1000 Solved Problems in Modern Physics

286 4 Thermodynamics and Statistical Physics

p=

1

3

ρ<ν^2 >

whereρis the mass density. In the case of photon gas, the speed of all photons

is identical being equal toc. Furthermore, from Einstein’s relation

u=ρc^2

whereuis the energy density. Replacing<ν^2 >byc^2

prad=

1

3

ρc^2 =

u

3

4.64 LetTandT 0 be the Kelvin temperatures of the body and the surroundings.
Then, by Stefan–Boltzmann law, the rate of loss of heat per unit area of the
body is
dQ
dt

=σ(T^4 −T 04 )

=σ(T−T 0 )(T+T 0 )(T^2 +T 02 )

If (T−T 0 ) be small, (T≈T 0 ), and

dQ

dt

=σ(T−T 0 )× 4 T 03

SinceT 0 is constant,

dT

dt

∝(T−T 0 ); (Newton’s law of cooling).

4.65 The energy densityuand pressurepof radiation are related by

p=

u

3

Furthermore,u= 4 σT^4 /c

Eliminatingu,

T=

(

3 cp

4 σ

) 1 / 4

=

(

3 × 3 × 108 × 4 × 108 × 1. 013 × 105

4 × 5. 67 × 10 −^8

) 1 / 4

= 2 × 107 K

4.66 (a)Power,P=σAT^4 = 4 πR^2 σT^4

= 4 π(7× 108 )^2 (5. 67 × 10 −^8 )(5,700)^4

= 3. 68 × 1026 W

Mass lost per second,m=P/c^2 =

3. 68 × 1026

(3× 108 )^2

= 4. 1 × 109 kg/s

(b) Time taken for the mass of sun (M) to decrease by 1% is

t=

M

100

×

1

m

=

2 × 1030

100

×

1

4. 1 × 109

= 4. 88 × 1018 s

=

4. 88 × 1018

3. 15 × 107

= 1. 55 × 1011 years