 # One of the statements of the second law is that no agency can be built whose sole effect is to convert some amount of heat entirely to work. But in case of boiling, the temperature being constant, entire heat supplied is converted into work, namely the work done by water against ambient pressure to expand to the vapour state. At which point am I going wrong? Luz Stokes 2022-07-19 Answered
One of the statements of the second law is that no agency can be built whose sole effect is to convert some amount of heat entirely to work.
But in case of boiling, the temperature being constant, entire heat supplied is converted into work, namely the work done by water against ambient pressure to expand to the vapour state.
At which point am I going wrong?
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When water boils, the heat mostly goes into breaking the bonds between water molecules. Suppose you boil mass m of water under constant pressure P. The work done due to expansion is
$A=P\left({V}_{\text{gas}}-{V}_{\text{liquid}}\right),$
and one can estimate
$P{V}_{\text{gas}}\simeq NkT,$
where N is the number of molecules in mass m. And so we get, for ${V}_{\text{gas}}\gg {V}_{\text{liquid}}$
$A\simeq NkT.$
That is, the work is kT per molecule. On the other hand, the hydrogen bond binding energy in water is on the order of 0.2 eV, if I rememeber correctly. That is equivalent to temperature$0.2\cdot {10}^{4}=2000$ K, and you have to break on the order of 2 hydrogen bonds. Therefore, at room temperature, only several percent of heat actually goes into work, while most of it goes into breaking the bonds between molecules.