Q115 / 200 · NEET 2008 (Screening) · Physics

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 $100^{\circ} \mathrm{C}$

  • Correct answer
    A
    $6.3 \mathrm{~min}$
  • B
    $8.4 \mathrm{~min}$
  • C
    $12.6 \mathrm{~min}$
  • D
    $4.2 \mathrm{~min}$

Solution

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 $\begin{gathered} I=4 \mathrm{~A}, V=220 \text { volt }, \quad m=1 \mathrm{~kg}, \\ \Delta t=(100-20)^{\circ} \mathrm{C}, s=4200 \mathrm{~J} / \mathrm{kg}^{\circ} \mathrm{C} \\ \therefore \quad t=\frac{1 \times 4200 \times 80}{220 \times 4}=6.3 \mathrm{~min} \end{gathered}$

Hence, the correct answer is option: 'A'.

Relevant Questions