501. Four rods P, Q, R and S are separately connected between the same two maintained temperatures. Their cross-sectional areas, lengths and thermal conductivities are:
| Rod | \(K\) | Area | Length |
|---|
| P | \(200\) | \(A\) | \(L\) |
| Q | \(100\) | \(2A\) | \(L\) |
| R | \(400\) | \(A/2\) | \(2L\) |
| S | \(50\) | \(4A\) | \(2L\) |
The numerical units of \(K\) are \(\text{W m}^{-1}\text{K}^{-1}\). Which ordering of steady heat currents is correct?
ⓐ. \(H_P\gt H_Q\gt H_R\gt H_S\)
ⓑ. \(H_P=H_Q=H_R=H_S\)
ⓒ. \(H_R=H_S\gt H_P=H_Q\)
ⓓ. \(H_P=H_Q\gt H_R=H_S\)
Correct Answer: \(H_P=H_Q\gt H_R=H_S\)
Explanation: For the common end-temperature difference,
\[
H\propto\frac{KA}{L}.
\]
Therefore only the geometrical-material factor \(KA/L\) needs to be compared.
For P,
\[
\frac{K_PA_P}{L_P}
=
\frac{200A}{L}.
\]
For Q,
\[
\frac{K_QA_Q}{L_Q}
=
\frac{(100)(2A)}{L}
=
\frac{200A}{L}.
\]
Thus,
\[
H_P=H_Q.
\]
For R,
\[
\frac{K_RA_R}{L_R}
=
\frac{(400)(A/2)}{2L}
=
\frac{100A}{L}.
\]
For S,
\[
\frac{K_SA_S}{L_S}
=
\frac{(50)(4A)}{2L}
=
\frac{100A}{L}.
\]
Hence,
\[
H_R=H_S.
\]
Comparing the common factors,
\[
H_P=H_Q=2H_R=2H_S.
\]
\(\textbf{Final answer:}\) \(H_P=H_Q\gt H_R=H_S\).
502. Two rods A and B are joined in series between a reservoir maintained at \(100^\circ\text{C}\) and a block of ice maintained at \(0^\circ\text{C}\) while melting. Both rods have area \(1.0\times10^{-4}\,\text{m}^2\), with
\[
K_A=200\,\text{W m}^{-1}\text{K}^{-1},\quad L_A=0.20\,\text{m},
\]
\[
K_B=100\,\text{W m}^{-1}\text{K}^{-1},\quad L_B=0.30\,\text{m}.
\]
Rod A is on the hot side. Side losses are negligible and steady state has been established. Take
\[
L_f=3.36\times10^5\,\text{J kg}^{-1}.
\]
What are the junction temperature and the mass of ice melted in \(30\,\text{min}\)?
ⓐ. \(50^\circ\text{C}\) and \(6.7\,\text{g}\)
ⓑ. \(75^\circ\text{C}\) and \(13.4\,\text{g}\)
ⓒ. \(75^\circ\text{C}\) and \(26.8\,\text{g}\)
ⓓ. \(25^\circ\text{C}\) and \(13.4\,\text{g}\)
Correct Answer: \(75^\circ\text{C}\) and \(13.4\,\text{g}\)
Explanation: This situation combines steady conduction with latent-heat transfer at the cold end.
\(\textbf{Thermal opposition of rod A:}\)
\[
R_A
=
\frac{L_A}{K_AA}
\]
Thus,
\[
R_A
=
\frac{0.20}
{(200)(1.0\times10^{-4})}
=
10\,\text{K W}^{-1}
\]
\(\textbf{Thermal opposition of rod B:}\)
\[
R_B
=
\frac{0.30}
{(100)(1.0\times10^{-4})}
=
30\,\text{K W}^{-1}
\]
Therefore,
\[
R_{\text{total}}=40\,\text{K W}^{-1}
\]
The maintained temperature difference is
\[
100\,\text{K}
\]
Hence the steady heat current is
\[
H=\frac{100}{40}=2.5\,\text{W}
\]
The drop across A is
\[
\Delta T_A=(2.5)(10)=25\,\text{K}
\]
Thus,
\[
T_J=100-25=75^\circ\text{C}
\]
\(\textbf{Energy delivered to the melting ice:}\)
\[
t=30\,\text{min}=1800\,\text{s}
\]
Therefore,
\[
Q=Ht=(2.5)(1800)=4500\,\text{J}
\]
At \(0^\circ\text{C}\), this energy produces melting:
\[
Q=mL_f
\]
Hence,
\[
m=
\frac{4500}{3.36\times10^5}
\approx1.34\times10^{-2}\,\text{kg}
\]
Thus,
\[
m\approx13.4\,\text{g}
\]
\(\textbf{Final answer:}\) Junction temperature \(=75^\circ\text{C}\); ice melted \(\approx13.4\,\text{g}\).
