Question

$1kg$ 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 $200V$ . Ignoring heat loss from the kettle, time taken for water to evaporate fully is close to

[ Specific heat of water $=4200Jk{g}^{-1}{}^{\circ }{C}^{-1}$ Latent heat of water $=2260kJk{g}^{-1}$ ]

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

Answer 1

Answer: (C)

Given specific heat of the water $4200Jk{g}^{-1}{}^{\circ }{C}^{-1}$
Latent heat of the water $=2260kJk{g}^{-1}$
$Q=P\times t$
$Q=mc\Delta T+mL$
$\Rightarrow P\times t=mc\Delta T+mL$
$\Rightarrow \frac{V_{rms}^2}{R}\times t=mc\Delta T+mL$
$\Rightarrow \frac{{\left(200\right)}^2}{20}\times t=4200\times 80+2260\times {\left(10\right)}^3$
$\Rightarrow t=1298\text{ s}\sim eq22min$