Question

A lightbulb manufacturer makes bulbs with different "color temperatures," meaning that the spectrum of light they emit is similar to a blackbody with that temperature. Assuming the emitting areas of the filaments in two bulbs with color temperatures of 2,000 K and 4,000 K are the same, which of the two is the brighter?

Answer 1

The bulb with higher temperature(4000 K) will be brighter

Step-by-step explanation:

From the question we are told that

The color temperature for first bulb is
`T_1 = 2000K`

The color temperature for second bulb is
`T_2 = 4000K`

Generally the emission power of black body radiation is mathematically represented as

`E = \sigma T^4`

Where
`\sigma` is the Stefan-Boltzmann constant with a value
`\sigma = 5.67 * 10^(-8) W m^(-2) K^(-4.)`

Now for
`T_1 = 2000K`

`E_1 = 5.67*10^(-8) * (2000)^4`

`E_1 = 907.2 \ W/m^2`

At
`T_2 = 4000K`

`E_2 = 5.67*10^(-8) * 4000`

`E_2 = 14515.2 \ KW/m^2`

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