I know that wind chill essentially works by 'wicking' more heat away from a substance, making it 'feel' cooler, but then why doesn't wind chill affect thermometers? Wouldn't the wind 'wick' heat away from the thermometer, making the thermometer's readings go down because of the reduction of heat? Any support is appreciated.
[Physics] Why doesn’t wind chill affect thermometers
temperaturethermodynamicsweather
Related Solutions
There are three processes to take into account:
- The warming of ice towards the melting point if it was originally below $0^{\circ} C$.
- The melting of ice itself
- The warming of the resulting water
The 1. and 3. part is addressed by heat capacity of ice and water respectively and the amount of heat will be directly proportional to temperature difference and weight of the water/ice. The proportionality constant (actually it also depends on the temperature but not very strongly so let's just ignore that) is called specific heat. For water it is about twice as large as that of ice at temperatures around $0^{\circ} C$.
As for the 2. part, this has to do with latent heat. Simply put, this is an amount of heat you need to change phases without changing temperature. Less simply put, when warming you are just converting the heat into greater wiggling of water molecules around their stable positions in the crystal thereby increasing their temperature. But at the melting point that heat will instead go into breaking chemical bonds between molecules in the ice lattice.
Now, latent heat is really big (you need lots of energy to break those bonds). To get a hang on it: you would need the same amount of heat to warm water from $0^{\circ} C$ to $80^{\circ} C$ as you would need to melt the same amount of ice.
Now, presumably you want your drink cold in the end so that temperature for 3. will be close to $0^{\circ} C$ and also the ice cubes should be pretty warm (no use in producing ice cubes of e.g. $-50^{\circ} C$, right?). This means that these processes won't contribute much cooling. It's fair to say that melting of the ice takes care of everything.
Note: we can also quickly estimate how much ice you need by neglecting the processes 1. and 3. Say you are starting with a warm drink of $25^{\circ} C$ and you want to get it to $5^{\circ} C$. So, reusing the argument about the $80^{\circ} C$ difference being equivalent to a latent heat of the same mass, we see that you need four times less ice than water to get the job done.
As BowlOfRed points out the incoming air has a certain temperature and the convective flow will tend to bring the temperature of the object towards the temperature of the incoming air.
In the case where the incoming air was going very fast it gets heated through adiabatic compression before it reaches the object, so the incoming air is at higher temperature than the free stream temperature, but it will still cool the object if it is at an even higher temperature.
Similarly, if you take something out of the freezer, a gentle breeze will actually have a warming effect on the object bringing it closer to the temperature of the incoming air.
If you want to know the relationship between the equilibrium temperature and the free stream temperature vs. velocity that's given by the isentropic flow equation:
$$T=T_0\,\left(1+\frac{\gamma-1}2 M^2 \right)$$
Where $T$ is the surface temperature (in Kelvin or Rankine), $T_0$ is the free stream temperature (in Kelvin or Rankine), $\gamma$ is the ratio of specific heats (1.4 for air), and $M$ is the Mach number.
Best Answer
The thermometer has no internal source of heat to be "wicked" away - so once it has reached the temperature of the surroundings there is no chilling effect.
If the thermometer is wet then the increased evaporation by the wind can cool it below ambient temperature, and the difference between the temperature of wet and dry thermometers can be used to measure humidity