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You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\frac1{\pi}\sigma T^4$, where $\sigma=5.670367(13)\times10^{−8}\ \mathrm{W\ m^{-2}\ K^{-4}}$. Plugging in $T=343\ \mathrm K$, we get $250\ \mathrm{W\ m^{-2}}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300\ \mathrm K$, we get $146\ \mathrm{W\ m^{-2}}$, for a net of $104\ \mathrm{W\ m^{-2}}$. This means that for every square centimeter, you'll get getting around $10\ \mathrm{mW}$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25\ \mathrm{mW\ cm^{-2}}$. (And if it's in an enclosed space, you may be getting less than $10\ \mathrm{mW}$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343\ \mathrm K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\sigma T^4$, where $\sigma=5.670367(13)\times10^{−8}\ \mathrm{W\ m^{-2}\ K^{-4}}$. Plugging in $T=343\ \mathrm K$, we get $785\ \mathrm{W\ m^{-2}}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300\ \mathrm K$, we get $459\ \mathrm{W\ m^{-2}}$, for a net of $326\ \mathrm{W\ m^{-2}}$. This means that for every square centimeter, you'll get getting around $33\ \mathrm{mW}$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $78\ \mathrm{mW\ cm^{-2}}$. (And if it's in an enclosed space, you may be getting less than $33\ \mathrm{mW}$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343\ \mathrm K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\frac1{\pi}\sigma T^4$, where $\sigma=5.670367(13)×10^{−8} W⋅m^{−2}⋅K^{−4}$. Plugging in $T=343$, we get $250 Wm^{-2}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^{-2}$, for a net of $104 Wm^{-2}$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^{-2}$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\frac1{\pi}\sigma T^4$, where $\sigma=5.670367(13)\times10^{−8}\ \mathrm{W\ m^{-2}\ K^{-4}}$. Plugging in $T=343\ \mathrm K$, we get $250\ \mathrm{W\ m^{-2}}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300\ \mathrm K$, we get $146\ \mathrm{W\ m^{-2}}$, for a net of $104\ \mathrm{W\ m^{-2}}$. This means that for every square centimeter, you'll get getting around $10\ \mathrm{mW}$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25\ \mathrm{mW\ cm^{-2}}$. (And if it's in an enclosed space, you may be getting less than $10\ \mathrm{mW}$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343\ \mathrm K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\frac1{\pi}\sigma T^4$, where $\sigma=5.670367(13)×10^{−8} W⋅m^{−2}⋅K^{−4}$. Plugging in $T=343$, we get $250 Wm^{-2}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^{-2}$, for a net of $104 Wm^{-2}$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^{-2}$. Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.

You seem to be wanting to make use of black body radiation. The formula for energy dissipated through black body radiation is $\frac1{\pi}\sigma T^4$, where $\sigma=5.670367(13)×10^{−8} W⋅m^{−2}⋅K^{−4}$. Plugging in $T=343$, we get $250 Wm^{-2}$. However, it would also be absorbing heat from the surroundings. If we model the surroundings as a black body emitting at temperature $300$, we get $146 Wm^{-2}$, for a net of $104 Wm^{-2}$. This means that for every square centimeter, you'll get getting around $10mW$ of cooling. This is probably going to be only a small fraction of the cooling you need. Most of your cooling is coming from conduction to the air, and then convection within the air, and as others have pointed out, a black coating will likely decrease the heat conductivity of the aluminum. It will likely be more productive to increase the surface area and air flow, and decrease the surrounding temperature.

The modelling of the surrounding as being a black body at room temperature is, of course, questionable, but even without it, you'll be getting only $25mWcm^{-2}$. (And if it's in an enclosed space, you may be getting less than $10mW$, as it will be heating up the surroundings, and that heat will just be radiated back.) Furthermore, it's quite possible that the effective black body temperature of the surroundings is greater than the temperature of your profile. Unless it's in a dark room, it will be absorbing heat from whatever lighting there is in the room. Thus, its net black body heat exchange may be positive, in which case making it darker would make things worse even without taking into account conductivity. Since we're not actually dealing with perfect black body radiation, there is a possibility that you can decrease its albedo in $343K$ range without significantly increasing it in the visible range, but that's rather advanced engineering.