Question

How much energy is required to heat a frozen can of juice (360 grams- mostly water) from 0 degrees Celsius ( the temperature of an overcooled refrigerator) to 110 degrees ( the highest practical temperature within a microwave oven)?

Answer 1

Answer:

1,100,160J or 262.94 kcal

Explanation:

The juice is frozen at 0 degrees Celsius and I assume that it will become gas at 100 degrees Celsius. So we change the form of the water from solid to liquid, then to gas. That means we have to find out how much heat needed to change water form too, not only the heat needed to increase its temperature.

The latent heat of water is 4.2J/g °C while the heat of fusion is 334 J/g and the heat of vaporization is 2260 J/g. The energy needed will be:

360g * 4.2J/g °C * (110-0°C ) + 360g * 334 J/g + 360g * 2260 /g = 1,100,160J or 262.94 kcal.