Steam at 100- C is passed into 1-1 kg of water contained in a calorimeter of water equivalent 0-02 kg at 15- C till the temperature of the calorimeter and its contents rises to 80- C- The mass of the steam condensed in kg isA- 0-130B- 0-260C- 0-135D- 0-065

The correct option is C 0.135 Heat required Q = (1.1 + 0.02) × 10^3 × 1 × (80 15) = 72800 cal. Therefore, mass of steam condensed (in kg) m = Q/L = 72800 ...

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Standard X

Physics

Latent Heat

Steam at 100∘...

Question

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

0.130

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0.065

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0.260

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0.135

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Solution

The correct option is D
0.135

Heat required
$Q = (1.1 + 0.02) \times 10^{3} \times 1 \times (80 - 15) = 72800 c a l .$
Therefore, mass of steam condensed (in kg)
$m = \frac{Q}{L} = \frac{72800}{540} \times 10^{- 3} = 0.135 k g .$

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Similar questions

Q. 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^{\circ} C

till the temperature of the calorimeter and its contents rise at

80^{\circ} C

. The mass of the stem condensed in kilogram is

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 = $540 c a l / g$ )

latent heat of vaporization of water = $540 c a l k g^{-} 1$ , specific heat capacity of water = $10^{3} c a l k g^{-} 1^{\circ} C^{-} 1$

Specific heat of water = $1 k c a l / k g^{\circ} C$ , Latent heat of Vaporisation = $540 k c a l / k g$

Q. Some 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

15^{\circ} C

so that the temperature of the calorimeter and its contents rises to

80^{\circ} C

. What is the mass of steam condensing (in

k g

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Steam at 100oC is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15oC till the temperature of the calorimeter and its contents rises to 80oC. The mass of the steam condensed in kg is

The correct option is D 0.135 Heat required Q=(1.1+0.02) × 103 × 1 × (80−15)=72800 cal. Therefore, mass of steam condensed (in kg) m=QL=72800540 × 10−3=0.135kg.

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

The correct option is D
0.135

Heat required
$Q = (1.1 + 0.02) \times 10^{3} \times 1 \times (80 - 15) = 72800 c a l .$
Therefore, mass of steam condensed (in kg)
$m = \frac{Q}{L} = \frac{72800}{540} \times 10^{- 3} = 0.135 k g .$