Theoretical Question 4 / Solutions Page 1/6

Theoretical Question 4: Heat conduction inside a solid sphere 1. Answers

(a) x

TTRjRJ inout22 4||4

−== σππ

(b) 2

)4( inout2 TTCDRQ

−= π

(c) 2(E)

4DCt

σ= = 3.0×10-3 s

(d) 2

cout

2(B)

)(24D

TTLD

Ct

−+=

ββ

β

σσ = 7.8×10-2 s.

(e) c)( TrT = for b0 Rr <<

b

cbout

b

outcb )(1)(RR

TRRTRR

TTRRr

rT−−

+−−

= for RrR <<b

(f) b

2b

coutb )(RRR

RL

TTdt

dR−

−= βσ .

(g) 2

cout )(6R

TTL

−=

βστ = 2.5 s.

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

xTT

rTj inout −−=∆∆

−= σσ . (a1)

Total heat flow rate J from the heat bath to the particle is

xTT

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 TTDCRQ −= π . (b1)

(c) [1 point] The change in Q is induced by J , that is

tJQ ∆=∆ , (c1) or,

tx

TTRx

TTCR ∆

−=∆

− inout2inout2 42

4 σππ . (c2)

Thus one obtains

xCdxdt

σ2= . (c3)

With the help of the integration formula, one obtains

2(E)

4DCt

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).

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

xTT

j cout −−= βσ (d1)

Total heat flow rate J from the heat bath to the particle is

xTT

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

LDRTT

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)

)(24D

TTLD

Ct

−+=

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 =drdTr σπ (e2)

By using the provided formula, one obtains

nn Y

rX

T += . (e3)

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

TRRTRR

TTRRr

rT−−

+−−

= for RrR <<b (e6)

(f) [2 points] According to the result in (e), jr 24π 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(

b2b

cout2

b

bb2

bb2

b

tRRR

RTTR

tRrjRrjRLRR

∆

−

−=

∆−=−+==∆

βσπ

ππ (f1)

Thus,

b2b

coutb )(RRR

RL

TTdt

dR−

−= βσ . (f2)

(g) [2 points] With the help of the provided formula (i), Eq. (f2) can be inverted to the following form,

RRRR

TTL

dRdt b

2b

inoutb )(−

−=

βσ, (g1)

which can be integrated by using the provided formula (ii) to produce

+−

−= constant

21

311

)(2b

3b

cout

RRRRTT

Ltβσ

. (g2)

The constant should be chosen so that 0=t for RR =b . Thus one finds

Theoretical Question 4 / Solutions Page 5/6

+−

−= 32

b3b

cout 61

21

311

)(RRRR

RTTLt

βσ. (g3)

When bR is replaced by 0, Eq. (g3) results in

2

cout )(6R

TTL

−=

βστ . (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)

)(2D

TTLt

−=

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).

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 jr 24π 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 jr 24π 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