Question

time taken by a 836 W heater to heat one litre of water from 10 degree Celsius to 40 degree Celsius is

Answer 1

hello friend...!!!

according to the given question we should find the time taken 

it is given that : 

⇒ P = 836W

⇒mass of water = 1L = 1000g

⇒temperature rise = ( Ф1 - Ф2 ) = ( 40 - 10 ) = 30°

⇒specific heat of water is C = 1 cal $g^{-1}$$c^{-1}$

now,

heat produced by the heater in time t is = P x T 

= 836 t joules ( we should convert to calories )

= 836 t / 4.2 calories---------------------(1)

now, heat taken by the water = mc(Ф1 - Ф2)

= 1000 x 1 x 30

=30000-----------(2)

we know that ,

heat produced by the heater in time t = heat taken by the water 

implies,

836 t / 4.2 = 30000

implies,

t = (30000 x 4.2) / 836

t = 126000 / 836

t = 150.7 seconds .

therefore the time taken is 150.7 seconds.

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hope it helps...!!

Answer 2

Explanation:

We know that heat required by water = heat taken by heater

msdelta T = W/t

t = W / msdelta T

t = 836 / [1*4.186*10^-3*(40-10)]

t = 150 sec