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Interconversion of State

Water of volu...

Question

Water of volume $2 L$ in a closed container is heated with a coil of $1 k W$ . 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 $27^{o} C$ to $77^{o} C$ ? (Specific heat of water is $4.2 k J / k g$ and that of the container is negligible)

8 m i n 20 s

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6 m i n 2 s

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7 m i n

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14 m i n

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Solution

The correct option is A $8 m i n 20 s$
Heat energy required to raise the temp of water to $77^{o} C$
$Q_{w} = 2 \times 4.2 \times (77 - 27) \times 1000 = 420000 J$
Heat lost for time t is
$Q_{L} = 160 t$

In time t, total energy supplied is 1000t. Therefore,
$1000 t = Q_{w} + Q_{L}$
Or,
$t = \frac{420000}{840} = 500 s = 8 m i n 20 s$

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2 L

1 k W

27^{o}

160

27^{o}

77^{o}

4.2

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\frac{J}{s}

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1 k W

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77^{o} C

4.2 k J / k g

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