At 1 atm, how much energy is required to heat 75.0 g H 2 O (s) at - 20.0 ∘ C to H 2 O (g) at 119.0 ∘ C?

238,485 Joules

Explanation:

The amount of energy required is a summation of heat of fusion, capacity and vaporization.

Q = mLf + mC∆T + mLv = m (Lf + C∆T + Lv)

m (mass of water) = 75 g

Lf (specific latent heat of fusion of water) = 336 J/g

C (specific heat capacity of water) = 4.2 J/g°C

∆T = T2 - T1 = 119 - (-20) = 119+20 = 139°C

Lv (specific latent heat of vaporization of water) = 2,260 J/g

Q = 75 (336 + 4.2*139 + 2260) = 75 (336 + 583.8 + 2260) = 75 (3179.8) = 238,485 J