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 eff

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?

in progress 0
Acacia 5 years 2021-08-30T03:50:38+00:00 1 Answers 13 views 0

Answers ( )

  1. mit
    0
    2021-08-30T03:52:20+00:00
    Reply

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )