Q. Time taken by a 836 W heater to heat one litre of water from $10℃\ to\ 40℃$ is

A) 50 s
B) 100 s
C) 150 s
D) 200 s
Explanation

(c) Let time taken in boiling the water by the heater is $t\ s.$ Then
$Q = ms\mathrm{\Delta}T\ \Rightarrow \frac{Pt}{J} = ms\mathrm{\Delta}T$
$\therefore\frac{836}{4.2}\ t = 1 \times 1000(40 - 10)$
$\ \ \ \ \ \frac{836}{4.2}t = 1000 \times 30$
$\ \ \ \ \ \ \ \ \ \ \ \ \ t = \frac{1000 \times 30 \times 4.2}{836} = 150s$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.