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An electric kettle takes 4 A current at $220 \mathrm{~V}$. How much time will it take to boil $1 \mathrm{~kg}$ of water from temperature $20^{\circ} \mathrm{C}$ ? The temperature of boiling water is $100^{\circ} \mathrm{C}$.
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Verified Answer
The correct answer is:
$6.3 \mathrm{~min}$
Heat taken by water when its temperature changes from $20^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$.
$$
\begin{aligned}
H_{1} &=m c\left(\theta_{2}-\theta_{1}\right)=1000 \times 1 \times(100-20) \mathrm{cal} \\
&=1000 \times 80 \times 4.2 \mathrm{~J}
\end{aligned}
$$
Heat produced in time $t$ due to current in resistor
$H_{2}=V i t=220 \times 4 \times t \mathrm{~J}$
According to question,
$$
\begin{array}{l}
220 \times 4 \times t=1000 \times 80 \times 4.2 \\
\Rightarrow t=\frac{1000 \times 80 \times 4.2}{220 \times 4}=381.8 \mathrm{~s}=6.3 \mathrm{~min}
\end{array}
$$
$$
\begin{aligned}
H_{1} &=m c\left(\theta_{2}-\theta_{1}\right)=1000 \times 1 \times(100-20) \mathrm{cal} \\
&=1000 \times 80 \times 4.2 \mathrm{~J}
\end{aligned}
$$
Heat produced in time $t$ due to current in resistor
$H_{2}=V i t=220 \times 4 \times t \mathrm{~J}$
According to question,
$$
\begin{array}{l}
220 \times 4 \times t=1000 \times 80 \times 4.2 \\
\Rightarrow t=\frac{1000 \times 80 \times 4.2}{220 \times 4}=381.8 \mathrm{~s}=6.3 \mathrm{~min}
\end{array}
$$
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