Question

Calculate the amount of heat required to change 35.0 g ice at -25.0ºC to steam at 125ºC. (Heat of fusion = 333 J/g; heat of vaporization = 2260 J/g; specific heats: ice = 2.09 J/g·K, water = 4.18 J/g·K, steam = 1.84 J/g·K)

Answer 1

There are five steps:
1) Bring the ice from -25 C to 0 C.

Heat required per gram = C * Dt

=2.09 J/g·K x 25 K

= 52.25 J/g
2) Melt the ice:

Heat of fusion = 333 J/g

3) Bring the water from 0 to 100 C:

= C * Dt

=4.18 J/g·Kx 100K = 418 J/g
4) Vaporize the water:

heat of vaporization = 2260 J/g

5) Bring the steam from 100 to 125 C:

= C * Dt

1.84 J/g·K X25K

= 46 J/g

Total: 3109 J/g means this heat is require to change 1.0 g ice at -25.0ºC to steam at 125ºC.

Now multiply it by 35.0 g and we get

=3109 J/g * 35.0 g

= 108815 J or 108.8 kJ which is require to change 35.0 g ice at -25.0ºC to steam at 125ºC.