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Standard X

Physics

Calorimeter

Water of volu...

Question

Water of volume $2 L$ in a container is heated with a coil of $1 k W$ at $27^{o}$ C. The lid of the container is open and energy dissipates at rate of $160$ J/s. In how much time temperature will rise from $27^{o}$ C to $77^{o}$ C?[Given specific heat of water is $4.2$ kJ/kg]

8 min 20 s

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6 min 2 s

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7 min

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14 min

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Solution

The correct option is B 8 min 20 s
$Q_{N E T} = Q_{S U P P L I E D} - Q_{R E J E C T E D} = 1000 - 160 = 840 J$

$D e n s i t y = \frac{M a s s}{V o l u m e}$

$M a s s = 2 K G$

We also know that $Q = m \times S \times (Δ T)$ and $Q = E \times t$
So, time taken will be:

$t = \frac{m \times S \times (Δ T)}{E}$

$t = \frac{2 \times 4200 \times 50}{840} = 500 s e c o n d$

Or we can say $1 m i n u t e = 60 s e c o n d$

So in $500 s e c o n d = 8.33 m i n u t e = 8 m i n u t e a n d 0.333 \times 60 s e c o n d = 8 m i n u t e a n d 20 s e c o n d$

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Water of volume 2L in a container is heated with a coil of 1kW at 27oC. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from 27oC to 77oC?[Given specific heat of water is 4.2kJ/kg]

Water of volume $2 L$ in a container is heated with a coil of $1 k W$ at $27^{o}$ C. The lid of the container is open and energy dissipates at rate of $160$ J/s. In how much time temperature will rise from $27^{o}$ C to $77^{o}$ C?[Given specific heat of water is $4.2$ kJ/kg]