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 27°C to 77°C?<p></p>

Heat energy required to raise the temperature of water to 77°C:
Qw = 2 × 4.2 × (77 - 27) × 1000 = 420000 J
Heat lost for time t is QL = 160t
In time t, total energy supplied is 1000t. Therefore,
1000t = Qw + QL
Or, 1000t = 420000 + 160t
Or, 840t = 420000
Or, t = 420000 / 840 = 500 s = 8 min 20 s

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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 27°C to 77°C?

Solution: