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. 8 min 20 sec
B. 6 min 2 sec
C. 7 min
D. 14 min

Answer

A

Explanation

Energy gained by water (in 1 s) = Energy supplied-energy lost = (1000 J -160 J) = 840 J
Total heat required to raise the temperature of water from 27 DC to 77 °C is ms `Deltatheta`. Hence, the required time,
`t = (ms Deltatheta)/("rate by which energy is gained by water")`
` = ((2)(4.2 xx 10^(3))(50))/840 = 500 s`
= 8 min 20 s
Hence, the time taken for temperature rise is 8 min and 20 sec.