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Standard XII

Physics

Calorimetry

Two litre of ...

Question

Two litre of water at initialled temperature of $27^{o}$ C is heated by a heater of power $1$ kW. 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 $4.2$ kJ.kg).

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 B $8$ min $20$ sec
By law of conservation energy.
Heat input=heat output
$p t = m c θ +$ heat lost to surrender
$\Rightarrow 1000 \times t = 2 \times 4200 \times 50 + 160 t$
$840 t = 2 \times 4200 \times 50$
$t = 500 s e c$
$= 8 m i n 20 s e c$

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Two litre of water at initialled temperature of 27oC is heated by a heater of power 1 kW. 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 4.2 kJ.kg).

The correct option is B 8 min 20 secBy law of conservation energy.Heat input=heat outputpt=mcθ+ heat lost to surrender⇒1000×t=2×4200×50+160t840t=2×4200×50t=500 sec=8 min 20 sec

The correct option is B $8$ min $20$ sec
By law of conservation energy.
Heat input=heat output
$p t = m c θ +$ heat lost to surrender
$\Rightarrow 1000 \times t = 2 \times 4200 \times 50 + 160 t$
$840 t = 2 \times 4200 \times 50$
$t = 500 s e c$
$= 8 m i n 20 s e c$