Practice question
Question
How much heat is required to raise 0.5kg of water from 20∘C to 80∘C and then convert 0.2kg of it to steam at 100∘C? (Specific heat of water = 4186J kg−1K−1, latent heat of vaporization = 2.256×106J kg−1)
Explanation
Q1 = 0.5×4186×(80−20) = 0.5×4186×60 = 125580J (to 80°C). Q2 = 0.5×4186×(100−80) = 0.5×4186×20 = 41860J (to 100°C). Q3 = 0.2×2.256×106 = 451200J (vaporization of 0.2 kg). Total: Q = 125580+41860+451200 = 618640J = 618.64kJ.