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)

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)

Natalie Anton · Accepted Answer

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.)

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

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"#.

Luna Chen · Answer

undefined To calculate the amount of energy required to boil the water, you can use the formula:

Energy = mass × heat of vaporization

Substitute the given values:

Energy = 46.0 g × 2.0934 J/g

Calculate:

Energy = 96.2564 J

So, it takes approximately 96.26 joules to boil the water.