Problems and Solutions on Thermodynamics and Statistical Mechanics

ThemcdyMmice 157

Solution:
(a) Because the material is uniform, we can assume the heat conduc-
tivity is uniform too. According to the formulas dQ = -k(dT/dr)sdt and
s = 27rlor, we have
dQ/dt = -27rlorkdT/dr.
Since dQ/dt is independent of r, we require dT/dr = A/r, where A is a
constant. Then T(r) = Alnr + B.
F'rom the boundary conditions, we have
T2-T1,
r2
In r2 - In -
rl rl

A=---- B= TI In r2 - T2 In rl

3

where TI = 373 K and T2 = 273 K, so that

1

T(r) = In rl - In r2

[(TI - T2) In r + T2 In rl - TI In r2]

(b) This is an irreversible adiabatic process, so that the entropy in-

crease s.

1159

When there is heat flow in a heat conducting material, there is an

increase in entropy. Find the local rate of entropy generation per unit vol-

ume in a heat conductor of given heat conductivity and given temperature

gradient.

(UC, Berkeley)

Solution:

then du = TdS. The heat conduction equation is

If we neglect volume expansion inside the heat conducting material,

duldt + V .q = 0.

Hence

dS/dt = -V. q/T = -V. (q/T) + 9. V(l/T) ,

where q/T is the entropy flow, and 9. V (i) is the irreversible entropy

increase due to the inhomogeneous temperature distribution. Thus, the

local rate of entropy generation per unit volume is