How much heat is absorbed by 73.0 grams of ice as it melts at 0°C? How much heat is absorbed by 73.0 grams of liquid as it vaporizes at 100°C?

24382 J

162790 J

Explanation:

At 0°C, ice requires a certain amount of heat/energy to break its certain bonds and change its state into liquid water. This amount of heat per gram of ice is known as the latent heat of fusion of ice. And this heat is already detrmined and is equal to 334 J/g. Similarly, For water to convert into vapor at 100°C, it requires a certain amount of heat/energy to break its certain bonds and change its state into gas/vapor. This amount of heat per gram of water is known as the latent heat of vaporization of water, and it is equal to 2230 J/g.

Now, for one gram ice, 334 J heat is required for melting, hence for 73g of ice, 73x334 = 24382 J of heat is required.

Now, for one gram water, 2230 J heat is required for melting, hence for 73g of water, 73x2230 = 162790 J of heat is required.