Ch 19, Temperature## Homework Solutions:

## Ch19: Questions 5, 9, *, 16, 17

## Ch19: Problems *, **, 18, 25, 27, 40, 47, 50, 64

19.Q5:Why should the amalgam used in dental fillings have the same average coefficient of thermal expansion as a tooth? What would occur if they were mismatched?Think of what would happen if the coefficients of thermal expansion for the tooth and the filling amalgam were

notthe same. Then they would expand differently as the temperature changed. This would cause stress -- andpain!

19.Q9:Markings to indicate length are placed on a steel tape in a room that has a temperature of 22^{o}C. Are measurements made with the tape on a day when the temperature is 27^{o}C greater than, less than, or the same length as the object's length?At 27

^{o}C, the tape is nowlongerand the distance or spacing between the markings arelongerorfarther apart.This means the tape will measure asmallernumer or alowernumber. Objects measured under these conditions will be measured to have asmallerlength.

19.*:The pendulum of a certain pendulum clock is made of brass. When the temperature increases, does the clock run too fast, run too slowly, or remain unchanged?Brass? Actually, the concept is exactly the same whether we are dealing with brass or steel or aluminum. As the temperature increses, the

lengthof the pendulum alsoincreases. With an increased length, theperiodof the pendulum is alsoincreased. As the pendulum oscillates more slowly, the clock keeps timemore slowly.

19.Q16:And automobile radiator is filled to the brim with water when the engine is cool. What happens to the water when the engine is running and the water is heated? What do modern automobiles have in their cooling systems to prevent the loss of coolants?As the temperature of the engine coolant increases, the volume also

increases. Today's automobiles have a plastic "overflow resevior" that the coolant flows into. Later, as the coolant cools, the pressure decreases and coolant is "pulled" our "suctioned" from this "overflow reservior" back into the radiator.

19.Q17:Metal lids on glass jars can often beloosened by running them under hot water. How is this possible?The coefficient of thermal expansion is

much greaterfor metals than for glass. Heating a metal lid on a glass jar will cause the metal lid to expand far more than the glass. This should allow you to unscrew the metal lid from the glass jar.

19.*Convert the following to equivalent temperaturs on the Celsius and Kelvin scales:(a) the normal human body temperature, 98.6

^{o}FT _{C}= (^{5}/_{9}) (T_{F}- 32)T

_{C}= (^{5}/_{9}) (98.6 - 32)T

_{C}= (^{5}/_{9}) (66.6)T

_{C}= 37^{o}C(b) the air temperature on a cold day, - 5

^{o}FT _{C}= (^{5}/_{9}) (T_{F}- 32)T

_{C}= (^{5}/_{9}) ( - 5 - 32)T

_{C}= (^{5}/_{9}) ( - 37)T

_{C}= - 20.6^{o}C

19.**There is a temperature whose numerical value is the same on both the Celsius and Fahrenheit scales. What is this temperature?T _{C}= (^{5}/_{9}) (T_{F}- 32)T

_{x}= (^{5}/_{9}) (T_{x}- 32)T

_{x}= (^{5}/_{9}) (T_{x}) - (^{5}/_{9})(32)T

_{x}= (^{5}/_{9}) (T_{x}) - (^{5}/_{9})(32)T

_{x}- (^{5}/_{9}) (T_{x}) = - (^{5}/_{9})(32)(

^{4}/_{9}) (T_{x}) = - (^{5}/_{9})(32)4 T

_{x}= - 5 (32)T

_{x}= - (^{5}/_{4})(32)T

_{x}= - 40meaning both

T _{x}= - 40^{o}Cand

T _{x}= - 40^{o}F

19.18A concrete walk is poured on a day when the temperature is 20.0^{o}C, in such a way that the ends are unable to move.(a) What is the stress in the cement on a hot day of 50.0

^{o}C?(b) Does the concrete fracture?

Take Young's modulus for concrete to be 7.00 x 10

^{9}N/m^{2}and the tensile strength to be 2.0 x 10^{9}N/m^{2}

If the ends were free to expand,how much would a section of length L expand?From Table 19.2, p 588, or from Problem 19.10, immediately above, we know that the coefficient of thermal expansion for concrete is

= 12 x 10 ^{ - 6}(C^{o}) - 1And the temperature difference is

T = 30 C ^{o}

L = [12 x 10

^{ - 6}(C^{o})^{ - 1}][ L ][30 C^{o}]L = 3.6 x 10

^{ - 4}LThink back, now, to PHY 1351 and recall what is meant by "stress" and "strain". Stress is the force per unit area and strain is the elongation or compression divided by the area.

