Current Electricity Charging Discharging Of Capacitors Question 88
Question: An immersion heater is rated 836 watt. It should heat 1 litre of water from $ 10{}^\circ C $ to $ 40{}^\circ C $ in about
[AIEEE 2004]
Options:
A) 200 sec
B) 150 sec
C) 836 sec
D) 418 sec
Show Answer
Answer:
Correct Answer: B
Solution:
$ W=JH $
Therefore $ P\times t=J\times m,s,\Delta \theta $
Therefore $ t=\frac{J\times m\times s\Delta \theta }{P} $ (For water 1 litre = 1kg)
Therefore $ t=\frac{4.2\times 1\times 1000\times (40-10)}{836}=150,sec $ Short Trick: use formula $ t=\frac{4200\times m\times \Delta \theta }{P} $
BETA
