = 6300+67000+83720+451200 = 608220J = 608.22kJ.