Current Electricity 3 Question 1
1. One kilogram 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 $=4200 J /\left(kg^{\circ} C\right)$, Latent heat of water $=2260 kJ / kg$ ]
(2019 Main, 12 April II)
(a) $16 min$
(b) $22 min$
(c) $3 min$
(d) $10 min$
Show Answer
Solution:
- Heat required by water for getting hot and then evaporated is
$$ \Delta Q=m s \Delta T+m L $$
Here, $\quad m=1 kg, \Delta T=100^{\circ}-20^{\circ}=80^{\circ} C$,
$s=4200 J kg^{-1}{ }^{\circ} C^{-1}, L=2260 \times 10^{3} J kg^{-1}$
So, heat required is
$$ \begin{aligned} \Delta Q & =1 \times 4200 \times 80+1 \times 2260 \times 10^{3} \\ & =336 \times 10^{3}+2260 \times 10^{3} \\ & =2596 \times 10^{3} J \end{aligned} $$
This heat is provided by a heating coil of resistance $R=20 \Omega$ connected with AC mains $V _{\text {rms }}=200 V$
So, heat supplied by heater coil is
$$ Q=P t=\frac{V _{rms}^{2}}{R} \times t $$
where, $P=$ power and $t=$ time.
$$ \begin{aligned} & =\frac{(200)^{2}}{20} \times t \\ & =2 \times 10^{3} \times t \end{aligned} $$
Substituting the value of heat from Eq. (ii), we get
$$ \begin{aligned} t & =\frac{2956 \times 10^{3}}{2 \times 10^{3}} s \\ & =\frac{2956}{2 \times 60} min=24.63 min \end{aligned} $$
Nearest answer is $22 min$.
