One kg of water, at 20oC, is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20 W. 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/(kgoC), Latent heat of water = 2260 kJ/kg]

Question:

One $\mathrm{kg}$ of water, at $20^{\circ} \mathrm{C}$, 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 $200 \mathrm{~V}$. Ignoring heat loss from the kettle, time taken for water to evaporate fully, is close to :

[Specific heat of water $=4200 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right)$, Latent heat of water $=2260 \mathrm{~kJ} / \mathrm{kg}$ ]

  1. 16 minutes

  2. 22 minutes

  3. 3 minutes

  4. 3 minutes


Correct Option: 2

Solution:

(2)

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