Chapter 17: Temperature and Heat Answers & Hints-Updated 10/17/10
The Constant-Volume Gas Thermometer and the Kelvin Scale
1. Thermometers measure their own temperature but they can be used to measure the temperature of other objects
because of the “Zeroth Law of Thermodynamics”.
a) Explain this statement. How does a thermometer affect the temperature that it’s measuring? Review the 0th
Law…thermometers come to thermal equilibrium with the object they are in contact with so that the thermometer
is at the same temp as the object. In the process, heat “flows” between the object and thermometer so that the
object’s temp changes from what it was before the thermometer was introduced. Hopefully this change is
negligible.
b) What would be the difficulty in measuring the temperature of a very small amount of fluid, such as a
thimbleful of water? Since the thermometer extracts energy from the fluid, if the fluid volume is small the energy
it losses (or gains) could significantly affect its original temperature…see above.
2. In principle, any gas can be used in a constant-volume gas thermometer.
a) What are the practical limitations of this principle? For example, is it possible to use oxygen for temperatures
as low as 15 K? (Look at the data on phase change temperatures in the textbook). Is there a gas that could be used
at such low temperatures? Practical limitation is that no substance is gaseous at all temps.. At some point the gas
will liquefy or solidify. Helium will remain gaseous the longest…
b) A constant-volume gas thermometer registers a pressure of 0.062 atm when it is at a temperature of 450 K. (i)
What is the pressure at the triple point of water (0.01 ºC)? (ii) What is the temperature when the pressure reads
0.015 atm? Set up proportions P1/P2= T1/T2, making sure the temps are in K. (i) Ptp=0.038 atm; (ii) 109K.
3. Practicing your (temperature) scales…Use the conversion formula: T(oF)= (9/5)T( oC) +32
a) A substance is heated from -12°F to 150°F. What is its change in temperature on (i) the Celsius scale and (ii)
the Kelvin scale? (i) 90 Co and (ii) 90 K.
b) How many degrees Fahrenheit (ºF) are there in one degree Celsius ºC)? 1 Co =1.8 Fo
c) At what temperature are the readings from a Fahrenheit thermometer and Celsius thermometer the same?
-40oC=-40 oF
Thermal Expansion of Solids and liquids
4. Markings to indicate length units are placed on a steel tape in a room at 22°C. Are measurements made with
the tape on a day when the temperature is 27°C too long, too short, or accurate? Defend your answer.
On the hotter day the marking would be further than they should be so the units are too large and therefore the
corresponding measurements will be too short.
5. One surprising feature of thermal expansion is the cavities inside materials expand at the same rate as the
surrounding material. Explain how that happens. The following questions all highlight that fact.
a) Metal lids on glass jars can often be loosened by running hot water over them. How is this possible?
The metal lid’s opening becomes larger…
b) A flat metal ring has a gap in it. If the ring is heated, will the gap get smaller, larger, or stay the same?
Any gap in any object grows at the same rate as the object…
c) In a class demonstration, when the metal ring and sphere are both at room temperature, the sphere fits through
the ring, but when heated, the sphere will not fit through the ring. After the ring is also heated, the sphere can be
passed through the ring again. Explain.
At first the sphere gets bigger and won’t fit through ring, but when ring is heated its hole grows and the sphere
can fit through…
d) If the ball is slightly bigger than the hole in the ring at room temperature, can BOTH be heated or cooled to
another temperature so that the ball can fit through the ring? Remember that both are made of the same material.
At first this appears possible, since expansion depends on the original size of the object, BUT thinking deeper so
can see that this would lead to a logical contradiction: If the diameter of the sphere and the hole in the ring can
be the same at some temperature, they should be the same at every other temperature, since they expand at the
same rate; conversely if the two don’t fit at some temperature they cannot be made to fit at any other
temperature.
6. At 20.0°C, an aluminum ring has an inner diameter of 5.000 cm and a brass rod has a diameter of 5.050 cm.
(a) To what temperature must the ring be heated so that it will just slip over the rod? (b) To what temperature
must both be heated so that the ring just slips over the rod? Would this latter process work?
You need Al=24x10-6 1/C0 and brass=20x10-6 1/C0. I got (a) T=417 C0T=437 0C; (b) T=2650 0C, which
seems too high, probably above the melting temp of the materials.
7. A copper rod and steel rod are heated. At O°C the copper rod has a length of Lc, the steel one has a length Ls.
When the rods are being heated or cooled, a difference of 5.0 cm is maintained between their lengths over a small
range of change (∆L<<L). Determine the values of LC and Ls.
Here the given condition is that (LS – LC)=5.0 holds true even as the rods expand, so LC = Ls. After writing
these in terms of ’s and T’s you can combine the equations to find LS =17 cm and LC =12 cm. I used
Cu=17x10-6 1/C0 and steel=12x10-6 1/C0.
