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Electric Current

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

Profile image of Saumya
12 Years agoGrade
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4 Answers

Profile image of Nirmal Singh.
12 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
Profile image of Jayant
10 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 
Profile image of KanishK Kumar
8 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.
Profile image of Kushagra Madhukar
6 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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