Electrical heat generated = quantity of heat required to change water from liquid to vapour
I2^22 R t = mL
I = current = 10A, R = resistance = 2Ω, t = time = 20s, m = mass of water = 5 X 10−3^{-3}−3kg and L = specific latent heat of vaporization.
102^22 x 2 x 20 = 5 X 10−3^{-3}−3L
4000 = 5 X 10−3^{-3}−3L
L = 40005×10−3\; \frac{4000}{5 \times 10^{-3}}5×10−34000 = 800000 ⇔ 8.0 x 105^55Jkg−1^{-1}−1