Note that this condition could not be sustained for a large change in length. See problem 14.
8. A pendulum clock with a brass suspension system has a period of 1.000 s at 20.0°C. If the temperature
increases to 30.0°C, (a) by how much does the period change and (b) how much time does the clock gain or lose
in one week? Hint: Recall that the period of a simple pendulum is T=2π(L/g)1/2 and take the derivative dT/dL.
Refer to the table in the textbook for expansion coefficients. Be careful with the symbol T which can stand for
period or temperature!
Using the hint: dT/dL=(2π/2√g)(L)-1/2 and combine with T=2π(L/g)1/2 to get that (dT/dL) =(T/2L) splitting the
derivative you get dT=(T/2L)dL and changing d’s to ’s to estimate small variations, you get the useful
expression that T=(T/2L)L. Finally, use thermal expansion (L/L)=(Ttemp) to arrive at the final
expression: T=(Tperiod /2)(brassTtemp)=~10-4 sec.
b) The clock loses ~10-4 sec for every 1-sec swing of the pendulum, so multiply 10-4 times the # of seconds in a
week 7(24)(3600)(10-4) ~60 sec lost in one week.
9. A rectangular plate has an area A equal to its length times its width (lw).
a) If the temperature increases by ∆T, show that the increase in area is ∆A = (2A∆T, where is the average
coefficient of linear expansion. What approximation does this expression assume? This was done in class…
b) Similarly, you can show that the volume expansion coefficient  = 3. Same procedure as above…
10. A liquid has a density . (a) Show that the fractional change in density (∆) for a change in temperature ∆T
is (∆) ~ - ∆T. What does the negative sign signify? (b) Fresh water has a maximum density of 1.000 g/cm3
at 4.0°C. At 10.0°C and its density is 0.9997 g/cm3. What is the  for water over this temperature interval?
Since =(mass/volume) and volume depend on temp d /dT=(-m/V2)(dV/dT)  d = -(m/V2)(dV)=(dV/V)
 Changing d’s to ’s and using the definition of , you get (∆) ~ - ∆T; the (-) indicates that when the
temp increases the density decreases.
For water using the givens:  =~(∆∆T)=~5x10-5 1/Co.
11. In designing structures engineers must make allowances for the thermal expansion of metals otherwise the
structures could experience catastrophic stresses. Consider this example: Steel rails for a rapid transit train form a
continuous track that is held rigidly in place in concrete. (a) If the track was laid when the temperature was O°C,
what is the stress (F/A) in the rails on a warm day when the temperature is 25°C? (b) What fraction of the yield
strength* of 52.2 X 107 N/m2 does this stress represent? Hint: You need to relate the linear expansion coefficient
 to the Young’s modulus (look up the definition!) through the fractional change in length (∆L/L). c) Does this
help you understand why gaps are part of the design of structures? Note: For steel Y=2.0 x1011 N/m2 and  = 12
x10-6 1/ºC.
* Yield strength is the maximum stress (F/A)max that a material can withstand before it begins to fail.
(a) Combine the definition of Y and of  to derive the expression: (F/A) = Y∆T=6x107 N/m2. This is less than
the yield stress…
b) 6x107 /52.2x107 =0.115. Since the ratio is less than one, the stress here is not catastrophic.
c) Without gaps there could catastrophic stress building up that would buckle the rails…
12. A cylindrical pot of cross-section A contains a volume V or water. The pot and water are heated by ∆T.
a) Derive a formula for the increase in the height of the water if the expansion of the pot is neglected.
b) Redo (a) including the expansion of the pot.
c) For a numerical version, assume the pot is made of aluminum. The pot is initially at 4.0°C, at which
temperature it has an inside diameter of 28.00 cm. Initially the pot contains 3.000 gal (1 gal ~ 3785 cm3) of water
at 4.0°C and then heated to 90.0 ºC. Allowing for the expansion of the water but ignoring the expansion of the
pot, what is the change in depth of the water? (The density of water is 1.000 g/cm3 at 4.0°C and 0.965 g/cm3 at
90.0°C.)
d) Modify your solution to allow for the expansion of the pot. Is it necessary to consider the expansion of the pot
in this problem?
a) The simplest way to find the increase in water level is to compare the initial and final heights of the water in
the container  ∆h =( hat 90º) (hat 4º) = [Vat 90º  Vat 4º]/A= ∆Vwater /A.
