Is the temperature of light affected by color filters? This video demonstrates how varying color filters alter the energy of photoelectrons emitted from a light source:
https://www.youtube.com/watch?v=kcSYV8bJox8
But I am confused by what these different colors having different "energies" means. Blue light is said to have more energy than red light. Does this mean that the blue light is hotter than the red light? If so, where does the heat go for lower energy colors - is it absorbed by the filter?
It seems to me this probably isn't true, and just filtering light wouldn't change it's temperature. But, if the colors are all the same temperature, then what does them having different "energy levels" mean? Is this a very specific definition relative to the emission of photoelectrons from a metal?
Just to clarify, I know that a light source will change colors as it gets hotter, I am not referring to a light source whose temperature is directly being changed like a fire, I am referring color filters like in the video.
 A: Light does not have a temperature but carries energy at different frequencies. The energy from the light causes matter to heat up when they interact. Energy that is filtered will cause the filter to heat up.
A: 
But I am confused by what these different colors having different "energies" means. Blue light is said to have more energy than red light.

The statement "blue light has more energy than red light" is an imprecise statement. A more precise statement is "blue light has more energy per photon than red light." Saying that blue light has more energy than red light is a bit like saying that gold weighs more than water.

If so, where does the heat go for lower energy colors - is it absorbed by the filter?

When a filter absorbs light, it absorbs its energy. So both a filter that absorbs blue light, and filter that absorbs red light, absorbs energy and decreases the energy of the light.
A: Temperature, colour and energy all mean different things.
A single 'photon' has an associated energy, but it's misleading to think of this as a colour.
Three things that are (for most intents and purposes) equivalent, are energy $E$, wavelength $\lambda$ and frequency $f$, which are related by
$$ E = hf = hc/\lambda$$
where $c$ is the speed of light and $h$ is Planck's constant. Note that as the wavelength gets smaller, $E$ gets bigger.
Humans have three kinds of cone (colour sensing) cells - red, green and blue, but these do not "only" see red green and blue respectively - they are excited to varying degrees by the other wavelengths as well.

To a human, a combination of 500nm and 660nm light appears yellow, but this is physically a different situation to e.g. a yellow sodium lamp, which emits primarily ≈580nm photons.
Light from something like the sun or an incandescent lamp is even more complicated - these are near-perfect blackbodies, meaning that they emit at all wavelengths to some extent:

