Time taken by a 836 W heater to heat one liter of water from 10ºC to 40ºC is:
Text Solution
Verified by ExpertsThe correct answer is:
C
Let time taken in boiling the water by the heater is t sec. Then
Q = ms Δ T ⇒ 
= ms Δ T
⇒ 
t = 1 × 1000 (40º – 10º)
t = 1000 × 30
⇒ t = 
= 150 second
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