But water’s expansion coefficient is not given, so you must use the water densities at 4º and 90º to determine the
increase in water volume: ∆Vwater=(Vw/0.965) – Vw.
b) The expansion of the container is reflected in its larger cross-sectional area [A’= A(1+2∆T)] at 90º, which
affects the height of the water at 90º. So now, ∆h = hat 90º  hat 4º = (Vat 90º /A) (Vat 4º/A’). It’s not necessary to
consider the rest of the container.
c) Using the given data: Vat 90º =3x3785 =11355 cm3; Vat 4º =11355 /0.965 = 11767 cm3; A=πR2=π(14)2=616
cm2; A’= A(1+2∆T)= 618 cm2. Ignoring the pot expansion: ∆h=412/616=0.67 cm. Including the pot
expansion: ∆h=(11767 /618) (11355 /616) =0.60 cm. The difference is about 10%, not negligible.
13. Explain why a column of mercury in a thermometer first descends slightly and then rises when placed in hot
water. Now try these problems, which have similarities to the one above.
a) A typical mercury thermometer is made up of a thin, cylindrical
capillary tube with a diameter of 0.0040 cm, and the spherical bulb with
∆h
a diameter of 0.25 cm. Neglecting the expansion of the glass, find the
change in height of the mercury column for a temperature change of 30°C.
b) If you don’t neglect the expansion of the glass, what is the change in
T
T+∆T
height of the mercury column above? Was it justified to ignore the glass?
-5
o
a) Similar to problem above. Expressions are the same. You need mercury=18x10 1/C and glass=2x10-5
1/Co. I got h=3.5 cm without correcting for the glass expansion.
b) Correcting for the expansion of the glass I got h=3.1 cm, ~12% difference. The correction is due mostly
the expansion of the larger spherical bulb, the expansion of the small tube’s cross-section is negligible.
14. The relationship L ~ Lo(1 + ∆T) is an approximation that works when the average coefficient of expansion
and/or the change in temperature is small. If  is sizable, the relationship dL/ dT =  L must be integrated to
determine the final length.
a) Assuming the “” is a constant, determine a general expression for the final length L.
b) Plot a graph of L vs. Temp. What does the slope of this graph represent? How would the curves for two rods
of dissimilar materials compare?
c) Given a rod of length 1.00 m and a "temperature change of 100.0 ºC, determine the % error caused by the
approximation when  = 2.00 X 10-5 (ºC)-1 (the normal value for common metals) and when  = 0.020 (ºC)-1 (an
unrealistically large value just for comparison purposes).
d) In problem 7 you were given a condition that the difference in length between two dissimilar rods remained
the same as the rods were being heated (or cooled). Explain why this condition is not possible over a long
temperature range.
a) L=Loe∆T. Note that if you use a binomial expansion approximation for the exponential part of this expression
(ex~ 1+x… for x<<1) you get the original approximate version of the expansion formula.
b) The graph is a rising exponential. Rods of different material would curve up at different rates, the one with the
higher would curve more steeply.
c) Comparing L=Loe∆T to L=Lo(1 +∆T), the difference is L= Lo[e∆T-∆T)]. For  = 2.00 X 10-5(ºC)-1
the difference is negligible (~2 X 10-4%), but for = 0.020 (ºC)-1 the difference was 440%!...
d) It should be clear from the curves in (b) that two rods of dissimilar metals must get closer of farther apart in
length over a sufficiently large temperature change. Problem 7 is only valid for small changes.
15. A bimetallic bar is made of two thin strips of dissimilar metals bonded together. As they are heated, the one
with the larger average coefficient of expansion expands more than the other, forcing the bar into an arc, with the
∆r
outer radius having a larger circumference.
(a) Derive an expression for the angle of bending ø in terms of the initial length
of the strips L, their average coefficients of linear expansion, the change in
temperature, and the separation of the centers of the strips (∆r=r2-r1)
(b) Show that the angle of bending goes to zero when ∆T goes to zero or when
ø
the two coefficients of expansion become equal.
(c) What happens if the bar is cooled rather than heated?
a) The arc length of each strip is the original length of the strip plus the expansion: r2 ø =L2=L(1+∆T), and
r1 ø =L1=L(1+∆T).
Now subtract the two expressions: (r2 - r1) ø =(L2- L1) = L(-∆T ø=L(∆∆T/∆r
b) Pretty obvious from the formula…
c) Strip would bend in the opposite direction…
Heat and Thermal Processes
You may need to refer to tables in the text for values of heat capacities and latent heats.
16. In an experiment you put heat into a 500 g solid sample of a material at the rate of 10 kJ/min, while
recording temperature as a function of time.
T(ºC)
40
You plot your data on a graph.
