Water of volume 2 litre in a container is heated with a coil of `1kW at 27^@C.` The lid of the container is open and energy dissipates at rate of `160J//s.` In how much time temperature will rise from `27^@C to 77^@C` Given specific heat of water is
`[4.2kJ//kg`]

Options

A. 7 min
B. 6 min 2 s
C. 8 min 20 s
D. 14 min

Answer

C

Explanation

As shown in fig the net heat absorbed by the water to raise its temperature`=(1000-160)=840J//s` . Now, the heat required to raise the temperature of water from `27^(@)C` to `77^(@)C` is `Q=mcDeltaT=2xx4200xx50J` . Therefore the time required
`t=(Q)/(840)=(2xx4200xx50)/(840)500s`
`=8min20s`