Q 11-10-122JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
$1$ 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 $= 4200\ \text{J kg}^{-1}{}^\circ\text{C}^{-1}$, latent heat of water $= 2260\ \text{kJ kg}^{-1}$]
Answer: (C) $22$ min
Power: $P = \dfrac{V^2}{R} = \dfrac{200^2}{20} = 2000$ W.
Heat needed: $1\times4200\times80 + 2.26\times10^6 = 2.596\times10^6$ J.
$$t = \frac{2.596\times10^6}{2000} \approx 1300\ \text{s} \approx 22\ \text{min}$$
Solution by Sreeraj P, M.Sc Physics