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 \(= 2260\, k\,J/kg\)]

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Answer & Solution

Correct Answer

(C) \(22\)

Step-by-step Solution

Detailed explanation

\(Q = P \times t\)

\(Q = mc\Delta T + mL\)

\(P = \frac{{V_{rms}^2}}{R}\)

\(4200 \times 80 + 2260 \times {10^3} = \frac{{{{\left( {200} \right)}^2}}}{{20}} \times t\)

\(t = 1298\,\sec \)

\(t \simeq 22\,\min \)

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