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Standard X

Physics

Specific Heat Capacity

Steam is pass...

Question

Steam is passed into $22$ gm of water at $20^{\circ}$ C. The mass of water that will be present when the water acquires a temperature of $90^{\circ} C$ (Latent heat of steam is $540$ cal/g) is

24.83 gm

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24 gm

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36.6 gm

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30 gm

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Solution

The correct option is B 24.83 gm
Let the mass of steam condensed be $m$ .
The heat released by steam is sum of heat released while condensing form stem to ice and heat released while temperature is dropped from $100^{o} C$ to $90^{o} C$ .
So $Q_{1} = m \times L + m \times (100 - 90) \times 1 = 540 m + 10 m = 550 m$
Heat absorbed by $22 g$ of water to increase its temperature from $20^{o} C$ to $90^{o} C$ is $Q_{2} = 22 \times (90 - 20) \times 1 = 1540 c a l$
Equating the two heats, $m = 1540 / 550 = 2.8 g$
So total mass is $22 + 2.8 = 24.8 g$

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Similar questions

Q. Steam is passed into 22 gm of water at

20^{o} C

. The mass of water that will be present when the water acquires a temperature of 90^oC (latent heat of steam is 540 cal/gm) is

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is passed into

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10 ° C

. When water acquires a temperature of

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, the mass of water present will be: [Take specific heat of water

= 1 {cal g}^{- 1} C^{- 1}

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= 540 {cal g}^{- 1}

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Steam at $100^{\circ} C$ is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at $15^{o} C$ till the temperature of the calorimeter and its contents rises to $80^{\circ} C .$

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Take:

Specific heat of water $= 1 c a l g m^{- 1}$

Latent heat of vaporization $= 540 c a l g m^{- 1}$

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Determine the final result when 200 gm of water and 20 gm at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm and 50 gm of steam is passed into it at $100^{\circ} C$ . Take latent heat of fusion 80 cal/gm and latent heat of vaporization as 540 cal/gm

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Steam is passed into 22 gm of water at 20∘C. The mass of water that will be present when the water acquires a temperature of 90∘C (Latent heat of steam is 540 cal/g) is

Steam is passed into $22$ gm of water at $20^{\circ}$ C. The mass of water that will be present when the water acquires a temperature of $90^{\circ} C$ (Latent heat of steam is $540$ cal/g) is