# [Physics] Drying up a wet object

fluid dynamicsphase-transitionwater

I am wondering whether an object which has been wet with hot water always dries up more easily and more quickly than one wet with cold water. How much do the shape, roughness, material, structure of the object matter?

Practically: if for example I needed to wash a can and a sweater, would in both cases hot water allow me to dry them up more quickly, just by means of natural ventilation?

I am wondering whether an object which has been wet with hot water always dries up more easily and more quickly than one wet with cold water.

Yes, it will. The more important question is, "How much better will it dry?".

Some simple intuition may be garnered from dishwashers. When a thick ceramic plate or bowl goes through a dishwasher, it winds up hot and covered in water by the end of its journey. When you open the dishwasher and wait for 15 minutes, you will find the thick plate or bowl will be almost completely dried. For a thin plastic dish, something entirely different happens: it still remains mostly wet by the time it has cooled down.

Why? The thick ceramic bowl has a greater thermal mass per surface area than the thin plastic plate, where thermal mass is defined to be $m C_p$ with $C_p$ being the constant pressure heat capacity. The thick bowl holds enough heat to evaporate its coat of water and dry in the presence of air; the thin plate simply doesn't hold enough heat to drive off its water.

Ignoring heat loss due to radiative effects, the ability of an object to hold enough heat to drive off its water is entirely determined by the ratio of its thermal mass to its water content. To be specific, the fraction of water which will be lost is approximately given by $$x=\frac{m_oC_p\Delta T}{m_w\Delta H}$$ where $m_o$ is the object mass, $m_w$ is the water mass adsorbed, $\Delta T$ is the difference between the object initial temperature and ambient temperature, $C_p$ is the heat capacity of the object (not the water), and $\Delta H$ is the latent heat of vaporization of water in units of energy per mass per Kelvin, which for water happens to be 2260 Joules per gram per Kelvin.

For an object with a typical heat capacity of 1 Joule/gram/Kelvin with 5% water content heated initially to 373K, a drop to room temperature will be achieved with about 60% water loss. So your object had better be less than 3% water if you want it to dry effectively just from its thermal heat.