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Standard XII

Physics

Calorimetry

Two litres of...

Question

Two litres of water (density =1 g/mL) in an open lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water
(Specific heat capacity $4.2 k J K^{- 1} k g^{- 1}$ ) from $20^{\circ} C$ to $75^{\circ} C$ is

340s

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550 s

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620 s

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760 s

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Solution

The correct option is B 550 s
Heat supplied = 1kW = 1000J/s
Heat lost = 160 J/s
Net heat supplied = 1000 - 160 = 840J/s
Heat required = mst
$= 2 k g \times 1200 J / k g K \times 55^{\circ} C$
$= 462000 J$
$∴$ Time taken $= \frac{462000}{840} = 550 s$

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Two litres of water (density =1 g/mL) in an open lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water (Specific heat capacity 4.2kJK−1kg−1) from 20∘ C to 75∘ C is

The correct option is B 550 sHeat supplied = 1kW = 1000J/s Heat lost = 160 J/s Net heat supplied = 1000 - 160 = 840J/s Heat required = mst =2kg×1200J/kg K×55∘C =462000J ∴ Time taken =462000840=550 s

Two litres of water (density =1 g/mL) in an open lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water
(Specific heat capacity $4.2 k J K^{- 1} k g^{- 1}$ ) from $20^{\circ} C$ to $75^{\circ} C$ is

The correct option is B 550 s
Heat supplied = 1kW = 1000J/s
Heat lost = 160 J/s
Net heat supplied = 1000 - 160 = 840J/s
Heat required = mst
$= 2 k g \times 1200 J / k g K \times 55^{\circ} C$
$= 462000 J$
$∴$ Time taken $= \frac{462000}{840} = 550 s$