30
a) What is the melting/freezing temperature and the latent heat of
20
fusion for this solid? Tmelt=10ºC; Lf=Q/m=10(1.5)/0.5=30 kJ/kg
b) What are the specific heats of the solid and liquid phases of the material? 10
csol=Q/m∆T=10/0.5x10= 2 kJ/kg·Cº;
0
1
2
3 t(min)
cliq= 5/0.5x30= 0.33 kJ/kg·Cº
c) Explain why the unit of specific heats can be equally expressed as
J/kg Cº or J/kg K. The c is related to temperature changes. Since the units in the Celsius and Kelvin scales
have the same size ∆T(Cº)=∆T(K)
17. A student inhales 22°C air and exhales 37°C air. The average volume of air in one breath is 200 cm3.
a) Neglecting evaporation of water into the air, estimate the amount of heat absorbed in one hour by the air
breathed by the student. The density of air is approximately 1.25 kg/m3, and the specific heat of air is 1.00
kJ/kg·Cº.
Using Q=c(V)∆T, I got 3.75 J/breath, then estimating 1 breath/3 secIn one hour Q=3.75x3600/3=4.5 kJ
b) Consider a classroom of dimension 10 m x10 m x4 m with 30 students. During the course of a two-hour
lecture, how much would the room’s temperature rise due to the breathing of the students?
Mass of air in the room =m=V= 1.25(400)=500 kg. Total Q from 30 students in 2 hours=4.5x2x30=270
kJ∆T=Q/cm=270/1x500=0.54 Cº.
c) If you consider that the human body is mostly water and a typical body mass is 60 kg, how much would
the body temperature drop in one hour due to the breathing? So how does our body keep its temperature?
In one hour a single student loses Q=4.5 kJ∆T=Q/cm=4.5/4.18x60=0.018 Cº. Our bodies need to eat
quite often to sustain this energy loss.
18. A 300-g aluminum vessel contains 200 g of water at 10°C. Assume the specific heat of aluminum is
0.215 cal/gCº.
a) 100 g of water at 100°C is poured into the container, what is the final equilibrium temperature of the
system? I got Tfinal=35ºC.
b) 100 g of steam at 100°C is poured into the container, what is the final equilibrium temperature of the
system and the amount of water at equilibrium? Tfinal=100ºC, the final state is water and steam at
equilibrium. The amount of steam condensing is found to be 44 g, so the final amount of water is 244 g.
c) What would have happened if the amount of steam in (b) were only 10g? All the steam would have
condensed and the equilibrium state would have contained only water at a temperature 30ºC.
d) Why can you get a more severe burn from steam at 100°C than from water at 100°C?
The greater amount of internal energy of the steam makes it more hazardous even though it’s at the same
temp.
19. A 1.0-kg block of copper and a 1.0-kg block of lead at 20°C are dropped into a large vessel of liquid
nitrogen at 77 K and some of the nitrogen boils away. (The specific heat of copper is 0.092 cal/g °C and of
lead it’s 0.031 cal/g °C. The latent heat of vaporization of nitrogen is 48 cal/g.)
a) Which one will boil away the greatest amount of nitrogen? Explain.
It took more heat to raise the temperature of the copper, so it will give up more heat as it cools down,
therefore the copper will boil away more of the nitrogen.
b) How many kilograms of nitrogen boil away by the time the blocks reach 77 K?
I got a total of 554 g of nitrogen (414g from the copper, and 140 g from the lead).
c) Which of these metals would be safer to hold if they were at 100ºC?
It would be safer to hold the lead block since it gives up less heat as it cools…
20A. In an insulated vessel, 250 g of ice at 0°C is added to 600 g of water at 18°C.
You need the heat of fusion of ice (L=80 cal/g) and the specific heat of ice (c=0.5 cal/gCº) for (d).
a) What is the final temperature of the system? The final state is an ice-water mixture at 0ºC
b) How much ice remains when the system reaches equilibrium? 135g of ice melt so 115 g of ice remain
c) Draw a T vs. ±Q graph for the ice and the water for the entire process
d) What if the ice started out at -10ºC?
Less ice (119g) would melt since ice would first have to rise to 0ºC to melt.
e) Would it be possible to freeze the added water?
No, the ice would have to start colder than absolute zero, which is impossible. The ice would have to be
lower than -470 ºC to freeze this much water.
20B. Redo the problem above but adding only 100 g of ice at 0ºC to the 600 g of water at 18°C.
a) In this scenario there’s not enough ice to cool the water to 0ºC, so all the ice will melt and the final state
will be water at 4ºC.
b) No ice will be left.
d) Even in this case all the ice will melt but the final temperature will be lower, ~3.3 ºC.
21. A cooking vessel on a slow burner contains 10.0 kg of water and an
T(ºC)
3.0
unknown mass of ice in equilibrium at 0ºC at time t = 0. The temperature
of the mixture is measured at various times, and the result is plotted in the
2.0
graph. During the first 50 min, the mixture remains at 0ºC. From 50 min
1.0
to 60 min, the temperature increases to 2.0°C. Neglecting the heat capacity
0.0
of the vessel, determine the initial mass of the ice.
