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Standard XII

Physics

Heat Capacity

Steam is pass...

Question

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

24.8 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 A 24.8 gm
$Solution :$
$m L_{S} + m s_{ω} (100 - 90) = 22 S_{ω} (90 - 20)$
$⟹ m \times 540 + m \times 1 \times 10 = 22 \times 70$
$⟹ m = \frac{22 \times 70}{550} = 2.8 g$
Total mass of water =22+2.8=24.8g

$Hence A is the correct option$

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Similar questions

Q. 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

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 $80^{\circ} C .$

What will be the mass of the steam condensed (in kg)?

Take:

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

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

Q. Steam at

100 ° C

is passed into

20 g

of water at

10 ° C

. When water acquires a temperature of

80 ° C

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

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

and latent heat of steam

= 540 {cal g}^{- 1}

]

Determine the final result when 200 gm of water and 20 gm of ice at $0^{\circ} C$ are in a calorimeter having a water equivalent of 30 gm. 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

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 20oC. 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

The correct option is A 24.8 gmSolution:mLS+msω(100−90)=22Sω(90−20)⟹m×540+m×1×10=22×70⟹m=22×70550=2.8g Total mass of water =22+2.8=24.8gHence A is the correct option

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

The correct option is A 24.8 gm
$Solution :$
$m L_{S} + m s_{ω} (100 - 90) = 22 S_{ω} (90 - 20)$
$⟹ m \times 540 + m \times 1 \times 10 = 22 \times 70$
$⟹ m = \frac{22 \times 70}{550} = 2.8 g$
Total mass of water =22+2.8=24.8g

$Hence A is the correct option$