An electric kettle with negligible heat capacity is rated at \(2000 W\), if \(2.0 kg\) of water is put in it, how long will take the temperature of water to rise from \(20^{\circ} C\) to \(100^{\circ} C\) ? [Specific heat capacity of water \(\left.=4200 Jkg ^{-1} K ^{-1}\right]\)
A. \(336 s\) B. \(240 s\) C. \(168 s\) D. \(84 s\)
Correct Answer: A
Explanation
Given, power \(=200 W\), Mass of water \(=2 kg\) \(\Delta \theta=100^{\circ} C -20^{\circ} C =80^{\circ} C\) \(c=4200 Jkg ^{-1} K ^{-1}\) Required time? \(Q=m c \Delta \theta\) \(=2 \times 4200 \times 80,=672000 J\) Power \(=\frac{\text { energy }}{\text { time }} ;\) Time \(=\frac{\text { energy }}{\text { power }}\) \(=\frac{672000}{2000}=336 s\)
