First
we need to calculate the energy needed to melt the ice:

specific
latent heat of fusion, L_{m}
= 334 000 J/kg.

DQ
= mL = 0.5 kg × 334 000 J kg^{1} =
167 000 J
Next
we need to find the energy supplied to bring the water to the boil:
Now
we need to find out the energy needed to boil the water away:

specific
latent heat of vaporisation is rather higher, L_{v}
= 2.3 × 10^{6} J kg^{1}

DQ
= mL = 0.5 kg × 2.3
× 10^{6}
J kg^{1} = 1 150 000 J
Add
them all together:
Total
energy = 167 000 + 1 150 000
+ 210000 = 1527000 J
How
long would this take a 2 kW kettle?
Power
= energy / time
Time
= energy / power = 1527000 J ÷ 2000 J/s = 763 s (about 13 minutes)