Stress = F / A Strain = L / L

Strain = [ 3.6 x 10

^{ - 4}L ] / LStrain = 3.6 x 10

^{ - 4}Y = Stress / Strain

Stress = [ Y ] [ Strain ]

Stress = [ 7.00 x 10

^{9}N/m^{2}] [ 3.6 x 10^{ - 4}]Stress = 2.52 x 10

^{ 6}N/m^{2}Tensile Strength = 2.0 x 10

^{9}N/m^{2}This thermal stress is much, much less than the tensile strength so the concrete does

notbreak.

19.25The concrete sections of a certain superhighway are designed to have a length of 25.0 m. The sections are poured and cured at 10.0^{o}C. What minimum spacing should the engineer leave between the sections to eliminae buckling of the concrete is to reach a temperature of 50.0^{o}C?How much witll the 25.0 m section expand as the temperature changes from 10.0

^{o}C to 40.0^{o}C?For concrete, the coefficient of thermal expansion is 12 x 10

^{ - 6}(C^{o})^{ - 1}(from Table 19.2, on p 588) ;= 12 x 10 - 6 (C ^{o}) - 1The temperature change, T, is 40 C

^{o};T = 40 C ^{o}Now we can put all the pieces together and evaluate the expansion, L ;

L = [12 x 10 - 6 (C ^{o}) - 1][25.0 m][40 C^{o}]L = 0.012 m

L = 1.2 cm

Since each concrete section will expand by 1.2 cm, there must be 1.2 cm between each section so they do not run into each other and cause compressive stresses.

19.27At 20.0^{o}C, an aluminum ring has an inner diameter of 5.000 0 cm and a brass rod has a diameter of 5.050 0 cm.(a) To what temperature must the ring be heated so that it will just slip over the rod?

We need to increase the diameter of the aluminum ring from L

_{o}= 5.000 0 cm (or L_{o}= 5 x 10^{ - 2}m) to 5.050 0 cm. That is an increase ofL = 0.050 cm = 5 x 10 ^{ - 4}mFrom Table 19.2, p 588, we find that the coefficient of thermal expansion for aluminum is

= 24 x 10 ^{ - 6}(C^{o})^{ - 1}L = L

_{o}TT = [L ] / [ L

_{o}]T = [ 5 x 10

^{ - 4}m ] / [ ( 24 x 10^{ - 6}(C^{o})^{ - 1}) ( 5 x 10^{ - 2}m )]T = 417 C

^{o}T

_{f}= 437^{o }C(b) To what common temperature must the two be heated so that the ring just slips over the rod? Would this latter process work?

Now we are going to heat

boththe aluminum ringandthe brass rod. Both will expand. We must heat them until the expansion of the aluminum ring diameter is 5 x 10 - 4 mgreaterthan the expansion of the brass rod diameter.

aluminum ring

_{Al}= 24 x 10^{ - 6}(C^{o})^{ - 1}L

_{Al}= L_{oAL}TL

_{Al}= ( 24 x 10^{ - 6}(C^{o})^{ - 1})(5 x 10^{ - 2}m) Tbrass rod

_{Br}= 19 x 10^{ - 6}(C^{o})^{ - 1}L

_{Br}= L_{oBr}TL

_{Br}= ( 19 x 10^{ - 6}(C^{o})^{ - 1})(5.05 x 10^{ - 2}m) TNow set

L _{Al}=L_{Br}+ 5 x 10 - 4 m( 24 x 10

^{ - 6}(C^{o})^{ - 1})(5.00 x 10^{ - 2}m) T = ( 19 x 10^{ - 6}(C^{o})^{ - 1})(5.05 x 10^{ - 2}m) T + 5 x 10^{ - 4}m( 24 x 10

^{ - 6}(C^{o})^{ - 1})(5.00) T = ( 19 x 10^{ - 6}(C^{o})^{ - 1})(5.05) T + 5 x 10^{ - 2}( 24 x 10

^{ - 6})(5.00) T = ( 19 x 10^{ - 6})(5.05) T + 5 x 10^{ - 2}C^{o}( 24)(5.00) T = (19)(5.05) T + 5 x 10

^{ + 4}C^{o}120 T = 96T + 5 x 10

^{ + 4}C^{o}24 T = 5 x 10

^{ + 4}= 50,000 C^{o}T = 2,080 C

^{o}T = 2,100

^{o}CIs this practical? It's difficult to even handle objects at this temperature! This may even be close to the melting temperature of either of the materials.

