1. Theoretical Question 4 / Solutions Page 1/6
Theoretical Question 4: Heat conduction inside a solid sphere
1. Answers
(a)
x
TT
RjRJ inout22
4||4
−
== σππ
(b)
2
)4( inout2 TT
CDRQ
−
= π
(c) 2(E)
4
D
C
t
σ
= = 3.0×10-3
s
(d) 2
cout
2(B)
)(24
D
TT
L
D
C
t
−
+=
ββ
β
σσ
= 7.8×10-2
s.
(e) c)( TrT = for b0 Rr <<
b
cbout
b
outcb )(1
)(
RR
TRRT
RR
TTRR
r
rT
−
−
+
−
−
= for RrR <<b
(f)
b
2
b
coutb
)(
RRR
R
L
TT
dt
dR
−
−
=
βσ
.
(g) 2
cout )(6
R
TT
L
−
=
βσ
τ = 2.5 s.

2. Theoretical Question 4 / Solutions Page 2/6
2. Solutions
(a) [0.5 points]
The heat flow rate per unit area at the outer surface of the particle is
x
TT
r
T
j inout −
−=
∆
∆
−= σσ . (a1)
Total heat flow rate J from the heat bath to the particle is
x
TT
RjRJ inout22
44
−
=−= σππ . (a2)
(b) [0.5 points]
Q =(heated volume)·C ·(average temperature change of the heated volume). To the
leading order of RD / ,
2
4 inout2 TT
DCRQ
−
= π . (b1)
(c) [1 point]
The change in Q is induced by J , that is
tJQ ∆=∆ , (c1)
or,
t
x
TT
Rx
TT
CR ∆
−
=∆
− inout2inout2
4
2
4 σππ . (c2)
Thus one obtains
x
C
dx
dt
σ2
= . (c3)
With the help of the integration formula, one obtains
2(E)
4
D
C
t
σ
= (c4)
The evaluation of this expression provides
E)(
t = 3.0×10-3
s (c5)
Alternative Solution:
Since Eq. (c3) is in the exactly same form as atdtdv =/ , where v corresponds to our
t , a corresponds to our σ2/C , and t corresponds to our x , students may use this
analogy to the accelerated motion to obtain (c4).
(*) Comment 1: Student may replace x in the result of (a2) by its average value 2/D
and divide Q by J to obtain Eq. (c4).

3. Theoretical Question 4 / Solutions Page 3/6
(*) Comment 2: Exact solution without the linear approximation indicates that the time
required to reach the slope DTT /)( inout − at Rr = is given by )/(2
πσCD ,
which is slightly larger than (E)
t since the linear approximation in (b) is bound to
underestimate Q .
(d) [2 points]
The outward heat flow per unit area at Rr = is
x
TT
j cout −
−= βσ (d1)
Total heat flow rate J from the heat bath to the particle is
x
TT
RjRJ cout22
44
−
=−= βσππ (d2)
as before. But the total amount of heat Q necessary to increase x from 0 to a finite
value D is changed to
LDR
TT
CDRQ )4(
2
)4( 2cout2
ππ β +
−
= (d3)
Here the second term arises due to the latent heat. In a similar way as in (c), one then
obtains
2
cout
2(B)
)(24
D
TT
L
D
C
t
−
+=
ββ
β
σσ
. (d4)
From the provided values of the relevant parameters, one obtains
(B)
t = 7.8×10-2
s. (d5)
(e) [2 points]
In the limit 0/ (B)(E)
→tt , the temperature profile within each phase is in its steady
state, that is,
)(4 2
rjrπ is independent of r within each phase (e1)
or
constant4 2
=
dr
dT
r σπ (e2)
By using the provided formula, one obtains
n
n
Y
r
X
T += . (e3)

