Physics !!!

Water at 20 ℃ must be heated to 100 ℃, then water can be boiled to vapour.

Power of the heater, P = 1 kW = 1000 W
Time taken, t = 10 mins = 600 s

Mass of water, m = 0.5 kg
Specific heat capacity of water, c = 4200 Jkg^-1℃^-1
Temperature change of water, ΔT = 100 - 20 = 80 ℃

Specific latent heat of vapourization of water, lv = 2.26 X 10^6 Jkg^-1

Let ma be the mass of water being boiled away.

Assume no energy is lost to the surroundings.

By the law of conservation of energy,

Energy supplied by the heater = Energy required to heat the water + Energy required to boil the water away

Pt = mcΔT + malv
(1000)(600) = (0.5)(4200)(80) + ma(2.26 X 10^6)
(2.26 X 10^6)ma = 432 000
ma = 0.191 kg


So, 0.191 kg of water is being boiled away. So the answer is A.