Question

At our distance from the Sun, the intensity of solar radiation is 1370 W/m^2. The temperature of the Earth is affected by the greenhouse effect of the atmosphere. This phenomenon describes the effect of absorption of infrared light emitted by the surface so as to make the surface temperature of the Earth higher than if it were airless. For comparison, consider a spherical object of radius r with no atmosphere at the same distance from the Sun as the Earth. Assume its emissivity is the same for all kinds of electromagnetic radiation and its temperature is uniform over its surface.
Compute its steady-state temperature. Is it chilly?

Answer 1

Answer:

The steady-state temperature is  $T = 4.85 ^oC$

Yes it is chilly

Explanation:

From the question we are told

       The intensity of solar radiation is  $I = 1320 \ \frac{W}{m^2}$

Generally the Stefan Boltzmann Law is mathematically represented as

          $P = A \epsilon \sigma T^4$

Where

         $P$ is the total power radiated

          $A$ is the surface area of the object

          $\epsilon$ is the emissivity

          T is the temperature of the object

           $\sigma$ is the Boltzmann constant with a value $\sigma = 5.670 *10^{-8} \frac{W}{m^2 K^4}$

Generally at steady state  the input power to the object  is equal to the output power from the object

           i.e    $P_A = P_B$

Now  $P_A$

which is the input power to the object is not dependent on the object temperature and on the Boltzmann constant

thus $P_A$  is mathematically represented as

               $P_A = \epsilon IA_a$

Where $A_a$  is  absorptive surface area mathematically represented as

            $A_a = \pi r^2$

Thus

        $P_A = \epsilon I \pi r^2$

And   $P_B$ which is the output  power to the object is mathematically represented a

                $P_B = A_s \epsilon \sigma T^4$

Where $A_s$ is the radiative surface area which is mathematically  as

           $A_s = 4\pi r^2$

So

            $P_B = 4\pi r^2 \epsilon \sigma T^4$

=>         $\epsilon I \pi r^2 = 4\pi r^2 \epsilon \sigma T^4$

=>          $T = \sqrt[4]{\frac{I}{4 \sigma } }$

substituting values

             $T = \sqrt[4]{\frac{1370}{4 * 5.670 *10^{-8} } }$

             $T = 278 \ K$

Converting to degrees

         $T = 278 - 273$

        $T = 4.85 ^oC$

This implies that at steady state it is chilly