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If the sample is at 100 0C, at standard pressure, no supplementary heating is necessary.

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A similar but more detailed take:

Water's latent heat of vaporization is listed as 2,260 kJ/kg = 2.26 J/gram

Since your sample of H2O is in equilibrium right now and sitting there with the uniform

temperature of 100°C, it's ready to vaporize the next time some energy is added,

no matter how little.

Let's say an additional 0.1 Joule comes along and is absorbed. Some little tiny

mass of water ... 0.1/2.26 gram in this example ... must change phase and turn

to vapor. There will be a small blurp in the beaker as the vapor rises, the

surface of the liquid will be disturbed, and anyone watching will report that

it has 'boiled'.

To change the phase of the entire 46.0 grams and turn it to vapor would require

46 x 2.26 = 103.96 Joules.

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11y ago
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10y ago

46 x 58 x 4.2 J/g K = 11,205.6 joules. but it will still be ice at 0oC. it requires additional energy to transform it into water

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7y ago

The heat is 5,624 kJ.

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14y ago

5.60 x 10^3

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14y ago

1.29 x 10^3

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Q: A sample of H2O with a mass of 46.0 grams has a temperature of -58.0 Celsius. how many joules of energy are necessary to heat the ice to 0 Celsius use 2.1 joules for the specific heat of ice?
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