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Q. One $kg$ of water, at $20^\circ C$, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of $20 \, \Omega$. The rms voltage in the mains is $200\, V$. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to :[Specific heat of water = $4200\, J/kg^\circ C$), Latent heat of water = $2260\, kJ/kg$]

JEE MainJEE Main 2019Current Electricity

Solution:

$Q = P \times t$
$Q = mc\Delta T + mL$
$P = \frac{V^2 _{rms}}{R}$
$4200 \times 80 + 2260 \times 10^3 \, \, = \, \frac{(200)^2}{20} \times t$
$t = 1298 \,sec$
$t \sim 22\, min$