A calorimeter of water equivalent 15g contains 165g of water at 25-C- Steam at 100-C is passed through the water for some time- The temperature is increased to 30-C and the mass of the calorimeter and its contents is increased by 1-5 g- Calculate the latent heat of vaporization of water- Heat capacity of water is 1 cal-g-C

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Byju's Answer

Standard XIII

Physics

Calorimetry

A calorimeter...

Question

A calorimeter of water equivalent 15g contains 165g of water at $25^{\circ} C$ . Steam at $100^{\circ} C$ is passed through the water for some time. The temperature is increased to $30^{\circ} C$ and the mass of the calorimeter and its contents is increased by 1.5 g. Calculate the latent heat of vaporization of water. Heat capacity of water is $1 c a l / g^{\circ} C$

630 cal/g

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530 cal/g

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1530 cal/g

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1500 cal/g

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Solution

The correct option is B 530 cal/g
Let L be the specific latent heat of vaporization of water. The mass of the steam condensed is 1.5g. Heat lost in condensation of steam is
$Q_{1} = (1.5 g) L$
The condensed water cools from $100^{\circ} C$ to $300^{\circ} C$ . Heat lost in this process is -
$Q_{2} = (1.5 g) (1 c a l / g^{\circ} C) (70^{\circ} C) = 105 c a l$
Heat supplied to the calorimeter and to the cold water during the rise in temperature from $25^{\circ} C$ to $30^{\circ} C$ is -
$Q_{3} = (15 g + 165 g) (1 c a l / g^{\circ} C) (5^{\circ} C) = 900 c a l$
If no heat is lost to the surrounding,
$(1.5 g) L + 105 c a l = 900 c a l$
or $L = 530 c a l / g$

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