503. Two rods P and Q are made of the same material and have the same volume. Rod P has length \(L\) and cross-sectional area \(A\). Rod Q has length \(2L\). The same temperature difference is maintained across each rod. If \(H_P,H_Q\) are their steady heat currents, and P is used for time \(t\) while Q is used for time \(4t\), which relation is correct?
ⓐ. \(H_Q=H_P/2\) and both transfer the same heat
ⓑ. \(H_Q=H_P/8\) and both transfer the same heat
ⓒ. \(H_Q=H_P/4\) and Q transfers twice as much heat
ⓓ. \(H_Q=H_P/4\) and both transfer the same heat
Correct Answer: \(H_Q=H_P/4\) and both transfer the same heat
Explanation: The rods have equal volume:
\[
AL=A_Q(2L)
\]
Therefore,
\[
A_Q=\frac A2
\]
For steady conduction,
\[
H=KA\frac{\Delta T}{L}
\]
\(\textbf{Compare the heat currents:}\)
\[
\frac{H_Q}{H_P}
=
\frac{A_Q}{A}
\frac{L}{2L}
\]
Thus,
\[
\frac{H_Q}{H_P}
=
\left(\frac12\right)\left(\frac12\right)
=
\frac14
\]
Hence,
\[
H_Q=\frac{H_P}{4}
\]
\(\textbf{Compare total heat transferred:}\)
\[
Q=Ht
\]
For Q operating through \(4t\),
\[
Q_Q
=
\frac{H_P}{4}(4t)
=
H_Pt
\]
Therefore,
\[
Q_Q=Q_P
\]
The equal-volume condition changes both area and length, so the heat current does not scale merely as \(1/L\).
\(\textbf{Final answer:}\) \(H_Q=H_P/4\), and both rods transfer the same total heat in the stated times.
504. A wall of area \(1.0\,\text{m}^2\) initially contains a layer A of thickness \(0.080\,\text{m}\) and thermal conductivity \(0.80\,\text{W m}^{-1}\text{K}^{-1}\). An insulating layer B of thermal conductivity \(0.040\,\text{W m}^{-1}\text{K}^{-1}\) is added in series on the colder side. The outer surfaces are maintained at \(40^\circ\text{C}\) and \(0^\circ\text{C}\). What thickness of B is required to reduce the steady heat current to \(16\,\text{W}\), and what is the A-B interface temperature?
ⓐ. \(4.8\,\text{cm}\) and \(32.0^\circ\text{C}\)
ⓑ. \(8.0\,\text{cm}\) and \(38.4^\circ\text{C}\)
ⓒ. \(9.6\,\text{cm}\) and \(38.4^\circ\text{C}\)
ⓓ. \(9.6\,\text{cm}\) and \(24.0^\circ\text{C}\)
Correct Answer: \(9.6\,\text{cm}\) and \(38.4^\circ\text{C}\)
Explanation: For layers in series, define the useful derived quantity
\[
R=\frac{L}{KA}
\]
The total steady heat current is
\[
H=\frac{\Delta T}{R_A+R_B}
\]
\(\textbf{Required total thermal opposition:}\)
\[
R_{\text{total}}
=
\frac{40}{16}
=
2.5\,\text{K W}^{-1}
\]
For layer A,
\[
R_A
=
\frac{0.080}{(0.80)(1.0)}
=
0.10\,\text{K W}^{-1}
\]
Therefore,
\[
R_B=2.5-0.10=2.40\,\text{K W}^{-1}
\]
\(\textbf{Required insulation thickness:}\)
\[
R_B=\frac{L_B}{K_BA}
\]
Hence,
\[
L_B=(2.40)(0.040)(1.0)
\]
Thus,
\[
L_B=0.096\,\text{m}=9.6\,\text{cm}
\]
\(\textbf{Temperature drop across layer A:}\)
\[
\Delta T_A=HR_A
\]
Therefore,
\[
\Delta T_A=(16)(0.10)=1.6\,\text{K}
\]
Since A is on the hot side,
\[
T_J=40-1.6=38.4^\circ\text{C}
\]
Most of the temperature drop occurs across the low-conductivity insulation.
\(\textbf{Final answer:}\)
\[
\boxed{L_B=9.6\,\text{cm},\quad T_J=38.4^\circ\text{C}}
\]
505. Two rods A and B of equal length and initially equal cross-sectional area are joined in series between \(100^\circ\text{C}\) and \(20^\circ\text{C}\). A is on the hot side and has thermal conductivity
\[
K_A=200\,\text{W m}^{-1}\text{K}^{-1}.
\]
The initial steady junction temperature is \(80^\circ\text{C}\). Rod B is then replaced by another rod of the same material and length but twice its original cross-sectional area, while A and the boundary temperatures remain unchanged. What are \(K_B\) and the new junction temperature?
