Question

How much heat (in kJ) is required to convert 420. g of liquid H2O at 24.9°C into steam at 145°C? (Assume that the specific heat of liquid water is 4.184 J/g·°C, the specific heat of steam is 2.078 J/g·°C, and that both values are constant over the given temperature ranges. The normal boiling point of H2O is 100.0°C.)

Answer 1

24.9°C -----------------> 100 oC -----------------> 145 oC

there are three heat conversions

1)   Q1 = m Cp dT = 420 x 4.184 x (100-24.9) = 1.32 x 10^5 J = 132 kJ

2) Q2 = m x DHvap = 420 x 2257 x 10^-3 kJ    = 948 kJ

3) Q3 = m Cp dT = 420 x 2.078 x (145-100) = 39.3 kJ

total heat = Q1 + Q2 + Q3

                    = 1119 kJ

heat required = 1119 kJ