Q. An electric heater of 1.08 Kw is immersed in water. After the water has reached a temperature of 100$℃$, how much time will be required to produce 100 g of steam?
A) 420 s
B) 210 s
C) 105 s
D) 50 s
Explanation
(b) Heat produced by heater per second $= 1.08 \times 10^{3}$J
Heat taken by water to form steam $mL$
$= 100 \times 540$ cal
$\ = 100 \times 540 \times 4.2$ J
$\therefore 1.08 \times 10^{3} \times t = 100 \times 540 \times 4.2$
or $t = \frac{100 \times 540 \times 4.2}{1.08 \times 10^{3}}$ = 210 s
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.