Power consumed = V*I = 12 *5.2 = 62.4 J/s.

Rate of evapoation = 21 mg/s = 2.1 * 10^-5  kg /s

Power consumed = Heat gaines by water per second

Heat of Vaporization, H  = 62.4 / (2.1 * 10^-5)  =  2971 KJ/kg =710 Kcal / kg.

heat generated in coil (H) = I²Rt

given V =12 and I = 5.2 A , by ohms law, R = V/I = 12/5.2 =2.3 ohms

dH/dt = I²R  = 62.2 J/sec

let specific heat of vapourisation is:  L J/g

rate of evaporation is : m' = 21x10^-3 g/sec

so, dH/dt = m' * L 

L = (dH/dt)/ m'

L= 62.2 / 21x10^-3

L = 2961.5 J/gram

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Naga Ramesh
IIT Kgp - 2005 batch

Heat generated per sec due to the current carrying coil,

                                                                                        H=VI  =>H =12*5.2 J/sec

 This heat will be equal to the latent heat of vapourization as the same is responsible for the vapourization of the liquid,

                                                       VI=mL      =>12*5.2=21/1000*L

                                                                 L=29.71*10^2 J/Kg