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Time taken by a 836 W heater to heat one litre of water from 10dgreeC to 40dgreeC is (A) 50 s (B) 100 s (C) 150 s (D) 200 s

Saumya , 11 Years ago
Grade
anser 4 Answers
Nirmal Singh.

Last Activity: 11 Years ago

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
Thanks & Regards,
Nirmal Singh
Askiitians Faculty

Jayant

Last Activity: 9 Years ago

heat required by water = heat emitted by heater
i.e ms/\T = power × time (since, heat = power × time)
=> time = ms/\T / power
=> time = {1000× 4.186 × (40-10)} / 836 (1 litre = 1kg = 1000 g)
=> time = 150 sec.
 method used in the previous post is wrong 

KanishK Kumar

Last Activity: 7 Years ago

/mass of water,m = volume × density
1000 ×1 = 1000g
Heat taken by water = heat produced by heater
hence, ms(02-01)=pt/j or
t = ms(02-01)j/p
t = 1000×1×(40-10)×4.18/836=150s.

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,
Please find the attached solution to your problem.
We know that heat required by water = heat taken by heater
msΔT = W/t
t = W / msΔT
t = 836 / [1*4.186*10^-3*(40-10)]
t = 150 sec
 
Thanks and regards,
Kushagra

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