Question

$1 \, kg$ of water, at $20 \,{}^\circ C$ 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 $200 \, V$ . Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to

[ Specific heat of water $=4200 \, J \, kg^{- 1} \,{}^\circ C^{- 1}$ Latent heat of water $=2260 \, kJ \, kg^{- 1}$ ]

• A. $3min$
• B. $16min$
• C. $22min$
• D. $10min$

Answer 1

Answer: (C)

Given specific heat of the water $4200 J kg ^{-1}{ }^{\circ} C ^{-1}$
Latent heat of the water $=2260 \, kJkg^{- 1}$
$Q=P\times t$
$Q=mc\Delta T+mL$
$\Rightarrow P\times t=mc\Delta T+mL$
$\Rightarrow \frac{V_{r m s}^{2}}{R}\times t=mc\Delta T+mL$
$\Rightarrow \frac{\left(200\right)^{2}}{20}\times t=4200\times 80+2260\times \left(10\right)^{3}$
$\Rightarrow t=1298\text{ s}\sim eq22 \, min$