Answer:
The bulb with higher temperature(4000 K) will be brighter
Explanation:
From the question we are told that
   The color temperature  for first bulb is  [tex]T_1 = 2000K[/tex]
    The color temperature  for second bulb is  [tex]T_2 = 4000K[/tex]
Generally the emission power of black body radiation is mathematically represented as
   [tex]E = \sigma T^4[/tex]
Where  [tex]\sigma[/tex] is the Stefan-Boltzmann constant  with a value [tex]\sigma = 5.67 * 10^{-8} W m^{-2} K^{-4.}[/tex]
Now for [tex]T_1 = 2000K[/tex]
   [tex]E_1 = 5.67*10^{-8} * (2000)^4[/tex]
    [tex]E_1 = 907.2 \ W/m^2[/tex]
At  [tex]T_2 = 4000K[/tex]
    [tex]E_2 = 5.67*10^{-8} * 4000[/tex]
    [tex]E_2 = 14515.2 \ KW/m^2[/tex]
Looking at the result we got we see that  the emission power  for the higher temperature bulb is higher, this means that its power to emit in the visible spectrum range would be higher Â
So the bulb with higher temperature will be brighter
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