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The heat of fusion of water is 79.9cal/g, the heat of vaporization of water is 540cal/g, and the specific heat of water is 1.00cal/deg/g. How much ice at 0 celsius could be converted to steam at 100 celsius by 9522cal of heat?

Sagot :

The amount of ice converted into steam is 13.23 gram.

What is Heat of Vaporisation ?

Heat of Vaporisation is the amount of heat required to convert 1 gm to liquid at its boiling point to vapour.

Let x gram of ice is there

At first x gm of ice will melt

The heat required is x * Heat of fusion = x * 79.9 = 79.9x

Then the heat required to rise the temperature of x gm of water from 0 degree to 100 degree Celsius .

= mc ΔT

= x * 1 * 100 = 100x

The heat required to convert x gm of liquid into x gm of steam

= x * 540 = 540x

The total heat supplied is 9522 cal

79.9x + 100x + 540x = 9522

719.9 x = 9522

x = 13.23 gm

The amount of ice converted into steam is 13.23 gram.

To know more about Heat of Vaporisation.

https://brainly.com/question/9967928

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