From the graph it should be clear that all of the ice melts since Tfinal>0.
20
40
60 t(min)
Start writing expression for the rate of heat flow into the system, dQ/dt,
both for the melting process: (dQ/dt)melt=Lmice/tmelt =80(mice/50) in cal/min,
and for the heating process: (dQ/dt)heat=c(mice+ mwater )T/theat ,= (1)( mice+ 104)(2)/10 in cal/min.
Since the heat flow dQ/dt is constant these expressions must be equal and you can combine to solve for the
mass of the ice. I got 1430 g=1.43 kg of ice.
22. A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique is to
measure the temperature difference between the input and output points of a flowing stream of the liquid
while adding heat at a known rate. In one particular experiment, a liquid of density 0.78g/cm3 flows through
the calorimeter at the rate of 4.0 cm3/s. At steady state, a temperature difference of 4.8°C is established
between the input and output points when heat is supplied at the rate of 30 J/s. What is the specific heat of
the liquid? We know that the heat that goes into heating the volume of the fluid that passes through the flow
calorimeter is Q= cmT =c(V)T, where is the density of the fluid. Per time this become the rate
dQ/dt=c(dV/dt)T. The only unknown is the specific heat “c”. Putting in the given values and matching
units (1cal=4.186 J), you can get the answer. I got 0.48 cal/gCo=2.0 kJ/kg Co.
23. In a famous experiment John P. Joule demonstrated that a change in potential energy could generate a
change in temperature in water, thus establishing the fact that “heat was a form of energy”. Joule was on the
right track although we now think of heat as a “transfer of energy”. In the following problems a change in
mechanical energy leads to a change in temperature or phase change. The key is to use “energy
conservation” and make the proper unit changes. Remember 1 J = 1 Nm and 1 cal= 4.186 J.
a) A 3.0-g copper penny at 25°C drops 50 m to the ground. If 60% of its initial potential energy goes into
increasing its internal energy, determine its final temperature. Does the result depend on the mass of the
penny? Explain.Basically 60% of the potential energy of the penny becomes heat that raises its
temperature: that is, (0.6)mgh= mcT. As you can see the mass drops out and has no bearing on the
answer, but you need to be careful with units. The specific heat of copper is (0.093)(4186)J/kgCo.
Solving for T, I got 0.77 Co.
b) A 3.0-g lead bullet at 25ºC is traveling at 240 m/s when it embeds in a block of ice at 0°C. If all the heat
generated goes into melting ice, what quantity of ice is melted? (The latent heat of fusion for ice is 80
kcal/kg, and the specific heat of lead is 0.030 kcal/kg Co). Here the kinetic energy of the bullet is turned
to heat to melt the ice (a negligible amount also goes into cooling the bullet to 0ºC). So,
mbulletv2/2=miceL. Substitute values and match units…I got mice=0.26 g.
24. An iron plate is held against an iron wheel so that there is a sliding frictional force of 50.0 N acting
between the two pieces of metal. The relative speed at which the two surfaces slide over each other is 40.0
m/s. (a) Calculate the rate at which mechanical energy is converted to thermal energy. (b) The plate and the
wheel have a mass of 5.00 kg each, and each receives 50% of the thermal energy. If the system is run as
described for 10.0 s and each object is then allowed to reach a uniform internal temperature, what is the
resultant temperature increase? You need ciron=470 J/kg Co.
a) The power generated by the friction force is P=(Force)(vel)=50x40=2000 watts. Friction generates heat
so P= (Q/time)= 2000 w.
b) Since Q=Pt = cm∆T  2000 (10)= (5+5)(470) ∆T  T = 4.25 Co.
Heat Transfer processes
25. It is important to distinguish between specific heat and heat conductivity. The following questions are
designed to clarify the difference.
a) A tile floor in a bathroom may feel uncomfortably cold to your bare feet, but a carpeted floor in an
adjoining room at the same temperature will feel warm. Why? Carpet has a very low heat
conductivity, even at the same temp, it will take more time for heat to flow out of the carpet.
b) So called “fire walkers” walk over burning coals seemingly unharmed. How are they able to do this?
Coal has low heat conductivity…in addition, “fire walkers” moistened their soles so that the heat
goes to evaporation rather than heating the soles of their feet. It is still risky to do this and some
people have been seriously burned performing this “stunt”.
c) Concrete has a higher specific heat than soil. Use this fact to explain (partially) why cities have a
higher average night-time temperature than the surrounding countryside. If a city is hotter than the
surrounding countryside, would you expect breezes to blow from city to country or from country to
city? Explain. The higher specific heat of concrete means that cities accumulate a lot of energy
during the day as the temperature rises, and they can release all this energy back to the air at night,
keeping the city warmer. Soil has less “energy content” per degree so it has less to give up at night.
