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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
We know that heat required by water = heat taken by heatermsdelta T = W/tt = W / msdelta Tt = 836 / [1*4.186*10^-3*(40-10)]t = 150 sec Thanks & Regards,Nirmal SinghAskiitians Faculty
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
/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.
Dear student,Please find the attached solution to your problem.We know that heat required by water = heat taken by heatermsΔT = W/tt = W / msΔTt = 836 / [1*4.186*10^-3*(40-10)]t = 150 sec Thanks and regards,Kushagra
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