This excites all three of our cones equally, which our brain processes as "white". The crucial point is that there is no such thing as a "white" photon, but there is such thing as a "blue" photon.
The changing colour of a blackbody as it gets hotter is due to the movement of the "peak" wavelength, i.e. the wavelength at which most light is emitted, which for the Sun is about 500nm. If the sun were to get hotter, the peak would shift to the left, exciting more blue cones and fewer red cones. Likewise, if it got colder, the peak would shift to the right (lower energy), making the Sun look redder.
The Bottom Line
All wavelengths of photons carry energy, "white" light is a mixture of many wavelengths, and filtering this mixture to only get e.g. red light will make the filter heat up / reflect the energy elsewhere.
A: Blue photons (= photons of ~420 nm wavelength) have higher energy than red ones (= photons of ~600 nm wavelength). That means that each individual absorbed blue photon will indeed heat the surface more, because it gives it more energy. But two red photons would heat the surface more than a single blue photon. Photons themselves don't "have" heat, heat is a collective consequence of random motion of many atoms/molecules/...
Photoelectric effect in the video is a bit different phenomenon from heating the material because photons need sufficient energy to eject electrons from the material. Red light might not eject any electrons no matter how many photons hit the material, while blue would even if you have few photons (and therefore lower heating).
Temperature of light, eg as on "3000K" light bulbs, is something different still. Blackbody gives out EM radiation. Spectrum of that radiation depends on the temperature - say something at 3000K has peak of radiation at approximately 1 micrometer wavelength (near-IR) while our sun at ~6000K has peak at green light. And that temperature designation of the light bulb simply means "this bulb gives out light that has approximately similar spectrum to the body at this temperature" (skipping details about color quality).
You can start with 3000K bulbs, then add filter to get 6000K light. First light is red-ish, second one is blue-ish. But even though after filter your photons flying out are more blue and therefore have more energy on average, this filter absorbed (or reflected; in video filter absorbed light) quite a lot of energy - many red photons had to get blocked. So, average photon has higher energy, but total energy of all photons in the light is lower. Shining unfiltered red-ish light on thermometer would heat it more than shining filtered blue-ish light on it.
This losing energy is true even if you have laser light and convert it to have half the wavelength (eg 1064 nm to 532 nm) - in that case all input photons have low energy and all output photons have high energy, but you lost many (at least half) photons in the process, so the total energy (that gets converted to heat by absorber) is the same or usually lower (with some energy lost during conversion and resulting in heating of the converter). In this case of up-conversion (photons flying out having higher energy) you can also get photoelectric effect with the converted light while original one doesn't kick single electron out of the material. This is unlike with filters that merely block some wavelengths - there, photoelectric effect of filtered light is at most the same as in original light.
A: Like @PabloH said, one of the big sources of confusion in your question is bringing the temperature to the discussion.
There are two very conflicting definitions: one being warm/cool color dichotomy tied to human perception only and association of reds and yellows with sun and therefore warmth and blue tint with shadows and cold. Color temperature goes the opposite way and is tied to black body radiation spectrum (as already explained in other answers).
Color theory is a very, very complex topic and has more to do with psychophysiology than physics, admittedly.
A: Short answer: Yes
Long answer: if you use a Blue filter you'll only let higher energy photons through. So the temperature registered will be higher.
A: The first thing you should know is, light does not have a 'temperature' and light does not carry 'heat'. Know that light and heat are two separate forms of energy. What you are confused with is the ability of light energy to convert to heat energy, that is how the sun heats up the Earth, through electromagnetic radiation (light). This is done through a process known as absorption.
Now coming to your original question, yes, blue light does carry more energy than red light. This does not mean that blue light has more 'temperature' but that it has more frequency (less wavelength), hence, it has a higher 'ability' to convert into heat energy on absorption. However, this also depends on one other condition- the intensity of light. If red light has a much higher intensity as compared to blue light, it is said to have higher energy. This is the case with lasers. To increase the intensity of light, you need to increase the photon flux, that is, increase the number of photons passing at a given time. So a revision to the original statement would be "blue light has more energy than red light at the same intensity level".
Coming to the colour filters point, the filter is able to show a particular colour because it absorbs the light of the opposite wavelength, and therefore the colour that is not being absorbed is visible. This can be seen through the wheel:

So for example, if the filter is for red light (that is, the final colour visible is red), green light is being absorbed, and therefore, red is visible. Now in this case, since the light passing through the filter is from the same source, the blue light in this case would have higher energy than red light (as the intensity remains the same).
A: Light does not have a temperature. "Temperature" is a property of mass, and since photons are massless, they have no temperature. Photons can be converted to heat when they are absorbed by some substance, but that's not the same as saying they have a temperature themselves.
Each photon of light does have a certain amount of energy, though, which is proportional to its wavelength and inversely proportional to its frequency. So, a low-frequency photon will carry more energy than one with a higher frequency. But the total energy in a beam of light only depends on how much energy the emitter put into that light beam. If the beam is composed primarily of low-frequency, high-energy photons, then there will be fewer of them, so that total energy is conserved.
Notice I haven't mentioned color yet. "Color" is created entirely in our eyes and brains, based on the combination of all the photons that strike one point on our retina in a short amount of time. It's actually a pretty poor way of characterizing frequency.
Imagine a beam composed entirely of 640nm "red" photons and 440nm "blue" photons. Such a beam would appear magenta to human eyes, even though it doesn't contain any "magenta" photons. Now put a blue filter in the path of the beam. The filter will absorb the "red" photons while allowing the "blue" photons to pass unchanged. Thus, the intensity of the beam is reduced, and the perceived color is shifted to a "pure blue". The target will experience less heating when it is struck by the beam, because some of the photons that would have hit it were already absorbed (and turned into heat) by the filter.
