# Is there a chemical compound can transform heat into light at 70 °C?

Is there is compound that can transform heat into light as phosphorescence / thermally generated photons at around 70 °C, in a reversible reaction?

For example, is there a phosphor which can be visible at 70'C and not at 20 degrees?

I'd very much appreciate some information. Thanks!

It might be possible in a few ways:

1. In the process of thermoluminescence, a previously-activated material gives off light when warmed. This is used to date materials such as ceramics, which accumulate energy from terrestrial radiation and cosmic rays. Though ceramics have to be heated quite a bit to release this stored energy, some substances exhibit thermoluminescence at room temperature or lower. $\ce{BeO:Zn}$ has peak emission at 307K, ~34oC.

N.B thermoluminescence is the release of energy previously stored in a material, not the direct transformation of heat energy into electromagnetic.

1. In an anti-Stokes process, a substance absorbs two or more photons of lower energy to release one of higher energy. This is used in IR beam visualization cards. However, it is unlikely that thermal energy would be sufficient to produce visible light.

2. Non-blackbody radiators can emit more energy at short wavelengths than would be expected according to Planck's law. "Limelight" owes its intense white light to candoluminescence, or non-blackbody radiation. Using a "heat mirror", the efficiency of incandescent lamps can be greatly improved. That said, though theoretically a material might exist that could prevent emission of almost all wavelengths longer than visible light, it is doubtful that a detectable amount of visible light would be emitted at 70oC.

The energy that can be transferred by collisions between particles at temperature $T = 343~\pu{K}$ is about

$$E_{therm.} = \frac{1}{2} k T = 0.015~\pu{eV} \tag{1}$$

The energy that corresponds to the lowest energy of visible light with a wavelength of $\lambda = 800~\pu{nm}$ is about

$$E_{h\nu} = h \nu = \frac{h \nu}{\lambda} = 1.55~\pu{eV} \tag{2}$$

This means that thermal energy at $70~\pu{°C}$ is at least two orders of magnitude too low for producing visible light. So the answer to your question is no.

• Could those who have downvoted my answer please explain why? Nov 29, 2016 at 18:39