Heat of Fusion
المؤلف:
GEORGE A. HOADLEY
المصدر:
ESSENTIALS OF PHYSICS
الجزء والصفحة:
p-283
2025-11-25
38
If heat is applied to a beaker of crushed ice, it will be noticed that while the ice is being melted the temperature of the resulting water is the same as that of the ice, i.e., zero. The effect of the heat is not to change the temperature, but to change the physical state from solid to liquid.
The number of heat units required to melt a unit mass of a substance without raising its temperature is called the heat of fusion of the substance. On solidification, the same amount of heat is given out.
Demonstration. - Pour 500 g. of water at 80° C. into a glass beaker weighing 100 g., and into this put 150 g. of cracked ice as dry as possible. Stir the ice until it is melted and take the resulting temperature of the water. It will be found to be about 43.8° С. The heat of fusion of ice can now be calculated as follows: The heat given out by the 500 g. of water and by the 100 g. of glass in losing 36.2° of temperature is used in melting the ice and raising the resulting water to 43.8°.
Heat given out by the water = 500 × 36.2 × 1 = 18,100 calories.
Heat given out by the beaker = 100 × 36.2 x.117 = 424 calories.
Total heat given out = 18,524 calories.
Heat taken up by ice in melting =150 x heat of fusion = 150 F calories.
Heat taken up by resulting water = 150 × 43.8 x 1 = 6570 calories.
The heat of fusion of ice, as found by careful experiments, is 80 calories. This means that it takes as much heat to melt one gram of ice as it would to raise one gram of water from 0° to 80°.
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