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
So the bulb with higher temperature will be brighter
