Answer:
The steady-state temperature is [tex]T = 4.85 ^oC[/tex]
Yes it is chilly
Explanation:
From the question we are told
The intensity of solar radiation is [tex]I = 1320 \ \frac{W}{m^2}[/tex]
Generally the Stefan Boltzmann Law is mathematically represented as
[tex]P = A \epsilon \sigma T^4[/tex]
Where
[tex]P[/tex] is the total power radiated
[tex]A[/tex] is the surface area of the object
[tex]\epsilon[/tex] is the emissivity
T is the temperature of the object
[tex]\sigma[/tex] is the Boltzmann constant with a value [tex]\sigma = 5.670 *10^{-8} \frac{W}{m^2 K^4}[/tex]
Generally at steady state the input power to the object is equal to the output power from the object
i.e [tex]P_A = P_B[/tex]
Now [tex]P_A[/tex]
which is the input power to the object is not dependent on the object temperature and on the Boltzmann constant
thus [tex]P_A[/tex] is mathematically represented as
[tex]P_A = \epsilon IA_a[/tex]
Where [tex]A_a[/tex] is absorptive surface area mathematically represented as
[tex]A_a = \pi r^2[/tex]
Thus
[tex]P_A = \epsilon I \pi r^2[/tex]
And [tex]P_B[/tex] which is the output power to the object is mathematically represented a
[tex]P_B = A_s \epsilon \sigma T^4[/tex]
Where [tex]A_s[/tex] is the radiative surface area which is mathematically as
[tex]A_s = 4\pi r^2[/tex]
So
[tex]P_B = 4\pi r^2 \epsilon \sigma T^4[/tex]
=> [tex]\epsilon I \pi r^2 = 4\pi r^2 \epsilon \sigma T^4[/tex]
=> [tex]T = \sqrt[4]{\frac{I}{4 \sigma } }[/tex]
substituting values
[tex]T = \sqrt[4]{\frac{1370}{4 * 5.670 *10^{-8} } }[/tex]
[tex]T = 278 \ K[/tex]
Converting to degrees
[tex]T = 278 - 273[/tex]
[tex]T = 4.85 ^oC[/tex]
This implies that at steady state it is chilly
