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Q.

A kettle with 2L water at 27oC is heated by operating coil heater of power 1 kW. The heat is lost to the atmosphere at constant rate 160 J/s, when it is open. In how much time will water be heated to 77oC (sp. heat of water = 4.2kJ/kg) with the lid open?

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a

8 min 20s

b

6 min 20s

c

14 min

d

7 min

answer is A.

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Detailed Solution

By the law of conservation of energy, energy given by heater must be equal to the sum of energy gained by water and energy lost from the lid.        Pt=msΔθ+ energy lost i.e., 1000t=2×4.2×102×50+160tor     840t=8.4×103×50or     t=500s=8min20s
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