One $kg$ of water, at $20\,^oC$, 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.......... $\min$ [Specific heat of water $= 4200\, J/kg\, ^oC$), Latent heat of water $= 2260\, k\,J/kg$]
JEE MAIN 2019, Diffcult
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$Q = P \times t$
$Q = mc\Delta T + mL$
$P = \frac{{V_{rms}^2}}{R}$
$4200 \times 80 + 2260 \times {10^3} = \frac{{{{\left( {200} \right)}^2}}}{{20}} \times t$
$t = 1298\,\sec $
$t \simeq 22\,\min $
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