Question

How much heat is required to completely boil away 0.200 kg of water (Hv = 2.26 × 103 kJ/kg)?

4.52 × 10^2 kJ

2.26 × 10^3 kJ

4.52 × 10^3 kJ

1.13 × 10^4 kJ

Answer 1

answer is (mass of water, kg) (heat of vaporization, kJ/kg)

= 0.2(2.26x10^3) = etc.