5 X 10\(^{-3}\)kg of liquid at its boiling point is evaporated in 20s by the heat generated by a resistor of 2\(\Omega\) when a current of 10A is used. The specific latent heat of vaporization of the liquid is
a
8.0 x 10\(^4\)Jkg\(^{-1}\)
b
8.0 x 10\(^5\)Jkg\(^{-1}\)
c
8.0 x 10\(^6\)Jkg\(^{-1}\)
d
8.0 x 10\(^7\)Jkg\(^{-1}\)
Explanation
Correct Option
bVideo Explanation
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Favekook
1 year ago
using the formula.... power×time=mas
s×latent
analysis parameter
mass=00.05kg
time=20sec
resistor=2ohms
current=10A
latent heat=??
remember in electricity.....power=I²R
I..current. R... resistor
power=10*10*2=200watt
using the formula.... power×time=mass×latent
200*20=0.005*latent
400=0.005latent
latent =400/0.005
latent=800000
hope it helps 
