Water of volume 2 L in a container is heated with a coil of 1 kW at 27$℃$ . The lid of the container is open and energy dissipates at rate of 160$\ {Js}^{- 1}$. In how much time temperature will rise from 27$℃$ to 77$℃$ [Given specific heat of water is 4.2 kJ ${kg}^{- 1}$]

Question

Dr. Abubacker · Accepted Answer

(a) 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$℃\ to\ 77℃\ is\ ms\mathrm{\Delta}	heta$. Hence, the required time,
$t = \frac{ms\mathrm{\Delta}	heta}{rate\ by\ which\ energy\ is\ gained\ by\ water}$
$= \ \frac{(2)\left( 4.2 	imes 10^{3} ight)(50)}{840} = 500\ s$
= 8 min 20 s.

Q. Water of volume 2 L in a container is heated with a coil of 1 kW at 27$℃$ . The lid of the container is open and energy dissipates at rate of 160$\ {Js}^{- 1}$. In how much time temperature will rise from 27$℃$ to 77$℃$ [Given specific heat of water is 4.2 kJ ${kg}^{- 1}$]

Explanation

Reviewed by Dr. Abubacker

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