Question

An electric kettle takes 4 A current at $220\, V$. How much time will it take to boil $1\, kg$ of water from temperature $20^{\circ} C ?$ The temperature of boiling water is $100^{\circ} C$.

• A. 12.6 min
• B. 4.2 min
• C. 6.3 min
• D. 8.4 min

Answer 1

Answer: (C)

Heat taken by water when its temperature changes from $20^{\circ} C$ to $100^{\circ} C$ is
$H_{1}=m s\left(\theta_{2}-\theta_{1}\right)=1000 \times 1 \times(100-20)$ cal
$=1000 \times 80 \times 4.2\, J$
Heat produced in time $t$ due to current in resistor is
$H_{2}=VI t=220 \times 4 t\, J$
As per question, $220 \times 4 \times t=1000 \times 80 \times 4.2$
or $t=\frac{1000 \times 80 \times 4.2}{220 \times 4}=382\, s =6.3$ min