In a calorimeter of water equivalent $20 g$ , water of mass $1.1 kg$ is taken at $288 K$ temperature. If steam at temperature $373 K$ is passed through it and temperature of water increases by $6.5^{\circ} C$ , then the mass of steam condensed is
[Take, $s = 1 cal /^{\circ} C g, L = 540 cal/g$ ]

17.5 g

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11.7 g

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15.7 g

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18.2 g

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Solution

The correct option is B $11.7 g$
The water equivalent of calorimeter is $20 g$ .

Initially, the temperature of water of mass $1.1 kg$ is,

$T_{i} = 288 K = 288 - 273 = 15^{\circ} C$

Let the mass of steam that get condensed, when it passes through water at $15^{\circ} C$ be $x g$ .

Heat released by steam,

$Q_{r e l e a s e d} = m L + m s Δ T$

$Q_{r e l e a s e d} = (x \times 540) + [x \times 1 \times (100 - 15 - 6.5)]$

$Q_{r e l e a s e d} = 540 x + 78.5 x . . . . . . (1)$

The heat released by the steam goes to the water-calorimeter system.

Heat gained by water,

$Q_{W} = m s Δ T = 1100 \times 1 \times 6.5$

$\Rightarrow Q_{W} = 7150 cal$

Similarly, heat gained by calorimeter,

$Q_{C} = m s Δ T = 20 \times 1 \times 6.5$

$\Rightarrow Q_{C} = 130 cal$

From energy conservation,

$Q_{r e l e a s e d} = Q_{a b s o r b e d}$

$\Rightarrow 540 x + 78.5 x = 7150 + 130$

$618.5 x = 7280$

$∴ x = \frac{7280}{618.5} = 11.77 g$

 Hence, $(C)$ is the correct answer.

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Similar questions

Q. In a calorimeter of water equivalent 20 g, water of mass 1.1 kg is taken at 288 K temperature 373 K is passed through it and temperature of water increases by

{6.5}^{\circ} C

then the mass of steam condensed is

Q. A steam at

100^{o} C

is passed into

1 k g

of water contained in a calorimeter of water equivalent

0.2 k g

9^{o} C

, till the temperature of the calorimeter and water in it is increased to

90^{o} C

. The mass of steam condensed in

k g

is nearly:

[sp. heat of water

= 1

c a l / g^{o}

C, latent heat of vaporisation

= 540 c a l / g

Steam at $100^{\circ} C$ is passed into $1.1 k g$ of water contained in a calorimeter of water equivalent $0.02 k g$ at $15^{\circ} C$ till the temperature of the calorimeter and its contents reaches $80^{\circ} C$ . Considering the latent heat of steam $= 540 c a l g^{- 1}$ $and s$ pecific heat capacity of water $= 4183 J k g^{- 1} K^{- 1}$ ${or}$ $= 1 c a l k g^{- 1} K^{- 1}$ ,
calculate the mass of the steam condensed.

Steam at $100^{o} C$ 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 $80^{o} C .$ The mass of the steam condensed in kg is
(Latent heat of vapor = $540 c a l / g$ )

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 $= 540 c a l g m^{- 1}$

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In a calorimeter of water equivalent 20 g, water of mass 1.1 kg is taken at 288 K temperature. If steam at temperature 373 K is passed through it and temperature of water increases by 6.5 ∘C, then the mass of steam condensed is [Take, s=1 cal/ ∘C g, L=540 cal/g ]

In a calorimeter of water equivalent $20 g$ , water of mass $1.1 kg$ is taken at $288 K$ temperature. If steam at temperature $373 K$ is passed through it and temperature of water increases by $6.5^{\circ} C$ , then the mass of steam condensed is
[Take, $s = 1 cal /^{\circ} C g, L = 540 cal/g$ ]