# Ques) A coil of current crrying Nichrome wire is immersed in aliquid. When the potetial difference across the coil is 12 V and the current through the coil is 5.2 A., the liquidm evaporated at the steady rate of 21mg/sec. Calculate the heat of  Vapourization of the liquid.

18 Points
14 years ago

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.

Ramesh V
70 Points
14 years ago

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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Vanya Saxena
18 Points
14 years ago

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