Current Electricity Ques 57
- 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
Answer:
Correct Answer: 57.(b)
Solution:
Formula:
- 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 \quad …….(i) \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 \quad …….(ii) \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} \text{min}=24.63 \text{ min} \end{aligned} $
Nearest answer is $22$ min.
BETA
