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 rise at 80-C- The mass of the stem condensed in kilogram is

Heat lost by steam = Heat gained by water + calorimeter ∴ mL+ms(100 80)=1.12 × s × (80 15) or m[540+(1 × 20)]=1.12 × 1 × 65 or m=1.12 × 1 × 65/560=65/500 kg ...

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Standard XIII

Physics

Calorimetry

Steam at 100∘...

Question

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

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 A 0.130
Heat lost by steam = Heat gained by water + calorimeter
$∴ m L + m s (100 - 80) = 1.12 \times s \times (80 - 15)$
or $m [540 + (1 \times 20)] = 1.12 \times 1 \times 65$
or $m = \frac{1.12 \times 1 \times 65}{560} = \frac{65}{500} k g o r m = 0.13 k g$

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(IIT JEE 1986)

Steam at 100⁰ C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15⁰ C till the temperature of the calorimeter and its contents rise to 80⁰ C. The mass of the steam condensed in kilogram is (latent heat of vaporization of water is 540 cal/gm)

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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 rise at 80∘C. The mass of the stem condensed in kilogram is

The correct option is A 0.130Heat lost by steam = Heat gained by water + calorimeter ∴ mL+ms(100−80)=1.12×s×(80−15) or m[540+(1×20)]=1.12×1×65 or m=1.12×1×65560=65500 kg or m=0.13 kg

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

The correct option is A 0.130
Heat lost by steam = Heat gained by water + calorimeter
$∴ m L + m s (100 - 80) = 1.12 \times s \times (80 - 15)$
or $m [540 + (1 \times 20)] = 1.12 \times 1 \times 65$
or $m = \frac{1.12 \times 1 \times 65}{560} = \frac{65}{500} k g o r m = 0.13 k g$