ⓐ. \(K_B=50\,\text{W m}^{-1}\text{K}^{-1}\), new \(T_J=60^\circ\text{C}\)
ⓑ. \(K_B=66.7\,\text{W m}^{-1}\text{K}^{-1}\), new \(T_J=80^\circ\text{C}\)
ⓒ. \(K_B=100\,\text{W m}^{-1}\text{K}^{-1}\), new \(T_J=60^\circ\text{C}\)
ⓓ. \(K_B=66.7\,\text{W m}^{-1}\text{K}^{-1}\), new \(T_J=68^\circ\text{C}\)
Correct Answer: \(K_B=66.7\,\text{W m}^{-1}\text{K}^{-1}\), new \(T_J=68^\circ\text{C}\)
Explanation: Initially, both rods have equal length and equal area.
The common steady heat current therefore requires
\[
K_A(100-80)=K_B(80-20)
\]
\(\textbf{Infer the conductivity of B:}\)
\[
(200)(20)=K_B(60)
\]
Hence,
\[
K_B=\frac{4000}{60}\approx66.7\,\text{W m}^{-1}\text{K}^{-1}
\]
\(\textbf{Changed geometry of B:}\)
Its new area is
\[
A_B'=2A
\]
Let the new junction temperature be \(T\).
Equality of heat currents gives
\[
K_AA(100-T)
=
K_B(2A)(T-20)
\]
Canceling the common area,
\[
200(100-T)
=
(66.7)(2)(T-20)
\]
Thus,
\[
20000-200T
=
133.4T-2668
\]
Hence,
\[
22668\approx333.4T
\]
Therefore,
\[
T\approx68^\circ\text{C}
\]
Increasing B's area reduces its temperature drop and shifts the junction temperature downward from \(80^\circ\text{C}\).
\(\textbf{Final answer:}\) \(K_B\approx66.7\,\text{W m}^{-1}\text{K}^{-1}\), new junction temperature \(\approx68^\circ\text{C}\).
506. Three materials have the following properties:
| Material | \(K\) \((\text{W m}^{-1}\text{K}^{-1})\) | \(c\) \((\text{J kg}^{-1}\text{K}^{-1})\) |
|---|
| P | 400 | 400 |
| Q | 0.050 | 800 |
| R | 50 | 900 |
Equal-sized slabs are placed between the same maintained temperatures, and separately equal masses of the materials receive the same heat without phase change. Which conclusion is correct?
ⓐ. P gives the smallest conductive heat current; Q gives the smallest temperature rise
ⓑ. Q gives the smallest conductive heat current; R gives the smallest temperature rise
ⓒ. R gives the smallest conductive heat current; P gives the smallest temperature rise
ⓓ. Q gives the largest conductive heat current; P gives the smallest temperature rise
Correct Answer: Q gives the smallest conductive heat current; R gives the smallest temperature rise
Explanation: For equal geometry and the same temperature difference,
\[
H\propto K.
\]
The smallest thermal conductivity is that of Q:
\[
K_Q=0.050\,\text{W m}^{-1}\text{K}^{-1}.
\]
Therefore Q gives the smallest steady conductive heat current.
For equal masses receiving equal heat,
\[
Q_{\text{heat}}=mc\Delta T.
\]
Thus,
\[
\Delta T\propto\frac1c.
\]
The largest specific heat capacity is that of R:
\[
c_R=900\,\text{J kg}^{-1}\text{K}^{-1}.
\]
Therefore R undergoes the smallest temperature rise.
Thermal conductivity and specific heat capacity describe different physical properties.
A good thermal insulator need not have the largest specific heat capacity.
Hence option B is correct.
507. A slab of material P has thickness \(20\,\text{cm}\) and thermal conductivity \(0.80\,\text{W m}^{-1}\text{K}^{-1}\). It is to be replaced by a slab of material Q having thermal conductivity \(0.040\,\text{W m}^{-1}\text{K}^{-1}\). The area and maintained temperature difference are unchanged. What thickness of Q gives the same steady conductive heat current?
ⓐ. \(0.5\,\text{cm}\)
ⓑ. \(1.0\,\text{cm}\)
ⓒ. \(4.0\,\text{cm}\)
ⓓ. \(10\,\text{cm}\)
Correct Answer: \(1.0\,\text{cm}\)
Explanation: For unchanged area and temperature difference,
\[
H=KA\frac{\Delta T}{L}
\]
Equal heat currents require
\[
\frac{K_P}{L_P}
=
\frac{K_Q}{L_Q}
\]
Therefore,
\[
L_Q
=
L_P\frac{K_Q}{K_P}
\]
Substitution gives
\[
L_Q
=
20\,\text{cm}
\left(\frac{0.040}{0.80}\right)
\]
Hence,
\[
L_Q
=
20\left(\frac1{20}\right)\text{cm}
=
1.0\,\text{cm}
\]
The much lower conductivity of Q allows a much thinner layer to provide the same reduction of conductive heat transfer.
\(\textbf{Final answer:}\)
\[
\boxed{1.0\,\text{cm}}
\]
508. Four experiments are claimed to use rods of the same material with
\[
K=100\,\text{W m}^{-1}\text{K}^{-1}.