19.40A tank having a volume of 0.100 m^{3}contains helium gas at 150 atm. How many balloons can the tank blow up if each filled balloon is a sphere 0.300 m indiameterat an absolute pressure of 1.20 atm?d = 0.300 m

r = d/2 = 0.150 m

First, the easy part; what is the volume of each balloon?

V _{bal}= (^{4}/_{3}) r^{3}V

_{bal}= (^{4}/_{3}) (0.150 m)^{3}V

_{bal}= 0.0141 m^{3}How many moles of helum are in each individual balloon?

P V = n R T n = P V / R T

n

_{bal}= [(1.2 atm)(0.0141 m^{3})] / [(0.082 L-atm/mol-K)(T)]While we do NOT know the temperature T, it won't make any difference since the tank and the balloons are at the

sametemperature.n

_{bal}= 0.207 moles/T(Remember, this has strange units simply because we do

notknowthe temperature T).Now let us return to the tank and determine how many moles of helum are in the tank.

P V = n R T n = P V / R T

n

_{tnk}= [(150 atm)(0.100 m^{3})] / [(0.082 L-atm/mol-K)(T)]n

_{tnk}= 183 moles/T(Again, these units look

strangesimple because we do not know the temperature T).The number of balloons N, then is

N = n

_{tnk}/ n_{bal}N = 183 / 0.207

N = 884 balloons

19.47An automobile tire is inflated with air originally at 10.0^{o}C and normat atmospheric pressure. During the process, the air is compressed to 28.0% of its original volume and its temperature is increased to 40.0^{o}C.(a) What is the tire pressure?

P V = n R T n R =P V / T = const

P

_{2}V_{2}/ T_{2}= P_{1}V_{1}/ T_{1}Remember, these temperatures must be

absolute temperatures, measured in kelvins!T _{1}= 10^{o}C = 283 KT

_{2}= 40^{o}C = 313 KV

_{2}= 0.28 V_{1}P

_{1}= 1.0 atmP

_{2}= ?P

_{2}V_{2}/ T_{2}= P_{1}V_{1}/ T_{1}P

_{2}= P_{1}[ (V_{1}/ V_{2}) (T_{2}/ T_{1}) ]P

_{2}= 1.0 atm [(V_{1}/0.28 V_{1}) (^{313 K}/_{283 K})]P

_{2}= 1.0 atm [(^{1}/_{0.28})(^{313}/_{283})]P

_{2}= 1.0 atm [3.95]

P_{2}= 3.95 atm(b) After the car is driven at high speed, the tire air temperature rises to 85.0

^{o}C and the interior volume of the tire increases by 2.00%. What is the new (absolute)tire pressure in pascals?The values T

_{1}, P_{1}, and V_{1}remain the same; but for this part T_{2}, V_{2}, and P_{2}are new.T _{2}= 85^{o}C = 358 KV

_{2}= (1.02)V_{2old}= (1.02)(0.28 V_{1})V

_{2}= 0.286 V_{1}P

_{2}= ?P

_{2}V_{2}/ T_{2}= P_{1}V_{1}/ T_{1}P

_{2}= P_{1}[ (V_{1}/ V_{2}) (T_{2}/ T_{1}) ]P

_{2}= 1.0 atm [(V_{1}/0.286 V_{1}) (^{358 K}/_{283 K})]P

_{2}= 1.0 atm [(^{1}/_{0.286})(^{358}/_{283})]P

_{2}= 1.0 atm [4.43]

P_{2}= 4.43 atm

19.50A cylinder is closed by a piston connected to a spring of constant 2.0 x 10^{3}N/m. While the spring is relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0^{o}C.(a) If the piston has a cross-sectional area of 0.010 0 m

^{2}and a negligible mass, how high will it rise when the temperature is increased to 250^{o}C?(b) What is the pressure of the gas at 250

^{o}C?

Caution:Remember "absolute pressure" and "gauge pressure"? When the pressure inside the cylinder is anabsolutepressure of 1.0 atm, the "gauge pressure" iszeroand the force the spring exerts iszero. This means the spring isnotcompressed.F _{spring}= k xF

_{outside}= (1.0 atm)A + F_{spring}F

_{outside}= F_{inside}= F_{piston}F

_{outside}= F_{piston}F

_{piston}= P A(1.0 atm)A + F

_{spring}= F_{piston}P V = n R T

n = P V / R T

n = (1.0 atm)(5.0 L) / [ (0.082 L-atm / mol-K) (293 K)]

n = 0.208 moles

P = n R T / V

What is the

newvolume? And what is thenewpressure? Be sure to use the new temperature in units of kelvins.V = 5.0 L + A x V = 5.0 L + ( 0.010 0 m

