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

Answer

C

Explanation

(c) As shown in the figure, the net
heat absorbed by the water to
raise its temperature
`=(1000-160)=840 J//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=500 sec=8 min 20 sec`