\]
| Experiment | Area \((\text{m}^2)\) | Length \((\text{m})\) | \(\Delta T\) \((\text{K})\) | Reported \(H\) |
|---|
| I | \(2.0\times10^{-4}\) | 0.50 | 50 | \(2.0\,\text{W}\) |
| II | \(4.0\times10^{-4}\) | 1.00 | 50 | \(2.0\,\text{W}\) |
| III | \(2.0\times10^{-4}\) | 0.25 | 25 | \(2.0\,\text{W}\) |
| IV | \(1.0\times10^{-4}\) | 0.50 | 100 | \(4.0\,\text{W}\) |
Which experiment is inconsistent with steady one-dimensional conduction?
ⓐ. I
ⓑ. II
ⓒ. III
ⓓ. IV
Correct Answer: IV
Explanation: Each experiment should satisfy
\[
H=KA\frac{\Delta T}{L}.
\]
For I,
\[
H
=
(100)(2.0\times10^{-4})\frac{50}{0.50}
=
2.0\,\text{W}.
\]
Thus I is consistent.
For II,
\[
H
=
(100)(4.0\times10^{-4})\frac{50}{1.00}
=
2.0\,\text{W}.
\]
Thus II is consistent.
For III,
\[
H
=
(100)(2.0\times10^{-4})\frac{25}{0.25}
=
2.0\,\text{W}.
\]
Thus III is consistent.
For IV,
\[
H
=
(100)(1.0\times10^{-4})\frac{100}{0.50}.
\]
Therefore,
\[
H=2.0\,\text{W},
\]
not \(4.0\,\text{W}\).
Hence experiment IV is inconsistent.
509. In an experiment to determine thermal conductivity, one end of a rod is maintained \(60\,\text{K}\) hotter than the other. The rod has length \(0.50\,\text{m}\) and area \(2.0\times10^{-4}\,\text{m}^2\). Heat conducted through the rod melts \(3.60\,\text{g}\) of ice in \(5.0\,\text{min}\). Take
\[
L_f=3.36\times10^5\,\text{J kg}^{-1}.
\]
Assume steady state and negligible side loss. What is the rod's thermal conductivity? If its area is then doubled and its length halved while the same temperature difference is maintained, how much ice melts in \(5.0\,\text{min}\)?
ⓐ. \(84\,\text{W m}^{-1}\text{K}^{-1}\) and \(7.2\,\text{g}\)
ⓑ. \(168\,\text{W m}^{-1}\text{K}^{-1}\) and \(7.2\,\text{g}\)
ⓒ. \(168\,\text{W m}^{-1}\text{K}^{-1}\) and \(14.4\,\text{g}\)
ⓓ. \(336\,\text{W m}^{-1}\text{K}^{-1}\) and \(28.8\,\text{g}\)
Correct Answer: \(168\,\text{W m}^{-1}\text{K}^{-1}\) and \(14.4\,\text{g}\)
Explanation: The melted ice measures the total conducted energy.
\(\textbf{Convert the melting observation into heat current:}\)
\[
Q=mL_f
\]
Thus,
\[
Q
=
(3.60\times10^{-3})(3.36\times10^5)
=
1209.6\,\text{J}
\]
The time is
\[
t=300\,\text{s}
\]
Therefore,
\[
H=\frac{Q}{t}
=
\frac{1209.6}{300}
=
4.032\,\text{W}
\]
\(\textbf{Infer the thermal conductivity:}\)
\[
H=KA\frac{\Delta T}{L}
\]
Hence,
\[
K
=
\frac{HL}{A\Delta T}
\]
Substitution gives
\[
K
=
\frac{(4.032)(0.50)}
{(2.0\times10^{-4})(60)}
\]
Therefore,
\[
K=168\,\text{W m}^{-1}\text{K}^{-1}
\]
\(\textbf{Changed geometry:}\)
The area doubles while the length halves:
\[
\frac{H'}{H}
=
2\times2
=
4
\]
For the same time and latent heat,
\[
m'\propto H'
\]
Thus,
\[
m'=4(3.60\,\text{g})=14.4\,\text{g}
\]
\(\textbf{Final answer:}\) \(K=168\,\text{W m}^{-1}\text{K}^{-1}\); new melted mass \(=14.4\,\text{g}\).
510. A uniform rod carries heat steadily between fixed end temperatures. Consider two independent modifications of the original rod.
Process P: the rod is cut transversely into two equal lengths, and one half is used between the same end temperatures.
Process Q: an identical original rod is cut lengthwise into two equal cross-sectional halves, and one half is used with the original length between the same temperatures.
If the original heat current is \(H\), what are the currents in P and Q?
ⓐ. \(H_P=H/2,\ H_Q=2H\)
ⓑ. \(H_P=H/2,\ H_Q=H/2\)
ⓒ. \(H_P=2H,\ H_Q=2H\)
ⓓ. \(H_P=2H,\ H_Q=H/2\)
Correct Answer: \(H_P=2H,\ H_Q=H/2\)
Explanation: For a uniform rod,
\[
H=KA\frac{\Delta T}{L}
\]
\(\textbf{Transverse cut:}\)
Using one half makes the new length
\[
L_P=\frac L2,
\]
while the area remains \(A\).
