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Question

A steam at 100oC is passed into 1 kg of water contained in a calorimeter of water equivalent 0.2 kg at 9oC, till the temperature of the calorimeter and water in it is increased to 90oC. The mass of steam condensed in kg is nearly:

[sp. heat of water=1 cal/goC, latent heat of vaporisation =540 cal/g

A
0.81
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B
0.18
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C
0.27
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D
0.54
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Solution

The correct option is B 0.18
Initial temperature of calorimeter and water Ti=9oC
Final temperature of calorimeter and water Tf=90oC
change in temperature: ΔT=909=81oC
Latent heat of vaporisation: Lv=540 cal/g
Mass of water present: mw=1kg=1000 g
Water equivalent of calorimeter is 0.2kg=200 g
mcSc=200×Sw=200×1=200

Let the mass of steam condensed be ms g
msLv=(mwSw+mcSc)ΔT
OR ms×540=(1000×1+200)×81 ms=180 g
Thus mass of steam condensed is 0.18 kg

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