Question

An electric kettle takes \(4\,A\) 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

Hint:

Always use \(P = VI\) for electrical power, and \(Q=mc\Delta T\) for heating water. Then time \(t=\dfrac{Q}{P}\).

Answer 1

The Correct Option is C

Step 1: Find electrical power supplied.
\[ P = VI = 220 \times 4 = 880\,W \] Step 2: Find heat required to raise water temperature.
Heat needed:
\[ Q = mc\Delta T \] where
\(m = 1\,kg\), \(c = 4200\,Jkg^{-1}K^{-1}\),
\[ \Delta T = 100-20 = 80^\circ C \] So,
\[ Q = 1 \times 4200 \times 80 = 336000\,J \] Step 3: Use relation \(Q = Pt\).
\[ t = \frac{Q}{P} = \frac{336000}{880} \approx 381.8\,s \] Step 4: Convert seconds to minutes.
\[ t = \frac{381.8}{60} \approx 6.36\,min \approx 6.3\,min \] Final Answer: \[ \boxed{6.3\ \text{min}} \]