How much heat is required to change 100 g of ice (H2O) at 253 K to vapor (steam) at 393 K?

Answer 1

One has to look at this problem in steps. First, we have to get the ice from 253K to 273.2K which is the melting point. Then we have to melt the ice into liquid water. Then we have to get the water from 273.2K to 373.2K. Then we have to evaporate the water into vapor. Then we have to get the vapor from 373.2K to 393K.

The specific heat capacity of ice is 2.108 kJ/kgK, or 2.108 J/gK. So to heat 100g of ice by 20.2K takes 2.108*100*20.2 J = 4,258.16 J.

The specific heat of melting of ice is 334 J/g, so melting 100g of ice will take 33,400 J.

The specific heat capacity of liquid water is 4.187 kJ/kgK, or 4.187 J/gK. So to heat 100g of water by 100K takes 4.187*100*100 J = 41,870 J

The specific heat of vaporization of water is 2230 J/g, so evaporating 100g of water will take 223,000 J.

The specific heat capacity of water vapor is 1.996 kJ/kgK, or 1.996 J/gK. So to heat 100g of water by 19.8Ktakes 1.996*100*19.8 J = 3,952.08 J

Adding up all the heat required for all these changes is 4,258.16+33,400+41,870+223,000+3952.08=306480.24 J.

So, changing 100g of ice at 253 K into 100g of water vapor at 393 K would take approximately 306.48 KJ of heat energy.