Q. What must be the efficiency of an electric kettle marked 500 W, 230 V, if it was found to bring 1 kg of water at 15$℃$ to boiling point in 15 min? (Given specific heat capacity of water $= 420\ $J/kg℃)
A) 79%
B) 81%
C) 72%
D) 69%
Explanation
(a) Heat produced $H = Vit = Pt$ J
Where, $P = Vi$ watt
$\therefore H = 500 \times 15 \times 60 = 45 \times 10^{4}$ J
Heat absorbed by water $= mass \times specific\ heaat\ capacity\ $
$\ \times rise\ in\ temperature$
$\ = 1 \times 4200 \times (100 - 15)$
$\ = 4.2 \times 85 \times 10^{- 3}$
Efficiency $= \frac{heat\ absorbed}{heat\ produced} \times 100$
$= \frac{85 \times 4.2 \times 10^{3}}{45 \times 10^{4}} \times 100 = 79\%$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.