4. Theoretical Question 4 / Solutions Page 4/6
The constants nX and nY should be chosen so that the expression gives the proper
value at each boundary, namely,
)0( =rT = finite, cb )( TRT = , and out)( TRT = . (e4)
This way, one obtains
c)( TrT = for bRr < (e5)
b
cbout
b
outcb )(1
)(
RR
TRRT
RR
TTRR
r
rT
−
−
+
−
−
= for RrR <<b (e6)
(f) [2 points]
According to the result in (e), jr2
4π is discontinuous at the phase boundary. Thus the
heat flux flowing into the phase boundary and the heat flux flowing out of the boundary
is not balanced, and this difference provides the necessary latent heat for the phase
transition. Therefore one obtains
( ) .4
)]0()0([4)4(
b
2
b
cout
2
b
bb
2
bb
2
b
t
RRR
R
TTR
tRrjRrjRLRR
∆





−
−=
∆−=−+==∆
βσπ
ππ
(f1)
Thus,
b
2
b
coutb
)(
RRR
R
L
TT
dt
dR
−
−
=
βσ
. (f2)
(g) [2 points]
With the help of the provided formula (i), Eq. (f2) can be inverted to the following form,
R
RRR
TT
L
dR
dt b
2
b
inoutb )(
−
−
=
βσ
, (g1)
which can be integrated by using the provided formula (ii) to produce






+−
−
= constant
2
1
3
11
)(
2
b
3
b
cout
RRR
RTT
L
t
βσ
. (g2)
The constant should be chosen so that 0=t for RR =b . Thus one finds

5. Theoretical Question 4 / Solutions Page 5/6






+−
−
= 32
b
3
b
cout 6
1
2
1
3
11
)(
RRRR
RTT
L
t
βσ
. (g3)
When bR is replaced by 0, Eq. (g3) results in
2
cout )(6
R
TT
L
−
=
βσ
τ . (g4)
Finally from the provided numerical values of relevant parameters, one obtains
τ = 2.5 s. (g5)
(*) Comment 1: If bR is replaced instead by DR − with RD << , Eq. (g3) results in
2
cout
(B)
)(2
D
TT
L
t
−
=
βσ
, which agrees with the result in Eq. (d4) except for the first
term, which is much smaller than the second term.
(*) Comment 2: The two expressions for τ and (B)
t are the same except for the factor,
1/6 vs. 1/2. This change of the numerical factor from 1/2 to 1/6 arises from the
reduction of the surface area of the phase boundary as bR approaches 0. While this
reduction is not important for RD << (and thus larger factor 1/2), it is not
negligible for D ~ R (and thus smaller factor 1/6).

6. Theoretical Question 4 / Solutions Page 6/6
3. Mark Distribution
No.
Total
Pt.
Partial
Pt.
Contents
0.2 −=J (surface area) j .
0.2 Correct j .
(a) 0.5
0.1 Correct expression for J .
0.2 CQ = (heated volume)∙(average temperature change).
0.2 Correct heated volume. Other answers, which differ in the second
leading order in RD/ , are also acceptable.
0.2 Correct average temperature change.
(b) 0.7
0.1 Correct expression for Q .
0.3 tJQ ∆=∆ .
0.3 )/(2/ Cxdtdx σ= .
0.5 Correct expression for E)(
t .
Integration may be replaced by analogy to accelerated motion.
(c) 1.4
0.3 Correct evaluation for E)(
t .
0.2 Correct expression for J .
0.6 Correct expression for Q .
0.4 Correct expression for B)(
t .
(d) 1.4
0.2 Correct evaluation for B)(
t .
1.0 jr2
4π is independent of r .
0.4 =T (const1)/ r +(const2).
0.5 Correct boundary conditions.
(e) 2.4
0.5 Correct expression for )(rT .
1.0 Latent heat is provided by the discontinuity in jr2
4π at the phase
boundary.
(f) 1.6
0.6 Correct expression for dtdR /b .
1.5 Correct expression for τ .(g) 2.0
0.5 Correct evaluation of τ .
Approximate result of (d) and
corresponding value: 1.0 pt
Total 10