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1 Calorie per degree C. If room temp is 72 degrees C then 28 calories to get to boiling then the it takes 540 calories to water from liquid to gas, called the heat of vaporization. So add 540 to 28 and you get 568 calories.

That's the energy to convert 1g of water at room temperature to 1g of steam at 100deg C. And if your room temperature is really 72 deg C you are in trouble - more probably degrees Fahrenheit.

Many times, the units of energy are given in joules (J) and 1 calorie is equivalent to 4.184 joules. The specific heat of water is 4.184 J/g/deg, so to raise the temp of 1 g water from 25ºC to 100ºC will take (4.184 J/g/deg)(75 deg)(1g) = 314 J. Then to convert that 1 g of 100ºC water into steam, you need to use the ∆Hvap which is 2260 J/g. Thus, 2260 J/g x 1 g = 2260 J to evaporate the water. Total energy required is 314 J + 2260 J = 2574 J.

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8y ago
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15y ago

The latent heat of vaporisation for water is 2260 Joules per gram

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14y ago

Heat of Vaporization for water = 540 Cal/g

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10y ago

580 calories

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6y ago

The needed heat is 40,65 kJ/mol.

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Q: How much energy is needed to evaporate 1g of water at room temperature?
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