Question

A sample of `H_2O` with a mass of 46.0 grams has a temperature of 100 °C. How many joules are necessary to boil the water? (use 2.0934 J/g for the heat of vaporization of water)

Answer 1

I get `"104000 J"` when I use the value given by many other textbooks.

(If I were to use your value, which is equal to `"0.0377 kJ/mol"`, I would only have gotten `"96.3 J"`, which is way too low.)

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Sorry, but I refuse to use an incorrect enthalpy of vaporization. The correct value is `"40.67 kJ/mol"`, or about `"2257.6 J/g"`, over one thousand times greater than what you have cited.

This can be found

• here
• here
• here
• ...and here.

Boiling a substance would be done at constant pressure and temperature, so we can equate the heat required with the enthalpy of vaporization:

`q = nDeltabarH_(vap)`

where `n` is the mols of liquid, `DeltabarH_(vap) = "40.67 kJ/mol"`, and `q` is the heat absorbed.

Thus, the heat absorbed is:

`color(blue)(q) = 46.0 cancel"g" xx cancel"1 mol"/(18.015 cancel"g") xx "40.67 kJ"/cancel"mol"`

`=` `color(blue)("104 kJ")`

or `"104000 J"`.