Hot air tends to rise and this allows air from the countryside to flow into the city as “breezes”.
d) If you hold water in a thin paper cup over a flame, you can bring the water to a boil without burning
the cup. How is this possible? As long as the paper is thin, heat will flow quickly into the water in the
cup. Water has a high heat capacity and large latent heat of vaporization, so it takes a lot of heat
before the temperature of the cup can rise higher than 100ºC. The ignition temp. of paper is higher
than that.
26. Two conceptual questions for those who drink coffee:
a) Suppose you pour hot coffee for your guests, and one of them chooses to drink the coffee after it has
been in the cup for several minutes. In order to have the warmest coffee, should the person add the cream
just after the coffee is poured or just before drinking? Explain. The goal is to lose as little heat as
possible during the waiting period. Since radiation is proportional to T4 (Stefan’s law) and conduction is
proportional to the ∆T (Newton’s law of cooling), you want to reduce the temperature of the coffee as
soon as possible to minimize the heat loss. So you should add the cream just after pouring the coffee, this
reduces the temperature difference between the coffee and air and reduces the heat losses through
radiation and conduction through the cup walls.
b) Two identical cups both at room temperature are filled with the same amount of hot coffee. One cup
contains a metal spoon, while the other does not. If you wait for several minutes which of the two will
have the warmer coffee? Which heat transfer process explains your answer? The metal spoon will absorb
heat from the coffee, cooling it down (conduction and specific heat), so initially the cup with the spoon
would be cooler. But the hotter cup is radiating heat more quickly, so it will catch up to the other cup in
temp pretty fast…
80 ºC
Au
Ag
30ºC
27. A bar of gold (Au) is in thermal contact with a bar of silver (Ag) of the
same length and area. One end of the compound bar is maintained at 80 ºC
Insulation
while the opposite end is at 30.0 ºC. I used kgold=314 W/mCº and ksilver=426 W/mCº.
a) When the heat flow reaches steady-state, find the temperature at the junction, in the middle.
H=A∆T/[(L/2k1) +(L/2k2)]=1.81x104A/L. Using the formula derived in class, you can show Tmid =51 ºC.
b) Determine the temperature gradient (dT/dx) in each metal in terms of L.
T(ºC)
80
Since the length is not given assume it is “L”, then: (dT/dx)=2∆T/L
59
 (dT/dx)gold =58/L and (dT/dx)silver =42/L (in Cº/m units).
51
30
c) Draw a graph of T vs. x from end to end. 
L/2
L x(m)
d) What would change if the two materials were interchanged?
The order of the materials doesn’t affect the overall H, but the temperature at the junction will change to
59º, and the temperature gradients will be interchanged. See dotted line in graph in (c).
e) If the two bars were replaced with a single bar of the same length and cross-section what would be the
thermal conductivity constant of the material that would have the same heat current as the two bars?
To find the equivalent k’ start with the equivalent R-factor: R’= R1+R2  (L/k’) = (L/2k1) + (L/2k2).
So the equivalent k’=2k1k2/( k1+k2)=362 W/mCº. You can show that in general k’=Lk1k2/(L2k1+ L1k2)
27B. For more practice, repeat problem above, but with the gold bar being L/3 and the silver being 2L/3.
T(ºC)
a) Tjunction=60ºC;
80
b) (dT/dx)gold =60/L and (dT/dx)silver =45/L (in Cº/m units).
60
c) Note that the slopes are inversely proportional to k and independent
30
of length but the lengths affect the mid-temperature.
L/3
L x(m)
d) Tjunction=67ºC
e) The equivalent k’=3k1k2/(2k1+k2).
27C. Two rods of the same length L, but different materials and cross-sectional areas
are placed side-by-side as shown. a) Determine the rate of heat flow in terms of the
thermal conductivities, areas, and ∆T’s. Generalize to several rods. b) Determine
Thot
1
Tcold
2
Insulation
the equivalent k’ of the material of a single rod that could replace the two given here.
a)These will act as two separate independent heat channels, so, in general, Htotal=H1+H2 = (kiAi)∆T/L.
b) k’(A1+A2)= (k1A1+ k2A2) k’ =(k1A1 + k2A2)/ (A1+A2)
27D. Derive general expressions for net H, and equivalent k’ for the following arrangements of heat
conductors over the same temperature difference: a) two different heat conductors in series that have the
same cross-sectional area but different lengths (L1 and L2). b) two different heat conductors in parallel that
have the same length but different cross-sectional areas (A1 and A2). Sketch heat gradient graphs for each
case.
