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 s
B. 6 min 2 s
C. 7 min
D. 14 min

Answer

A

Explanation

1 kW = 1000 J/s
In one second net heat absorbed by water = 1000 J/s - 160 J/s = 840 J/s
In t seconds total heat absorbed by water = Q = (840 x t) J
Total heat required to raise the temperature of water (volume 2L) from `27^@C` to `77^@C`
`Q = M_(water) xx C xx Delta T [C = 4.2 xx 10^3 J//kg ]`
` = 2 xx 4.2 xx 10^3 xx 50`
[ mass = `1000 xx 2 xx 10^(-3) = 2 kg ` ]
`849 xx t = 2 xx 10^3 xx 4.2 xx 50`
` t = (2 xx 10^3 xx 4.2 xx 50)/(840)`
= 500 s = 8min 20 s