-1
$\begingroup$

A message has been floating around saying that it is better to wash your hands with warm water, as it helps the soap create more foam. Is it true? If it is true, why is it so? Is there a temperature dependant factor in the soap's cleansing mechanism?

$\endgroup$
  • 1
    $\begingroup$ Foam is hardly relevant at all. It might be true that warm water cleans better, though. $\endgroup$ – Ivan Neretin Mar 28 at 6:58
  • $\begingroup$ All chemical processes have a temperature dependance, higher temperatures increase reation rates. The breakdown of fats (including virus particles) by soap is a chemical process. $\endgroup$ – Waylander Mar 28 at 8:37
  • 1
    $\begingroup$ @Waylander A virus particle is hardly a fat, and soap does not chemically break down fats either. ??? $\endgroup$ – Karl Mar 28 at 8:57
  • $\begingroup$ The lipid bilayer is similar enough for someone who is not a chemist, as is the dispersal of fats by detergent $\endgroup$ – Waylander Mar 28 at 9:13
  • $\begingroup$ At high temperature, the dissolution rate of any substance in water is greater than at low temperature. This has nothing to do with the nature of the solute, soap or not soap. $\endgroup$ – Maurice Mar 28 at 10:26
0
$\begingroup$

Is there a temperature dependant factor in the soap's cleansing mechanism?

Yes, because hydrophilic-hydrophobic interactions are always driven by entropy. Entropy contributes to Gibbs Energy which is the final quantity determining whether a reaction takes place or not (or in which direction it takes place for that matter). But your question implies other questions that should be answered.

Does soap create more foam with warm water?

I don't know. In general, mixtures exhibit phase separation at a lower temperature bound of miscibility and a upper temperature bound of miscibility. Some mixtures do not have a temperature range where no phase separation occurs, some are miscible in all circumstances, you get evaporation/freezing before you get phase separation, think water and acetone.

Does more foam in my soap help clean my hands?

No. Foam/bubbles are small sheets of water with surfactant towards the gaseous phase on the in- and outside. Hence the oily look and texture. What you cleans your hands is the surfactant, because it also helps solubilize hydrophobic stuff like certain proteins. The cleaning is done by hydrophilic stuff being dissolved in the water, and hydrophobic stuff being covered inside micelles of the surfactant, which are soluble in the water.

Does more soap help against viruses?

If your hands are really dirty, maybe. Surfactants are known to denature proteins. You wish to denature the virus proteins so that they can no longer carry out their functions. This would also rob your skin of its protective fatty outside layer. You need not worry to denature your own proteins unless you are trying to drink soap water. Also, aggressive surfactants like SDS work better than your standard fatty acid soap, being even more damaging to your skin as well.

How can I optimally protect myself against viruses on my skin?

Use a desinfectant like ethanol or iso-propanol with at least 70% concentration. It denatures proteins quite effectively. Usage for at least 30 seconds is the official recommendation to kill most common viruses and bacteria. I personally don't wash off, but wait for it to evaporate, which is a good heuristic for the 30 seconds. Make sure to use some sort of lipid replenishing cream every now and then to recover your skin.

| improve this answer | |
$\endgroup$
  • $\begingroup$ The T-dependence of a reaction (whether it shifts to the left or right, and by how much, as the temperature is changed) is determined by the T-dependence of the equilibrium constant (Keq), and the T-dependence of Keq is determined by the sign and magnitude of the enthalpy, not the entropy. Yes, the T-dependence of the Gibbs free energy is determined by the sign and magnitude of the entropy, but Keq is related to the Gibbs free energy through a factor of T, which shifts the T-dependence from the entropy (in the Gibbs free energy) to the the enthalpy (in Keq). $\endgroup$ – theorist Mar 28 at 16:02
  • $\begingroup$ @theorist higher temperature means higher entropic contribution to reaction directionality, no matter how you put it. Either it increases the factor in front of dS or it acts as a divisor for dH. $\endgroup$ – tillyboy Mar 28 at 20:27
  • $\begingroup$ Nope, that's incorrect. The entropy has no effect upon the T-dependence of the equilibrium constant (i.e., upon the first derivative of the equilibrium constant with respect to T). The T-dependence of the equilibirum constant is given by the van't Hoff equation: $\frac{d ln K_{eq}}{dT} = \frac{\Delta H}{RT^2}$ Note that $\Delta S$ is not present in that equation. To understand why $\Delta S$ disappears, you should go through the van't Hoff equation's derivation. This can be found in any good pchem textbook. $\endgroup$ – theorist Mar 28 at 20:41
  • $\begingroup$ If something is not present in the first derivative of something, it could as well have a constant contribution to an quantity, right? The van't Hoff equation shows that with rising temperature, the contribution of enthalpy to the equilibrium constant approaches zero. $\endgroup$ – tillyboy Mar 28 at 21:03
  • $\begingroup$ Now you're confusing two different things, the T-dependence of Keq, and the value of Keq. At the beginning of your answer, you were saying there is a T-dependence in Keq because of the entropy, which is wrong, as shown by the van't Hoff eqn. Now you've switched to talking about the value of Keq itself, which does depend on the entropy change, but misapplying the van't Hoff equation to the value of Keq, when in fact the van't Hoff equation is for the T-dependence of Keq. $\endgroup$ – theorist Mar 28 at 21:23

Not the answer you're looking for? Browse other questions tagged or ask your own question.