Therefore,
\[
H_P
=
KA\frac{\Delta T}{L/2}
=
2H
\]
\(\textbf{Longitudinal cut:}\)
Using one lengthwise half leaves the length unchanged but gives
\[
A_Q=\frac A2
\]
Hence,
\[
H_Q
=
K\frac A2\frac{\Delta T}{L}
=
\frac H2
\]
Both pieces may contain the same material, but changing the cut direction changes different geometrical factors in the conduction relation.
\(\textbf{Final answer:}\)
\[
\boxed{H_P=2H,\ H_Q=H/2}
\]
511. A uniform metal rod has length \(1.0\,\text{m}\) at a uniform reference temperature of \(20^\circ\text{C}\) and coefficient of linear expansion
\[
\alpha=2.0\times10^{-5}\,\text{K}^{-1}.
\]
Its ends are then maintained at \(100^\circ\text{C}\) and \(20^\circ\text{C}\). After steady one-dimensional conduction is established, its temperature varies linearly along its length and the rod remains free to expand. Neglect the tiny effect of expansion on the temperature profile. What is its total increase in length? How does this compare with the increase if the whole rod were instead uniformly at \(100^\circ\text{C}\)?
ⓐ. \(0.4\,\text{mm}\), one-fourth as large
ⓑ. \(0.8\,\text{mm}\), one-half as large
ⓒ. \(1.6\,\text{mm}\), equal
ⓓ. \(1.6\,\text{mm}\), twice as large
Correct Answer: \(0.8\,\text{mm}\), one-half as large
Explanation: In steady conduction through a uniform rod, the temperature profile is linear.
Relative to the \(20^\circ\text{C}\) reference, the local temperature rise varies linearly from
\[
80\,\text{K}
\]
at one end to
\[
0
\]
at the other.
\(\textbf{Average thermal rise along the rod:}\)
For a linear profile,
\[
\Delta T_{\text{avg}}
=
\frac{80+0}{2}
=
40\,\text{K}
\]
The total free expansion is obtained by summing the local expansions along the rod:
\[
\Delta L
=
\alpha L\Delta T_{\text{avg}}
\]
Thus,
\[
\Delta L
=
(2.0\times10^{-5})(1.0)(40)
\]
Therefore,
\[
\Delta L=8.0\times10^{-4}\,\text{m}=0.8\,\text{mm}
\]
\(\textbf{Uniform 100^\circ\text{C} comparison:}\)
Then the whole rod has
\[
\Delta T=80\,\text{K}
\]
Hence,
\[
\Delta L_{\text{uniform}}
=
(2.0\times10^{-5})(1.0)(80)
=
1.6\,\text{mm}
\]
Thus the steady-gradient expansion is one-half as large.
\(\textbf{Final answer:}\) \(0.8\,\text{mm}\), which is half the uniform-\(100^\circ\text{C}\) expansion.
512. A metal layer and an insulating layer of equal area are in series between fixed temperatures. Initially their values of \(L/K\) are in the ratio
\[
\left(\frac{L}{K}\right)_{\text{metal}}:
\left(\frac{L}{K}\right)_{\text{insulator}}
=
1:3.
\]
The thickness of only the insulating layer is then halved. All conductivities and boundary temperatures remain unchanged. By what factor does the steady heat current change?
ⓐ. \(4/3\)
ⓑ. \(5/2\)
ⓒ. \(2\)
ⓓ. \(8/5\)
Correct Answer: \(8/5\)
Explanation: For equal area, the quantity controlling the temperature drop of each layer is proportional to
\[
\frac{L}{K}
\]
Initially the total is proportional to
\[
1+3=4
\]
Heat current varies inversely with this total:
\[
H\propto\frac1{4}
\]
\(\textbf{After halving only the insulating thickness:}\)
Its \(L/K\) contribution becomes
\[
\frac32
\]
The new total is therefore
\[
1+\frac32=\frac52
\]
Thus,
\[
\frac{H'}{H}
=
\frac{4}{5/2}
\]
Hence,
\[
\frac{H'}{H}
=
\frac85
\]
The heat current does not double because the unchanged metal layer still contributes to the total opposition to heat flow.
\(\textbf{Final answer:}\) The heat current increases by a factor \(8/5\).
513. A cooking vessel is to be made using one material for its base and another for a handle. The following data are available:
| Material | Thermal conductivity \(K\) \((\text{W m}^{-1}\text{K}^{-1})\) | Specific heat \(c\) \((\text{J kg}^{-1}\text{K}^{-1})\) |
|---|
| P | 380 | 390 |
| Q | 0.12 | 1500 |
| R | 45 | 460 |
If the design priority is rapid conduction through the base and suppression of conduction along the handle, which selection is most appropriate on the basis of the listed data?
