# Does a glass of water at room temperature emit (infrared?) radiation

While reading the introduction to Feynman's lectures, it's mentioned how a glass of water cools down through evaporation, when some molecules get a bit extra energy and break free. If it's not a closed system, energy will be gradually taken away from the cup, hence blowing at the soup helps move those molecules away so that they don't reenter the surface.

But I thought that all bodies also radiate heat? Does a cup of water also emit low frequency radiation, or is my understanding incorrect?

• Bodies can lose heat by more than one method at the same time : conduction, convection and radiation (as well as evaporation). Aug 19, 2017 at 10:19
• @sammygerbil: it was just strange that it isn't mentioned at all in the article, so I thought that perhaps there is a minimum temperature before radiation happens, I am obviously a newbie.
– Lou
Aug 19, 2017 at 10:21
• When describing a topic such as evaporation physics textbooks usually ignore all other complications which might affect the situation. Aug 19, 2017 at 10:25

Yes, all matter above absolute zero emits radiation. To quote wiki:

When the temperature of a body is greater than absolute zero, inter-atomic collisions cause the kinetic energy of the atoms or molecules to change. This results in charge-acceleration and/or dipole oscillation which produces electromagnetic radiation, and the wide spectrum of radiation reflects the wide spectrum of energies and accelerations that occur even at a single temperature.

This continuous release of energy would eventually cool the source to a lower and lower temperature except your glass of water is in contact with a heat reservoir (the room) which compensates for the energy loss.

• Sitting in a room, the glass of water is not just emitting infrared radiation, but also receiving it from all the other surfaces/objects in the "room-temperature" room. If the glass of water starts out hot, the exchange will be un-balanced, so the glass cools off as it emits more than it receives; if it starts out chilled, it will warm up as it receives more than it emits; at the same temperature as its surroundings, as much energy is absorbed as radiated, so it is at equilibrium. Aug 19, 2017 at 21:08
• @AnthonyX While the physical mechanism you describe is accurate, the equilibration of water with its surroundings is dominated by conduction and convection mechanisms, not radiative transfer. Aug 19, 2017 at 21:36
• True. Just wanted to make the point that everything is radiating, the water is not just radiating but also absorbing, and that heat loss (or gain) will continue up to a point of equilibrium. Aug 19, 2017 at 22:40

As @lemon explained, every body emits electromagnetic radiation if its temperature is above absolute zero

The peak wavelength of the radiation emitted by a blackbody is given by Wien's law: $$\lambda=2900/T$$ where T is the temperature in K, and the wavelength ($\lambda$) is in $\mu m$.

For a temperature of 300K (27C), approximately room temperature, the peak wavelength is $9.7\mu m$.

More details can be found at this site. The site has a calculator that shows the peak wavelength, given the temperature, as well as a graph that shows the spectrum at the selected temperature.