1 kg of water, at 20 ° C is heated in an electric kettle whose heating element has a mean (temperature averaged) resistance of 20   Ω . 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 kg - 1 ° C - 1 Latent heat of water = 2260 k J k g - 1 ]

Step-by-step solution

Given specific heat of the water 4200   J   kg - 1   ° C - 1 Latent heat of the water = 2260     k J   k g - 1     Q = P × t Q = m c Δ T + m L ⇒ P × t = m c Δ T + m L ⇒ V r m s 2 R × t = m c Δ T + m L ⇒ 200 2 20 × t = 4200 × 80 + 2260 × 10 3 ⇒ t = 1298  s ≃ 22   m i n