ⓐ. Base Q, handle P
ⓑ. Base P, handle Q
ⓒ. Base R, handle P
ⓓ. Base Q, handle R
Correct Answer: Base P, handle Q
Explanation: For a given geometry and temperature difference,
\[
H\propto K.
\]
A cooking-vessel base intended to conduct thermal energy rapidly should therefore use a material with large thermal conductivity.
Among the listed materials,
\[
K_P=380\,\text{W m}^{-1}\text{K}^{-1}
\]
is the largest.
A handle intended to suppress conduction toward the hand should use a material with small thermal conductivity.
Here,
\[
K_Q=0.12\,\text{W m}^{-1}\text{K}^{-1}
\]
is by far the smallest.
The listed specific heat capacities do not reverse these conduction-based design choices.
Specific heat describes the energy required for a temperature change of a given mass; it is not the same property as thermal conductivity.
Therefore P is appropriate for the conducting base and Q for the insulating handle.
514.
For a family of rods made from the same material, all having length \(0.50\,\text{m}\) and maintained end-temperature difference \(40\,\text{K}\), a graph of steady heat current \(H\) against cross-sectional area \(A\) is a straight line through the origin with slope \(8.0\times10^3\,\text{W m}^{-2}\).
What is the thermal conductivity of the material? If a rod of this material has area \(3.0\times10^{-4}\,\text{m}^2\), length \(0.25\,\text{m}\), and end-temperature difference \(20\,\text{K}\), what heat current does it carry?
ⓐ. \(K=50\,\text{W m}^{-1}\text{K}^{-1}\), \(H=1.2\,\text{W}\)
ⓑ. \(K=100\,\text{W m}^{-1}\text{K}^{-1}\), \(H=4.8\,\text{W}\)
ⓒ. \(K=200\,\text{W m}^{-1}\text{K}^{-1}\), \(H=4.8\,\text{W}\)
ⓓ. \(K=100\,\text{W m}^{-1}\text{K}^{-1}\), \(H=2.4\,\text{W}\)
Correct Answer: \(K=100\,\text{W m}^{-1}\text{K}^{-1}\), \(H=2.4\,\text{W}\)
Explanation: For the graph conditions,
\[
H=KA\frac{\Delta T}{L}
\]
Therefore the graph slope is
\[
\frac{H}{A}
=
K\frac{\Delta T}{L}
\]
\(\textbf{Infer K from the slope:}\)
\[
8.0\times10^3
=
K\frac{40}{0.50}
\]
Thus,
\[
8.0\times10^3=80K
\]
Hence,
\[
K=100\,\text{W m}^{-1}\text{K}^{-1}
\]
\(\textbf{Changed rod dimensions and temperature difference:}\)
\[
H
=
KA\frac{\Delta T}{L}
\]
Substitution gives
\[
H
=
(100)(3.0\times10^{-4})
\frac{20}{0.25}
\]
Therefore,
\[
H=2.4\,\text{W}
\]
The graph provides a material property only after its geometrical and thermal factors are separated.
\(\textbf{Final answer:}\)
\[
\boxed{K=100\,\text{W m}^{-1}\text{K}^{-1},\quad H=2.4\,\text{W}}
\]
515.
A student claims that a uniform rod of constant thermal conductivity and constant cross-sectional area has reached steady one-dimensional conduction with perfectly insulated sides and no internal heat source. However, the measured temperature-versus-position graph is smoothly curved rather than a straight line.
Which conclusion is most appropriate?
ⓐ. The curved graph confirms all the stated assumptions
ⓑ. At least one of the stated ideal assumptions is not satisfied
ⓒ. The thermal conductivity must be zero
ⓓ. A curved graph is required because heat current increases continuously along a uniform rod
Correct Answer: At least one of the stated ideal assumptions is not satisfied
Explanation: For a uniform rod in steady one-dimensional conduction,
\[
H=KA\left|\frac{dT}{dx}\right|.
\]
Under the stated assumptions, \(H\), \(K\) and \(A\) are constant along the rod.
Therefore,
\[
\left|\frac{dT}{dx}\right|
\]
must also be constant.
A constant temperature gradient produces a linear temperature-position graph.
A smoothly curved measured profile is therefore incompatible with the complete idealised claim.
It may indicate, for example, violation of uniformity, side heat loss, internal heat production, non-steady behaviour, or temperature-dependent properties.
The observation alone does not identify which assumption failed.
It only establishes that the full stated ideal model cannot simultaneously be correct.
Hence option B is the defensible conclusion.
516. Two rods A and B are to be joined in series between \(100^\circ\text{C}\) and \(20^\circ\text{C}\). The designer wants the junction temperature to be exactly \(60^\circ\text{C}\). Rod A has
\[
L_A=0.20\,\text{m},\quad
K_A=200\,\text{W m}^{-1}\text{K}^{-1},\quad
A_A=1.0\times10^{-4}\,\text{m}^2.
\]
Which choice for rod B will achieve the required midpoint junction temperature?
