# 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). – sammy gerbil Aug 19 '17 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 '17 at 10:21
• When describing a topic such as evaporation physics textbooks usually ignore all other complications which might affect the situation. – sammy gerbil Aug 19 '17 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. – Anthony X Aug 19 '17 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. – lemon Aug 19 '17 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. – Anthony X Aug 19 '17 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.