Question

At 1 bar, how much energy is required to heat 55.0 g of H,O(s) at -24.0°C to H,O(g) at 131.0°C?

Answer 1

Here thing to be understood is the process

When the ice at -240C is converted to steam at 1310C

Then,

Heat is required to bring ice at   -240C to 00C with Cice  = 2.09 J/g/deg

Heat is required to convert ice to water with latent heat of fusion = 334 j/g

Heat is required to bring water at 00C to 1000C with CH2O = 4.184 J/g/deg

Heat is required convert water to steam latent heat of fusion= 2260 j/g

Heat is required to take temperature from 100 to 131 0C with Csteam = 1.89 J/g/deg

So

The Heat required to bring ice at   -240C to 00C = mCp$\Delta$t

= 55 g * 2.09 J/g/deg * 24 0C = 2.758 kJ

The Heat required to for converting ice to water = mL (WHERE L IS LATENT HEAT OF FUSION )

= 55 g * 334 j/g

= 18.37 KJ

The Heat required to convert Water at 00C to 1000C = mCp$\Delta$t

= 55 g * 4.184 J/g/deg * 100 deg

= 23.012 KJ

The Heat required for converting Water to steam = mL ( latent heat of vaporization = 2260 j/g )

= 55 g * 2260 j/g

= 124.3 KJ

The Heat required to convert steam at 100 to 131 0C =  mCp$\Delta$t \

= 55 g * 1.89 J/g/deg * 31

= 3.22 KJ

The amount of total heat required = sum of All the steps =  2.758 kJ + 18.37 KJ +  23.012 KJ + 124.3 KJ + 3.22 KJ

=  171.49 KJ

The Amount heat required is 171.49 KJ/mole