JEE Mains · Physics · STD 12 - 3. current electricity
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 \(= 2260\, k\,J/kg\)]
- A \(3\)
- B \(10\)
- C \(22\)
- D \(16\)
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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