ⓐ. \(L_B=0.10\,\text{m},\ K_B=100\,\text{W m}^{-1}\text{K}^{-1},\ A_B=1.0\times10^{-4}\,\text{m}^2\)
ⓑ. \(L_B=0.20\,\text{m},\ K_B=100\,\text{W m}^{-1}\text{K}^{-1},\ A_B=1.0\times10^{-4}\,\text{m}^2\)
ⓒ. \(L_B=0.10\,\text{m},\ K_B=200\,\text{W m}^{-1}\text{K}^{-1},\ A_B=2.0\times10^{-4}\,\text{m}^2\)
ⓓ. \(L_B=0.20\,\text{m},\ K_B=400\,\text{W m}^{-1}\text{K}^{-1},\ A_B=2.0\times10^{-4}\,\text{m}^2\)
Correct Answer: \(L_B=0.10\,\text{m},\ K_B=100\,\text{W m}^{-1}\text{K}^{-1},\ A_B=1.0\times10^{-4}\,\text{m}^2\)
Explanation: The desired junction temperature is exactly halfway between the boundary temperatures:
\[
100-60=60-20=40\,\text{K}
\]
Thus the two rods must have equal temperature drops.
Because the heat current is common in steady state,
\[
\Delta T=H\frac{L}{KA}
\]
Equal temperature drops therefore require
\[
\frac{L_A}{K_AA_A}
=
\frac{L_B}{K_BA_B}
\]
\(\textbf{Required value from rod A:}\)
\[
\frac{L_A}{K_AA_A}
=
\frac{0.20}
{(200)(1.0\times10^{-4})}
=
10\,\text{K W}^{-1}
\]
For option A,
\[
\frac{L_B}{K_BA_B}
=
\frac{0.10}
{(100)(1.0\times10^{-4})}
=
10\,\text{K W}^{-1}
\]
Thus option A gives equal temperature drops and the required junction temperature.
The other choices give different values of \(L/(KA)\).
\(\textbf{Final answer:}\) Option A.
517. Consider the following statements about steady conduction.
Statement I: Thermal conductivity is a material property, whereas the heat current also depends on geometry and temperature difference.
Statement II: Two series-connected sections carrying the same steady heat current can have different temperature gradients.
Statement III: Doubling the thickness of a single uniform insulating slab while keeping all other quantities fixed halves its steady conductive heat current.
Statement IV: For two different materials, the one with smaller thermal conductivity must necessarily have smaller specific heat capacity.
Statement V: Total heat transferred during a steady interval is \(Q=Ht\).
Which statements are correct?
ⓐ. I, II, III and V only
ⓑ. I, III and IV only
ⓒ. II, IV and V only
ⓓ. I, II, III, IV and V
Correct Answer: I, II, III and V only
Explanation: Statement I is correct because
\[
H=KA\frac{\Delta T}{L}
\]
contains both the material property \(K\) and geometrical/thermal factors.
Statement II is correct.
For a common steady heat current,
\[
\left|\frac{dT}{dx}\right|=\frac{H}{KA},
\]
so different \(K\) or \(A\) values produce different gradients.
Statement III is correct because
\[
H\propto\frac1L
\]
for a single uniform slab under fixed \(K\), \(A\) and \(\Delta T\).
Statement IV is incorrect.
Thermal conductivity and specific heat capacity are distinct physical properties and do not have the asserted necessary ordering.
Statement V is correct because steady heat current is energy transferred per unit time.
Hence Statements I, II, III and V only are correct.
518. Two cylindrical rods of the same material and same length are maintained between the same two temperatures. Rod P has diameter \(d\), while rod Q has diameter \(2d\). In a second comparison, Q is made twice as long while retaining diameter \(2d\). Which statement is correct?
ⓐ. Initially \(H_Q=2H_P\); after doubling Q's length, \(H_Q=H_P\)
ⓑ. Initially \(H_Q=4H_P\); after doubling Q's length, \(H_Q=2H_P\)
ⓒ. Initially \(H_Q=4H_P\); after doubling Q's length, \(H_Q=4H_P\)
ⓓ. Initially \(H_Q=8H_P\); after doubling Q's length, \(H_Q=4H_P\)
Correct Answer: Initially \(H_Q=4H_P\); after doubling Q's length, \(H_Q=2H_P\)
Explanation: For a cylindrical rod,
\[
A=\frac{\pi d^2}{4}
\]
Therefore,
\[
A\propto d^2
\]
When the diameter doubles,
\[
A_Q=4A_P
\]
For equal material, length and temperature difference,
\[
H\propto A
\]
Hence initially,
\[
H_Q=4H_P
\]
\(\textbf{Changed length of Q:}\)
Heat current also varies as
\[
H\propto\frac1L
\]
Doubling Q's length therefore halves its previous current:
\[
H_Q'
=
\frac12(4H_P)
=
2H_P
\]
The diameter change acts through area, so treating \(H\) as directly proportional to diameter would miss the square dependence.
