Question

An immersion heater is rated 418 watt. It should heat a liter of water from 10 deg. Celcious to 30 deg. Celcious in nearly

Answer 1

Answer:

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}g

−1

c^{-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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