Weight Of Ice Melted By Heat Liberated From The Condensation Of Steam

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Weight Of Ice Melted By Heat Liberated From The Condensation Of Steam

Determine the weight of ice melted at 0^{0}C by the heat liberated when 100 grams of steam at 100^{0}C condenses to liquid.

Heat of vaporization = 540 cal/g

Heat of fusion = 80 cal/g.

**The strings**:

S_{7}P_{1}A_{15} (containership - mass)
**The math**:

Pj Problem of Interest is of type *containership*. It is the mass of the melted ice that is of interest eventhough heat energy is operative.

**Heat Of Vaporization**: the quantity of heat required to vaporize 1 gram of a liquid substance at its boiling point at constant temperature.
**Heat Of Fusion**: the quantity of heat required to liquefy 1 gram of a solid substance at its melting point at constant temperature.

Amount of heat liberated by condensation 100 grams of steam at 100^{0}C

= 540 x 100 = 54000 cal

Amount of heat required to melt 1 g of ice at 0^{0}C = 80 cal

So, weight of ice melted at 0^{0}C by 5400 cal = 54000/80 = 6.75 grams.

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