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Standard X

Physics

Specific Heat Capacity

40 g of water...

Question

40 g of water at 60 $^{0} C$ is poured into a vessel containing 50 g of water at 20 $^{0} C$ . The final temperature recorded is 30 $^{0} C$ . Calculate the thermal capacity of the vessel. (Take specific heat capacity of water as 4.2 $J g {- 1}^{0} C^{- 1}$ ).

500 J ⁰C^-1

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3500 J ⁰C^-1

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2942 J ⁰C^-1

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3870 J ⁰C^-1

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Solution

The correct option is C
2942 J ⁰C^-1

Let the thermal capacity of the vessel be C’.

According to the principle of mixture,

Heat given by hot water = Heat taken by cold water and vessel

$40 \times 4.2 J g^{- 1}^{0} C^{- 1} \times (60 - 30)^{0} C = 50 g \times 4.2 J g^{- 1}^{0} C^{- 1} \times (30 - 20)^{0} C + C^{'} \times (30 - 20)^{0} C$
$C^{'} = 2942 J^{0} C^{- 1}$

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40 g of water at 60°C is poured into a vessel containing 50 g of water at 20°C. The final temperature recorded is 30°C. The thermal capacity of the vessel is: [Take specific heat capacity of water as 4.2J g^{-1 o}C^-1]

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40 g of water at 600C is poured into a vessel containing 50 g of water at 200C. The final temperature recorded is 300C. Calculate the thermal capacity of the vessel. (Take specific heat capacity of water as 4.2 Jg−10C−1​).

The correct option is C 2942 J 0C-1 Let the thermal capacity of the vessel be C’. According to the principle of mixture, Heat given by hot water = Heat taken by cold water and vessel 40×4.2Jg−1 0C−1×(60−30)0C=50g×4.2Jg−1 0C−1×(30−20)0C+C′×(30−20)0C C′=2942 J 0C−1

40 g of water at 60 $^{0} C$ is poured into a vessel containing 50 g of water at 20 $^{0} C$ . The final temperature recorded is 30 $^{0} C$ . Calculate the thermal capacity of the vessel. (Take specific heat capacity of water as 4.2 $J g {- 1}^{0} C^{- 1}$ ).