T(ºC)
T(ºC)
a) H=A(∆T)/[(L1/k1) +(L2/k2)]; (L1+L2)/k’=(L1/k1) +(L2/k2)…
b) H= [(A1k1) +(A2k2)]∆T/L; (A1+A2)k’=(A1k1) +(A2k2)…
L1
(L1+L2) x(m)
L x(m)
28. A box with a total surface area of 1.20 m2 and a wall thickness of 4.00 cm is made of an insulating
material. A 10.0-W electric heater inside the box maintains the inside temperature at15.0°C above the·
outside temperature.
a) Find the thermal conductivity k of the insulating material. Just plug into the formula…I got 0.022
W/mCº.
b) If you were using two or more layers of insulation, would the order of the layers matter? For example,
would the heat flow rate be different if the higher k material were closer to the colder temperature or
vice-versa? If you grasped the meaning of the formula derived in class H=A(Th-Tc)/Ri, you should
realize that the order of the insulating layers makes no difference in the H value.
29. More and more windows are made with “double-pane” glasses to save energy. Calculate the “R” value
of (a) a window made of a single pane of glass 1/8 in. thick and (b) a thermal pane window made of two
single panes each 1/8 in. thick and separated by a l/4-in. air space. (c) By what factor is the heat loss reduced
by using the thermal window instead of the single pane window?
The only difference between this problem and the example worked out in class is the units…changing inches
to cm gives 1/8 in=.32 cm and ¼ in=.64 cm. The rest is the same with kglass=0.8 and kair=0.024 in W/mCº.
a&b) R1=0.0040 m2Cº/W and R2=0.28 m2Cº/W
c)H2/H1= R1/R2=0.015 (1/69)…double pane transmits 69 times less heat.
The following problems require calculus.
30. When a warm object cools to room temperature its temperature decreases exponentially over time. This
is called “Newton’s law of cooling” : T= (To – Troom)e-Ct + Troom , where To is the initial temperature of the
body and “C” here is a constant that depends on the physical properties of the object (we will be testing this
law in the lab). a) Show that this law can be derived from the formula for the heat current. b) What
properties does “C” depend on?
a)We know H=dQ/dt=Ak(∆T/∆x) and here dQ is the heat loss by the body as it cools, so dQ=cm(-dT).
Combing the two dT/dt=-(constant)(T-Troom)=-C(T-Tr) Set up the integral To∫T dT/(T-Tr)=-C ∫dt
T=(To – Tr)e-Ct + Tr.
b) C =Ak/cm∆x is this case. The geometry of the conductor would alter this formula somewhat (see the
problems below).
31. The following problems have similar solutions to the example worked out in class for a thermal
conductor with a varying cross-section. In these problems the heat current flows radially outward and the
solution starts with the differential form of the heat current formula: (dQ/dt)= kA(-dT/dr), where (dT/dr) is
the temperature gradient. The varying area “A” is identified and expressed in terms of “r”. Once the proper
r1
r2
integral is set-up the derivation is straightforward.
a) The example worked out in class featured a heat conductor of length L and
L
H
conductivity constant k that is shaped like a truncated cone. The radii of the ends
T1
T2
are r1 and r2. Show that the heat current is H= kπ(r1r2)∆T/L.
b) A vessel in the shape of a spherical shell has an inner radius a and outer radius b.
The wall has a thermal conductivity k. If the inside is maintained at a temperature Ta
and the outside is at a temperature Tb , show that the heat current between the
dQ 4kab
spherical surfaces is:
 
Ta  Tb 
dt  b  a 
a
Ta
b
Tb
c) The inside of a hollow cylinder is maintained at a temperature T1 while the outside
is at a lower temperature, T2. The wall of the cylinder has a thermal conductivity k.
 that the rate current from the inner wall (radius r1)
T2
r2
Neglecting end effects, show
 T  T 
dQ
T1 r1
to the outer wall (radius r2) in the radial direction is:
 2Lk 1 2 
dt
ln r2 /r1 
2
The basic integral set-up is: H ∫dr/A=-k ∫dT. In (b) A=4πr and in (c) A=2πrL. Solving the integrals produce
the given formulas.
Challenge problems

32. For a numerical version of the problem above consider a Thermos bottle in the shape of a cylinder that
has an inner radius of 4.0 cm, outer radius of 4.5 cm and length of 30.0 cm. The insulating walls have a
thermal conductivity equal to 2.0 X 10-5 cal/s·cm·Cº. One liter of hot coffee at 90ºC is poured into the bottle.
If the outside wall remains at 20°C, how long does it take for the coffee to cool to 50°C? (Assume that
coffee has the same properties as water.)