^{2}) xT = 250

^{o}C = (250 + 273) K = 523 KP = n R T / V

P = ( 0.208 moles) (0.082 L-atm / mol-K) (523 K)/ [ 5.0 L + ( 0.010 0 m

^{2}) x]P = 8.92 L-atm / [ 5.0 L + ( 0.010 0 m

^{2}) x](1.0 atm)A + F

_{spring}= F_{piston}F

_{spring}= k x = ( 2.0 x 10^{3}N/m ) xF

_{piston}= P A = { 8.92 L-atm / [ 5.0 L + ( 0.010 0 m^{2}) x ] } { 0.010 0 m^{2}}(1.0 atm)A + F

_{spring}= F_{piston}(1.0 atm)A + (2.0 x 10

^{3}N/m) x = { 8.92 L-atm / [ 5.0 L + ( 0.010 0 m^{2}) x ] } { 0.010 0 m^{2}}

Be carefulwith the units! We were given a volume as5.0 Land an initial pressure of1.0 atmso it seems "reasonable" to use R in units ofL-atm/mol-K. But we also have a spring constant k in units ofN/mso it would also be "reasonable" to use R in units ofJ/mole-K. Ineither case, conversions will be required!(1.0 atm)A + (2.0 x 10 ^{3}N/m) x = 0.0892 L-atm-m^{2}/ [ 5.0 L + ( 0.010 0 m^{2}) x ](1.0 atm)(0.010 0 m

^{2}) + (2.0 x 10^{3}N/m) x = 0.0892 L-atm-m^{2}/ [ 5.0 L + (0.010 0 m^{2}) x ]From page 465, we know that

1.0 atm = 1.013 x 10 ^{ + 5}Pa = 1.013 x 10^{ 5}N/m^{2}(1.013 x 10

^{ 5}N/m^{2})(0.010 0 m^{2}) + (2.0 x 10^{3}N/m) x = 0.0892 L-atm-m^{2}/ [ 5.0 L + (0.010 0 m^{2}) x ](1.013 x 10

^{ 3}N) + (2.0 x 10^{3}N/m) x = 0.0892 L-atm-m^{2}/ [ 5.0 L + (0.010 0 m^{2}) x ](1.013 x 10

^{ 3}N) + (2.0 x 10^{3}N/m) x = 0.0892 L(1.013 x 10^{ 5}N/m^{2})m^{2}/ [ 5.0 L + (0.010 0 m^{2}) x ]1 013 N + (2.0 x 10

^{3}N/m) x = 9 036 L-N/ [ 5.0 L + (0.010 0 m^{2}) x ]1.0 L = 1,000 cm

^{3}= 0.001 m^{3}= 10^{ - 3}m^{3}1 013 N + (2.0 x 10

^{3}N/m) x = 9 036 N/ [ 5.0 (10^{ - 3}m^{3}) + (0.010 m^{2}) x ]1 013 N + (2.0 x 10

^{3}N/m) x = 9 036 (10^{ - 3}m^{3})N/ [ 5.0 x10^{ - 3}m^{3}+ (0.010 m^{2}) x ]1 013 N + (2.0 x 10

^{3}N/m) x = 9 .036 m^{3}-N/ [ 5.0 x10^{ - 3}m^{3}+ (0.010 m^{2}) x ]1 013 + (2.0 x 10

^{3}1/m) x = 9 .036 m^{3}/ [ 5.0 x10^{ - 3}m^{3}+ (0.010 m^{2}) x ]1 013 + (2.0 x 10

^{3}1/m) x = 9 .036 / [ 5.0 x10^{ - 3}+ (0.010 m^{ -1}) x ]If we require that the distance x also be measured in meters, then we can write this equation as

1 013 + (2.0 x 10 ^{3}) x = 9 .036 / [ 5.0 x10^{ - 3}+ 0.010 x ][1 013 + 2 000 x] [0.005 + 0.010 x ] = 9 .036

5.065 + 10.13 x + 10 x + 20 x

^{2}= 9.03620 x

^{2}+ 20.13 x + 5.065 = 9.03620 x

^{2}+ 20.13 x - 3.97 = 0Finally, we have something that looks familiar. Now we can use the

quadratic formulawith a = 20, b = 20.13, and c = - 3.97 .x = { - b SQRT[ b ^{2}- 4 a c] }/ 2 ax = { - 20.13 SQRT[ (20.13)

