Current Electricity Charging Discharging Of Capacitors Question 130
Question: Water of volume 2 litre in a container is heated with a coil of 1 kW at 27 oC. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27oC to 77oC
[Given specific heat of water is 4.2 kJ/kg] [IIT-JEE (Screening) 2005]
Options:
A) 8 min 20 s
B) 6 min 2 s
C) 7 min
D) 14 min
Show Answer
Answer:
Correct Answer: A
Solution:
Heat gained by water = Heat supplied by container - heat lost
Therefore mSDq = 1000t ? 160t
Therefore $ t=\frac{2\times 4.2\times 1000\times 50}{840} $ = 8 min 20 sec
BETA
