Question

what heat is required to change 20 g of 26 ∘C water to 100 ∘C steam. Express your answer in calories.

Answer 1

Let M be the mass of water present:

ΔQ = M (T2 - T1) Sw + M Ss using specific heat of water and latent heat of vaporization

ΔQ = 20 g * 74 deg C * 1 cal / (g & deg C) + 20 g * 540 cal / g)

ΔQ = 12,300 cal