This problem is more than a numerical substitution; it requires setting up an integral to find the time.
Given the units in this problem it’s unnecessary to change units and substituting in the formula in (b)
dQ/dt= 0.032 (T-20) in cal/sec where T is the temperature of the coffee which changes from 90ºC to 50ºC.
Because the coffee temperature is changing you must set up an integral relating temp. and time. One liter of
coffee has a mass of 1000 g, so dQ=cm(-dT)=-(1)(1000)dT in cal. Combining with conductivity formula
above and separating the variables gives 103[90∫ 50 dT/(T-20)]=-0.032 ∫dt After solving the integral and
solving for “t”, I got t=2.65 x 104 sec~7.4 hrs.
33. A pond of water at 0°C is covered with a layer of ice 4.0 cm thick. Heat flows from the warmer water
below to the colder air above through the ice layer. If the air temperature stays constant at -l0°C, how long
will it be before the ice thickness is 8.0 cm? (Hint: To solve this problem, write the conductivity formula for
the heat current through the ice as dQ/dt= kA(∆T/x) and note that the incremental heat dQ extracted from the
water through the thickness x of the ice is the amount required to freeze a thickness dx of ice. That makes dQ
= LAdx, where is the density of the ice, A is the surface area of the ice layer, and L is the latent heat of
freezing.) You will need the following: kice=2 W/mCº, L=335 kJ/kg, ice=917 kg/m3.
Air at -10ºC
This is a tricky problem that requires some thought before launching into
Ice
x
a solution. As heat flows from the water to the air, water freezes and the
ice layer grows at the water-ice boundary. As indicated in the hint:
freezing layer of ice
dx
dQ/dt= kA(∆T/x) from the conductivity formula. The mass of the freezing
Water at 0ºC
layer at any time is dm=(Adx) and requires a heat loss dQ= Ldm =LAdx.
Combining these two formulas and rearranging: LA dx/x = kA(∆T)dt, and the only variables are “x” and
“t” so we can integrate: LA ∫xdx = kA(∆T)∫dt L(x2- xo2)/2 = k(∆T)t. The area A of the layer drops out,
and you can substitute the given numbers to solve for the time “t”. I got t=37x103 sec=~10 hrs
Note of interest: This shows how ice actually insulates water below from the colder temperatures above in
the air, and keeps the bodies of water on Earth from completely freezing over, which would be highly
detrimental to all life on our planet.
34. Two problems about radiation: Stefan’s law can be used to determine the heat flow in or out of bodies
due to electromagnetic radiation.
a) Consider the human body. Typically the skin surface area is ~1.2 m2 and the skin temperature is ~30ºC.
Considering the body and the surrounding air at ~20ºC, determine the net heat loss rate of our warm
bodies. Assume the emissivity of our bodies “e”=1. If you eat 2000 kcal/day, what % of that amount is
heat loss?
The Y&F textbook does this problem as an example. Basically plug into the formula…Hnet=72 w, about
74%.
b) The intensity of sunlight (the “solar constant”) arriving at the earth’s surface has been measured to be
~956 W/m2 (this takes into account the 30% or so reflected sunlight due to clouds, etc…; above the
atmosphere the solar constant is larger, ~1366 W/m2). Only half the earth’s surface can absorb sunlight
at one time and, and since the earth is round, you have to consider the perpendicular area component to
the sunlight. This amounts to a disk-shape collection area of πR2 (were R is the radius of the earth, 6.4 x
106 m). However, the earth radiates energy outward in all directions over its entire surface so the
radiation area is 4πR2. If the earth is in thermal equilibrium with sunlight (so the earth is not heating up
or cooling down), what is the earth’s average temperature over its surface? Assume that the emissivity of
our earth is 0.6.
Habsorbed=IA=956 π(6.4 x 106)2 and Hradiated=(5.67x10-8)(0.6) 4π(6.4 x 106)2T4. Equating these two and
solving for T, I got T=290 K= 17ºC, (which is not too far off the official value of ~15ºC).
Note: Because the earth absorbs sunlight primarily in the visible range of the electromagnetic spectrum
(=0.5 µm) but radiates primarily in the infrared (=10 µm) the “emission” and “absorption” factors
are not the same (e~0.6 for emission and e~0.7 for absorption). This difference gives rise to the
“greenhouse effect” which keeps our planet warmer than it would otherwise be if the two factors were
closer in value. Present concerns about “global warming” and “climate change” are due to the fact that
rising CO2, methane gas, and other “greenhouse gases” could widen the gap even further by lowering
the “emissivity” factor without much altering the “absorption” factor. In that case, Earth’s temperature
would rise until the balance between absorbed and emitted energy is restored.