One 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° C), Latent heat of water = 2260 kJ/kg]

Question

One 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° C), Latent heat of water = 2260 kJ/kg] (A) 3 minutes (B) 22 minutes (C) 10 minutes (D) 16 minutes

Tardigrade Admin · Accepted Answer

Q = P × t Q = mcΔ T + mL P = (V2 rms/R) 4200 × 80 + 2260 × 103 = ((200)2/20) × t t = 1298 sec t ∼ 22 min

3 minutes

22 minutes

10 minutes

16 minutes

Q=P×t Q=mcΔT+mL P=RVrms2​​ 4200×80+2260×103=20(200)2​×t t=1298sec t∼22min

$Q = P \times t$
$Q = m c Δ T + m L$
$P = \frac{V _{r m s}^{2}}{R}$
$4200 \times 80 + 2260 \times 1 0^{3} = \frac{( 200 ) ^{2}}{20} \times t$
$t = 1298 sec$
$t \sim 22 min$