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An electric kettle takes $4 \mathrm{~A}$ 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:
$6.3 \mathrm{~min}$
Key Idea : Heat eyolved due to Joule's effect is used up in boiling water.
As per key idea
$\begin{aligned}
V I t & =m s \Delta t \\
\text{or } t & =\frac{m s \Delta t}{V}
\end{aligned}$
Putting under given values
$\begin{gathered}
I=4 \mathrm{~A}, V=220 \text { volt }, \quad m=1 \mathrm{~kg}, \\
\Delta t=(100-20)^{\circ} \mathrm{C}, s=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \\
\therefore \quad t=\frac{1 \times 4200 \times 80}{220 \times 4}=6.3 \mathrm{~min}
\end{gathered}$
As per key idea
$\begin{aligned}
V I t & =m s \Delta t \\
\text{or } t & =\frac{m s \Delta t}{V}
\end{aligned}$
Putting under given values
$\begin{gathered}
I=4 \mathrm{~A}, V=220 \text { volt }, \quad m=1 \mathrm{~kg}, \\
\Delta t=(100-20)^{\circ} \mathrm{C}, s=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \\
\therefore \quad t=\frac{1 \times 4200 \times 80}{220 \times 4}=6.3 \mathrm{~min}
\end{gathered}$
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