Given specific heat of the water 4200J{\mathrm{kg}}^{-1}^{\circ}{C}^{-1}
Latent heat of the water =2260kJkg−1
Q=P×t
Q=mcΔT+mL
⇒P×t=mcΔT+mL
⇒RVrms2×t=mcΔT+mL
⇒20(200)2×t=4200×80+2260×103
⇒t=1298 s≃22min
JEE Main 2019 — Physics Thermodynamics
1kg of water, at 20∘C is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20Ω. The rms voltage in the mains is 200V. Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to
[ Specific heat of water =4200 J{\mathrm{kg}}^{-1}^{\circ}{C}^{-1} Latent heat of water =2260kJkg−1 ]
Held on 12 Apr 2019 · Verified 6 Jul 2026.
3min
16min
22min
10min
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