Question

One kg of water, at ${{20}^{0}}C$ , 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 the heat loss from the kettle, time taken for water to evaporate fully is close to:
[Specific heat of water =$4200J/kg{}^{0}C$ , latent heat of water =$2260kJ/kg$ ]
A) 3 minutes
B)22 minutes
C)10 minutes
D)16 minutes

Answer 1

Hint: The kettle will supply the necessary heat to evaporate the water. We need to equate the heat produced by the electric kettle in the required time as the heat required to evaporate the water. From that equation we can get the required time. We need to go in this way to solve the question.
Formula used: $H=mst+mL$ ,$H=\dfrac{{{V}^{2}}}{R}T$

Complete answer:
Let H is the heat required to evaporate the water. Let mass of the water is $m$, specific heat of the water is $s$ ,latent heat of water is $L$ and If the change in temperature is $t$, then we can write
$H=mst+mL$
Putting the values of all quantities we get
$\begin{align}
  & H=1\times 4200\times (100-20)+1\times 2260000J \\
 & orH=2596000J............(1) \\
\end{align}$
Now if $V$ is applied voltage,$R$ is the resistance of the heating element and $T$ is the time required then from Joule’s heating formula to generate the heat then we can write
$H=\dfrac{{{V}^{2}}}{R}T$
Now putting the value of $H$ from $(1)$ and all other quantities we have
\[\begin{align}
  & 2596000=\dfrac{{{200}^{2}}}{20}\times T \\
 & or T=\dfrac{2596000\times 20}{{{200}^{2}}}s=1298s=\dfrac{1298}{60}minutes=21.63minutes \\
\end{align}\]

So the correct answer is B.

Note:
In this kind of problem we must first convert all the quantities in the same system units. Like in this problem, we need to convert the latent heat given in $kJ/kg$ to $J/kg$ . Also when we are applying Joule’s heating formula we should remember that the time will come in seconds. We have to convert the time in accordance with the question.