\(\textbf{Final answer:}\) Initially \(H_Q=4H_P\); after doubling Q's length, \(H_Q=2H_P\).
519. Two insulating materials P and Q are available in equal masses \(m\) to form plane slabs of the same face area \(A\). Their densities and thermal conductivities are
\[
\rho_P=100\,\text{kg m}^{-3},\quad K_P=0.040\,\text{W m}^{-1}\text{K}^{-1},
\]
\[
\rho_Q=300\,\text{kg m}^{-3},\quad K_Q=0.020\,\text{W m}^{-1}\text{K}^{-1}.
\]
Each entire mass is formed into a slab and placed between the same maintained temperatures. Which material gives the smaller conductive heat current, and by what factor?
ⓐ. P; \(H_P/H_Q=2/3\)
ⓑ. P; \(H_P/H_Q=1/3\)
ⓒ. Q; \(H_Q/H_P=1/2\)
ⓓ. Q; \(H_Q/H_P=2/3\)
Correct Answer: P; \(H_P/H_Q=2/3\)
Explanation: Equal masses and equal face areas do not give equal slab thicknesses.
For a slab,
\[
m=\rho AL
\]
Therefore,
\[
L=\frac{m}{\rho A}
\]
The conduction relation is
\[
H=KA\frac{\Delta T}{L}
\]
Substituting the mass-based thickness,
\[
H
=
KA\Delta T
\frac{\rho A}{m}
\]
Thus, for common \(m\), \(A\) and \(\Delta T\),
\[
H\propto K\rho
\]
\(\textbf{Compare the two materials:}\)
For P,
\[
K_P\rho_P=(0.040)(100)=4
\]
For Q,
\[
K_Q\rho_Q=(0.020)(300)=6
\]
Hence,
\[
\frac{H_P}{H_Q}
=
\frac46
=
\frac23
\]
Although Q has the smaller conductivity, its much larger density makes an equal-mass slab thinner.
Therefore P gives the smaller heat current under the stated equal-mass design constraint.
\(\textbf{Final answer:}\) P, with \(H_P/H_Q=2/3\).
520. A composite wall is made of two layers A and B in series. At steady state, the temperature drop across A is \(15\,\text{K}\) and across B is \(45\,\text{K}\). The layers have equal area. Layer A has
\[
L_A=0.10\,\text{m},\qquad K_A=0.20\,\text{W m}^{-1}\text{K}^{-1}.
\]
Layer B has thickness \(0.15\,\text{m}\). The entire wall is then rebuilt with B's thickness doubled, while the boundary temperatures and all material properties remain unchanged. What is \(K_B\), and what fraction of the original heat current flows through the rebuilt wall?
ⓐ. \(K_B=0.20\,\text{W m}^{-1}\text{K}^{-1}\), new current \(=1/2\) original
ⓑ. \(K_B=0.10\,\text{W m}^{-1}\text{K}^{-1}\), new current \(=4/7\) original
ⓒ. \(K_B=0.10\,\text{W m}^{-1}\text{K}^{-1}\), new current \(=2/3\) original
ⓓ. \(K_B=0.05\,\text{W m}^{-1}\text{K}^{-1}\), new current \(=4/7\) original
Correct Answer: \(K_B=0.10\,\text{W m}^{-1}\text{K}^{-1}\), new current \(=4/7\) original
Explanation: In steady series conduction, the same heat current passes through both layers.
Therefore their temperature drops are proportional to
\[
R=\frac{L}{KA}
\]
Since the areas are equal,
\[
\frac{\Delta T_B}{\Delta T_A}
=
\frac{L_B/K_B}{L_A/K_A}
\]
\(\textbf{Infer the conductivity of B:}\)
\[
\frac{45}{15}
=
\frac{0.15/K_B}{0.10/0.20}
\]
Thus,
\[
3
=
\frac{0.15/K_B}{0.50}
\]
Hence,
\[
\frac{0.15}{K_B}=1.50
\]
Therefore,
\[
K_B=0.10\,\text{W m}^{-1}\text{K}^{-1}
\]
\(\textbf{Compare total thermal opposition before and after rebuilding:}\)
Using the common area only as a common factor,
\[
R_A\propto\frac{0.10}{0.20}=0.50,
\]
\[
R_B\propto\frac{0.15}{0.10}=1.50
\]
Thus initially,
\[
R_{\text{old}}\propto2.00
\]
After doubling B's thickness,
\[
R_B'\propto\frac{0.30}{0.10}=3.00
\]
Hence,
\[
R_{\text{new}}\propto0.50+3.00=3.50
\]
For fixed boundary-temperature difference,
\[
H\propto\frac1R
\]
Therefore,
\[
\frac{H_{\text{new}}}{H_{\text{old}}}
=
\frac{2.00}{3.50}
=
\frac47
\]
\(\textbf{Final answer:}\) \(K_B=0.10\,\text{W m}^{-1}\text{K}^{-1}\), and the rebuilt wall carries \(4/7\) of the original heat current.
