Question

A block of ice (-30°C) with a mass of 1 lb is heated. How much heat (kJ) is needed to turn the ice into stream (vapor) at 200°C? Which part of the process contributes (requires) the most energy?

Answer 1

1 lb = 453.59 gm

Heating ice

the specific heat of ice is 0.50 cal/g-^oC

The ice temperature must be raised 30 degrees to reach 0^oC

30 x 453.59 x 0.50 cal = 6803.85 Cal

Melting ice

The latent heat for melting ice is 80 cal/g

453.59 x 80 cal = 36287.2 Cal

Heating water

the specific heat of water is 1.00 cal/g-^oC

100 x 453.59 x 1.00 = 45359 cal

Boiling water

The latent heat for boiling water is 540 cal/g

453.59 x 540 = 244938.6 cal

Heating steam

the specific heat of steam is 0.48 cal/g-^oC

100 x 453.59 x 0.48 = 21772.32 cal

Total

The total heat required to change 1lb ice into stream (vapor) at 200°C is:

6803.85+ 36287.2+ 45359 + 244938.6+ 21772.32 = 355160.97 cal. or 355.160 Kcal or 1485.989 kJ

Boiling water contributes (requires) the most energy

Everything i have calculated in calories finally converted into kJoules (1 Kcal = 4.184 kJ)