Question

If 100 gm of steam at 150°c is cooled and Frozen into 100 gm of ice at 0°c ( sp. heat of steam =2.01 kj/kg K and that of water = 4.18kj/ kg K ). The removed heat is:

Answer 1

Given

Mass of steam = 100gm = 373 K

Temperature of the steam = 150°C = 423 K

Specific heat of steam = 2.01 kJ/Kg K

Specific heat of water = 4.18 kJ/Kg K

To Find :

The heat removed in the process

Solution :

• For the steam to convert into ice , several phase changes xoccur duting the process

• Steam at 150°C changes into steam at 100°C

              $E_{1} =mc\Delta T\\ E_{1} =0.1\times 2.01\times (423 - 373)\\ E_{1} = 10.05\:kJ$

• Steam at 100°C changes into water at 100°C

              $E_{2} = mL_{steam} \\ E_{2} =0.1\times 2257\\ E_{2} =225.7 \:kJ$        

• Water at 100°C changes into water at 0°C

              $E_{3} = mc\Delta T\\ E_{3} = 0.1\times4.18\times(373-273)\\ E_{3} = 41.8\:kJ$

• Water at 0°C changes into ice at 0°C

              $E_{4} = mL_{ice} \\ E_{4} =0.1\times 334\\ E_{4} =33.4\:kJ$

• Total energy removed =  E₁ + E₂ + E₃ + E₄

                                              = 10.05 + 225.7 + 41.8 + 33.4

                                              = 310.95 kJ

 The heat energy removed is 310.95 kJ.