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With two sources of light:

  • an incandescent light approximate color temperature of 3000 K,
  • sun with an approximate color temperature of 6000 K. W

Which source produces warmer tones?

Consider taking a properly-exposed photograph of a white sheet of paper in each of these two lighting situations with your camera set at a white balance of 4500 K; in which photo would the white sheet of paper appear warmer?

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Here is where some people get color temperature and thermal temperature confused. If the camera is set at 4500K, the picture taken with the 3000K incandescent light will be warmer. That is, it will be more on the orange/yellow end of the spectrum. The picture taken with the 6000K light source will be cooler, or more towards the blue end of the spectrum. Lower color temperature light sources produce light on the warm end of the spectrum. Higher color temperature light sources produce light on the cool end. From warm to cool the visible color spectrum runs red->orange->yellow->green->blue->indigo->violet.

Even though a blue flame is generally hotter than an orange one (it really depends on the chemical composition of the burning material and the oxidizer), we generally associate orange with heat and blue with cooler things like water. It gets even more confusing for some people because when you are editing RAW files and you change the color temperature value from 3000K to 6000K the picture gets more orange and less blue. This is because in the first case when the convertor is told the light source is 3000K it boosts the blue values to compensate for the orange light you told the convertor was illuminating the scene. When the color temperature is set for 6000K it will boost the orange/red end of the spectrum to compensate for the bluer light that illuminated the scene.

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By convention, yellow-red colors (like the flames of a fire) are considered warm, and blue-green colors (like light from an overcast sky) are considered cool. Confusingly, higher Kelvin temperatures (3600–5500 K) are what we consider cool and lower color temperatures (2700–3000 K) are considered warm. Cool light is preferred for visual tasks because it produces higher contrast than warm light. Warm light is preferred for living spaces because it is more flattering to skin tones and clothing. A color temperature of 2700–3600 K is generally recommended for most indoor general and task lighting applications. Color Temperature is not an indicator of lamp heat.

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Photographers use the terms "warm" and "cool" in exactly the opposite way physicists use the term.

Physicists are referring to thermal energy (specifically, the spectral emission of a black-body radiator) at some particular Kelvin, while photographers are talking about the yellowness or blueness of the light imparting a "fireside warmth" or "arctic glacier chill" feel to an image.

To answer your questions:

1) The incandescent light produces warmer (feeling) tones.

2) With the camera set to any specific setting (including 4500K), a white sheet of paper will be rendered warmer (feeling) when exposed under incandescent light relative to daylight.

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  • \$\begingroup\$ Yet even in physics the hottest stars are blue and the coolest ones are red... and with thermal energy a blue flame is generally much hotter than an orange one. \$\endgroup\$
    – Michael C
    Apr 28, 2013 at 4:33
  • \$\begingroup\$ Michael, correct, physicists use the term in the opposite way that photographers use it. I thought that was clear from my answer? If not, I'm open to suggestions on how to make it clearer. \$\endgroup\$
    – U007D
    Apr 28, 2013 at 12:55
  • \$\begingroup\$ Just pointing out that blue is usually indicative of a higher temperature than orange to a physicist as well as to a photographer. \$\endgroup\$
    – Michael C
    Apr 28, 2013 at 13:01
  • \$\begingroup\$ To a physicist, blue is warmer than orange. To a photographer, orange is warmer than blue (the opposite). \$\endgroup\$
    – U007D
    Apr 28, 2013 at 13:09
  • \$\begingroup\$ But blue is a higher temperature than orange to both a physicist and a photographer. \$\endgroup\$
    – Michael C
    Apr 28, 2013 at 13:44

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