Question

How much heat must be removed when 102 g of steam at 144°C is cooled and frozen into 102 g of ice at 0°C. (Take the specific heat of steam to be 2.01 kJ/kg·K.)

I got 67 but its wrong.

Answer 1

Heat removed will be

Q = Q1 + Q2 + Q3 + Q4

Q1 = heat removed from 144 C steam to 100 C steam

Q2 = Heat removed from 100 C steam to 100 C water

Q3 = heat removed from 100 C water to 0 C water

Q4 = heat removed from 0 C water to 0 C ice

Q = m1*Cs*dT1 + m1*Lv + m1*Cp*dT2 + m1*Lf

given values ae:

m1 = 102 gm = 0.102 kg

Cp = 4186 kJ/kg-K

Cs = 2010 kJ/kg-K

Lv = 2.26*10^6 kJ/kg

Lf = 3.34*10^5 kJ/kg

Using these values:

Q = 0.102*(2010*44 + 2.26*10^6 + 4186*100 + 3.34*10^5)

Q = 3.163*10^5 J