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Standard XII

Physics

Conduction Law

Two litre wat...

Question

Two litre water at initial temperature of $27^{o} C$ is heated by a heater of power $1$ $k W$ . If the lid of kettle is opened, then heat is lost at the constant rate of $160$ $J / s$ . Find the time required to raise the temperature of water to $77^{o} C$ with the lid open :
(Specific heat of water $4.2$ $k J$ / $k g$ )

5

min 40 sec

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14

min 20 sec

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8

min 20 sec

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16

min 10 sec

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Solution

The correct option is C $8$ $min 20 sec$
Effective rate of heat gain is
$Q = 1000 - 160 = 840 J / s$
Thus, to raise the temperature,
$840 t = (2) (4.2) \times 1000 (77 - 27)$
$t = 500 s = 8 m i n 20 s e c$

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Two litre water at initial temperature of 27oC is heated by a heater of power 1kW. If the lid of kettle is opened, then heat is lost at the constant rate of 160 J/s. Find the time required to raise the temperature of water to 77oC with the lid open :(Specific heat of water 4.2kJ/kg)

The correct option is C 8 min20secEffective rate of heat gain isQ=1000−160=840J/sThus, to raise the temperature,840t=(2)(4.2)×1000(77−27)t=500s=8min 20sec

The correct option is C $8$ $min 20 sec$
Effective rate of heat gain is
$Q = 1000 - 160 = 840 J / s$
Thus, to raise the temperature,
$840 t = (2) (4.2) \times 1000 (77 - 27)$
$t = 500 s = 8 m i n 20 s e c$