290
4 Thermodynamics and Statistical Physics
The negative sign in (3) is omitted because as λ increases v decreases.
4.76 Power radiated from the sun = σ × (surface area) × T
4
s
P s = σ 4π R
2
s T
4
s
Power received by the earth,
P E =
π R
2
e
4πr 2 .P s
The factor π R
2
e represents the effective (projected) area of the earth on
which the sun’s radiation is incident at a distance r from the sun. The factor
4πr
2 is the surface area of a sphere scooped with the centre on the sun. Thus
π R
2
e /4πr
2 is the fraction of the radiation intercepted by the earth’s surface
area.
Now power radiated by earth,
P E = σ 4π R
2
E T
4
E
For radiation equilibrium, power radiated by the earth=power received by
the earth.
σ 4π R
2
E T
4
E = σ 4π R
2
s T
4
s .
π R
2
E
4πr 2
or T E = T s
R s
2r
1/2
= 5,800
7 × 10
8
2 × 1.5 × 10 11
1/2
= 280 K = 7
◦ C
Note that the calculations are approximate in that the earth and sun are not
black bodies and that the contribution of heat from the interior of the earth has
not been taken into account.
4.77 Power radiated by the sun, P s = σ 4π R
2
s T
4
s
Power received by 1 m
2 of earth’s surface,
S =
σ 4π R
2
s T
4
s
4πr 2
=
(5.7 × 10
−8 )(7 × 10
8 )
2 (5,800)
4
(1.5 × 10 11 ) 2
= 1,400 W/m
2
4.78 P = 4πr
2
σ T
4
= 4π (0.3)
2 (5.67 × 10
−8 )(10
7 )
4
= 6.4 × 10
20 W