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Standard XI

Physics

Calorimetry

Two litres of...

Question

Two litres of water at initial temperature of $27^{\circ} C$ is heated by a heater of power $1 kW$ in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of $160 J/s$ . The time in which the temperature will rise from $27^{\circ} C$ to $77^{\circ} C$ is (specific heat of water $= 4.2 kJ/kg$ ).

5 min 20 sec.

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8 min 20 sec.

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10 min 40 sec.

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8 min 50 sec.

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Solution

The correct option is B $8 min 20 sec.$
From principle of calorimetry,
Heat gained by water = Heat supplied - Heat loss
$m s Δ θ = 1000 t - 160 t \Rightarrow t = \frac{2 \times 4200 \times 50}{840} = 8 min. 20 sec.$

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Two litres of water at initial temperature of 27∘C is heated by a heater of power 1 kW in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of 160 J/s. The time in which the temperature will rise from 27∘C to 77∘C is (specific heat of water =4.2 kJ/kg).

The correct option is B 8 min 20 sec.From principle of calorimetry, Heat gained by water = Heat supplied - Heat loss msΔθ=1000t−160t⇒t=2×4200×50840=8 min.20 sec.

The correct option is B $8 min 20 sec.$
From principle of calorimetry,
Heat gained by water = Heat supplied - Heat loss
$m s Δ θ = 1000 t - 160 t \Rightarrow t = \frac{2 \times 4200 \times 50}{840} = 8 min. 20 sec.$