How much heat is required to convert 422 g of liquid h2o at 23.5 °c into steam at 150 °c?

Heat is given by multiplying the specific heat capacity of a substance by mass and the change in temperature. The heat capacity of water is Approximately 4184 J/K/C.

Therefore, heat = mc0 mass in kg

= (422/1000) * 4184 * (100-23.5)

= 135072.072 J

Latent heat of vaporization is 2260 kJ/kg

Thus the heat will be 0.422 * 2260000 = 953720 J

Heat to raise steam from 100 to 150

2000 * 0.422 * 50 = 42200 J

Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules or 1330 kilo joules