^{2}- 4 (20)( - 3.97 ) ] }/ [ 2 (20) ]x = { - 20.13 SQRT[ 405.2 + 317.6 ] }/ 40

x = { - 20.13 SQRT[ 722.8] }/ 40

x = { - 20.13 26.88 }/ 40

x

_{1}= 0.17 m = 17 cm or x_{2}= - 1.175 mThe

positivedistance is the physically significant one. The negative value satisfies the mathematics but not the Physics; it is an extraneous solution.x = 17 cmNow, what about the new

pressure? Wecouldsolve for that with the Ideal Gas Law. Let's use that as a "check". It seems easier to use the compression of the spring to find the new pressure.F _{outside}= (1.0 atm)A + F_{spring}F

_{outside}= F_{inside}= F_{piston}F

_{outside}= F_{piston}F

_{piston}= P AP A = (1.0 atm)A + k x

P = 1.0 atm + k x / A

P = 1.0 atm + (2.0 x 10

^{3}N/m)(0.17 m) / 0.010 m^{2}P = 1.0 atm + 3.4 x 10

^{4}N/m^{2}P = 1.0 atm + 3.4 x 10

^{ 4}Pa [ 1 atm / 1.013 x 10^{5}Pa]

P = 1.33 atmWe can check this -- and our entire calculation -- by checking this result with the result we get for the

new pressureusing theIdeal Gas Law. Now that we have solved for x, the linear distance of expansion of the piston, we can easily solve for the new volume.V = A x = (0.010 m ^{2}) (0.17 m)V = 0.001 7 m

^{3}V = [ 0.001 7 m

^{3}] [ 1.0 L / 0.001 m^{3}]V = 1.7 L

Thus, the

new volumeisV = V _{o}+VV = 5.0 L + 1.7 L

V = 6.7 L

P V = n R T

P = n R T / V

P = (0.208 moles)(0.082 L-atm/mol-K) (523 K) / 6.7 L

P = 1.33 atm

19.64A steel ball bearing is 4.000 cm in diameter at 20.0^{o}C. A bronze plate has a hole in it that is 3.994 cm in diameter at 20.0^{o}C. What common temperature must they have so that the ball just squeezes through the hole?

steel ball bearing

_{Al}= 11 x 10^{ - 6}(C^{o})^{ - 1}L

_{St}= L_{oSt}TL

_{St}= L_{oSt}+ LL

_{St}= L_{oSt}( 1 + T )L

_{St}= [4.000 cm] [ 1 + ( 11 x 10^{ - 6}(C^{o})^{ - 1})T ]bronze plate

_{Br}= 19 x 10^{ - 6}(C^{o})^{ - 1}L

_{Br}= L_{oBr}TL

_{Br}= L_{oBr}+ LL

_{Br}= L_{oBr}( 1 + T )L

_{Br}= [3.994 cm] [ 1 + ( 19 x 10^{ - 6}(C^{o})^{ - 1})T ]L _{St}= L_{Br}[4.000 cm] [ 1 + ( 11 x 10

^{ - 6}(C^{o})^{ - 1})T ] = [3.994 cm] [ 1 + ( 19 x 10^{ - 6}(C^{o})^{ - 1})T ]4.000 cm + [4.000 cm] [ ( 11 x 10

^{ - 6}(C^{o})^{ - 1})T ] = 3.994 cm + [ 3.994 cm] [ ( 19 x 10^{ - 6}(C^{o})^{ - 1})T ]0.006 cm + (4.000 cm) ( 11 x 10

^{ - 6}(C^{o})^{ - 1})T ] = (3.994 cm) ( 19 x 10^{ - 6}(C^{o})^{ - 1})T ]0.006 cm + 4.4 x 10

^{ - 5}T cm (C^{o})^{ - 1}= 7.58 x 10 - 5 T cm (C^{o})^{ - 1}0.006 cm = [(7.58 x 10

^{ - 5}) - ( 4.4 x 10^{ - 5})] [ T cm (C^{o})^{ - 1]}0.006 cm = [(3.18 x 10

^{ - 5})] [ T cm (C^{o})^{ - 1]}0.006 = [(3.18 x 10

^{ - 5})] [ T (C^{o})^{ - 1]}(3.18 x 10

^{ - 5}) [ T (C^{o})^{ - 1}] = 0.006(3.18 x 10 - 5)T = 0.006 C

^{o}T = 0.006 C

^{o}/ 3.18 x 10^{ - 5}

T = 189^{o}C

Return to Temperature ToC(c) Doug Davis, 2002; all rights reserve