The tropospheric lapse rate (the rate at which temperature decreases with altitude) is approximately 6°C per kilometer. Given that the mean surface temperature of Earth is 288 K and the effective radiating temperature is 255K, from what altitude does most of the emitted radiation derive?

33 C/(6 C/km) = 5.5 km above the earth's surface is where the atmospheric temp is 255 K.

To determine the altitude from which most of the emitted radiation derives, we can calculate the temperature difference between the mean surface temperature and the effective radiating temperature and then divide it by the lapse rate.

The temperature difference is: 288 K - 255 K = 33 K

Therefore, the altitude from which most of the emitted radiation derives can be calculated using the lapse rate formula:

Altitude = (Temperature Difference) / (Lapse Rate)
= 33 K / 6 °C per kilometer

Converting 33 K to °C gives: 33 K ≈ 33 °C

Altitude = 33 °C / 6 °C per kilometer
= 5.5 kilometers

So, most of the emitted radiation derives from an altitude of approximately 5.5 kilometers.

To determine the altitude from which most of the emitted radiation derives, we can use the concept of the lapse rate and the temperatures provided.

The tropospheric lapse rate is given as approximately 6°C per kilometer. However, temperature is usually measured in Kelvin (K) as it is an absolute scale.

First, let's convert the lapse rate to Kelvin (K) per kilometer:
6°C per kilometer = 6 K per kilometer (since a change of 1°C is equivalent to a change of 1K)

Next, we can calculate the temperature difference between the surface temperature and the effective radiating temperature:
Temperature difference = Mean surface temperature - Effective radiating temperature
= 288 K - 255 K
= 33 K

Now, let's divide the temperature difference by the lapse rate to find the altitude:
Altitude = Temperature difference / Lapse rate
= 33 K / 6 K per kilometer
≈ 5.5 kilometers

So, from an altitude of approximately 5.5 kilometers, most of the emitted radiation derives.