Question

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

Answer 1

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

Answer 2

The heat that is required to convert 422 g of liquid H₂O at 23.5 °c into steam at 150 °c can be calculated as follows:

The heat capacity of water = 4184 J K⁻¹ C⁻¹

Therefore, heat required to warm the water from

23.5 °C to 100.0 °C.

                         = $m\times c\times \Delta T$

Here,

m=0.422 kg

c=4184 J K⁻¹ C⁻¹

$\Delta T$ =100-23.5

   so, heat required to warm the water from  23.5 °C to 100.0 °C

                 = 0.422 × 4184 × (100-23.5)

                        = 135072.072 J

Latent heat of vaporization of water is 2260 kJ/kg

Thus the heat will be 0.422 × 2260000 = 953720 J

Heat required 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