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Question

Water of volume 2L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27oC to 77oC? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible)

A
8 min 20 s
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B
6 min 2 s
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C
7 min
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D
14 min
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Solution

The correct option is A 8 min 20 s
Heat energy required to raise the temp of water to 77oC
Qw=2×4.2×(7727)×1000=420000J
Heat lost for time t is
QL=160t

In time t, total energy supplied is 1000t. Therefore,
1000t=Qw+QL
Or,
t=420000840=500 s=8min 20 s

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