Question

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

Answer 1

Correct Answer - A
Heat gained by the water = (Heat supplied by the coil) - (Heat dissipated to enviroment)
`rArr mc Delta theta = P_("Coil") t - P_("Loss")t`
`rArr 2 xx 4.2 xx 10^(3) xx (77-27) = 1000t = 160t`
`rArr t = (4.2xx10^(5))/(840) = 500 sec = 8 min 20 sec`.