Tutorial 5 Solutions PDF

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Tutorial 5 Solutions 1) Obtain values for the force constant  and the angular frequency of vibration of the atoms, ω, for diamond copper and lead using the following data: For diamond Y = 9 1011 N-1, a0 = 0 nm, m = 12 amu For copper Y = 1 1011 N-1, a0 = 0 nm, m = 64 amu For diamond Y = 1 1011 N-...

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Tutorial 5 Solutions 1) Obtain values for the force constant  and the angular frequency of vibration of the atoms, ω, for diamond copper and lead using the following data: For diamond Y = 9.5 1011 N.m-1, a0 = 0.154 nm, m = 12 amu For copper Y = 1.3 1011 N.m-1, a0 = 0.265 nm, m = 64 amu For diamond Y = 1.5 1011 N.m-1, a0 = 0.350 nm, m = 207 amu. The force constant  is given by  = Ya0. So, for diamond the force constant is given by  = Ya0 = (9.5  1011 N.m-1)  (0.154  10-9 m) = 146 N.m-1. We can then find the vibrational frequency of the atoms from 

 146 Nm  1   8.54 1013 rad.s  1 . m 12 1.66 10  27 kg 

Similarly, substituting values for copper gives Cu = 33.3 N.m-1, Cu = 1.77  1013 rad.s-1, and for lead Pb = 52.5 N.m-1, Pb = 1.24  1013 rad.s-1,

2) Given that the value of the force constant  for copper is 100 N.m-1 and the atomic spacing is 0.265nm, estimate the amplitude of vibration of the atoms at 300 K as a percentage of the equilibrium spacing. For any harmonic oscillator the potential energy at a distance x from the equilibrium position is ½x2, where  is the force constant. At the maximum amplitude, xmax, all of the energy is potential energy, therefore 1 2 x k BT 2 max

Rearranging gives 12

 2k B T   xmax    

this, for copper the maximum amplitude is xmax

 2k B 300      100 N.m 1 

12

9.10 10  12 m

As a fraction of the atomic spacing this is 9.1010  12 m 100% 3.5% 0 .256 10  9 m

3) Both ends of a steel girder of length 20m are clamped in position at a temperature of 5C. Calculate the change in length that would occur if the temperature is increased to 35C and determine the stress that is produced in the girder as a result of clamping the ends. (The linear coefficient of thermal expansion of steel is 1.1 10-5 K-1 and Young’s modulus is 2.1 1011 N.m-2).

The change in length of the girder is L L 0 T

where  is the thermal coefficient of linear expansion, L0 is the original length, L is the change in length and T is the change in temperature.





L  1. 110  5 K  1  20m 30K 6. 6 10  3 m

We can now calculate the strain in the girder: 

L 6.6 10 3 m   3.3 10 4 L0 20m

and then we can find the stress







  Y  2. 11011 N.m  2 3. 3 10  4 6.93 107 N.m  2

Since this value exceeds that yield stress it will produce a permanent deformation of the girder.

4) In A Bragg x-ray diffraction experiment using aluminium at 300 K a sharp peak is observed at an angle of 15.46, at 800 K the same peak has a Bragg angle of 15.27 . Use this information to calculate the coefficient of linear expansion of aluminium. Using the Bragg equation for the diffraction of x-rays, the atomic spacing a at 300K is n n d 300K    1.8757n 2 sin  2 sin 15.46  where  is the wavelength of the x-rays. Similarly at 800 K we have d 800k  

n n  1.8985n  2 sin  2 sin 15.27  

The difference is therefore

 d  d 300K  d 300K  1.8757n  1.8985n  0.0228n

To find the expansion coefficient we assume that the fractional change in the length of the sample is equal to the fractional change in the atomic spacing, then L d  L0 d 300K 

therefore, using the definition of  

1 0.0228n 1 d L 1  2.43 10 5 K 1   L 0 T d  300K  T 1.8757n 500K

5) Estimate the energy required to raise the temperature of 5 kg of the following materials from 20C to 150C: aluminum (specific heat = 900 J.kg-1.K-1), brass (specific heat = 375 J.kg-1.K-1), aluminum oxide (alumina) (specific heat = 775 J.kg1 .K-1). The energy, E, required to raise the temperature of a given mass of material, m, is the product of the specific heat, the mass of material, and the temperature change, ΔT, as E Cm T

The ΔT in this problem is equal to 150°C – 20°C = 130°C (= 130 K), thus E(aluminum) = (900 J/kg-K)(5 kg)(130 K) = 5.85 x 105 J E(brass) = (375 J/kg-K)(5 kg)(130 K) = 2.44 x 105 J E(alumina) = (775 J/kg-K)(5 kg)(130 K) = 5.04 x 105 J

6) For copper, the heat capacity at constant volume, Cv, at 20 K is 0.38 J/mol-K, and the Debye temperature is 340 K. Estimate the specific heat (a) at 40 K and (b) at 400 K. (a) For copper, C at 20 K may be approximated by C AT 3

since this temperature is significantly below the Debye temperature (340 K). The value of C at 20 K is given, and thus, we may compute the constant A as 

A

C 0.38J.mol 1 .K  T3  20K 3

Therefore, at 40 K



1

 4.7510  5 J.mol  1.K  4 .



C AT 3  4.75 10  5 J.mol  1 .K  4  40K 3 3.02J.mol  1 .K  1 47.8J.kg  1 .K  1

(b) Since 400 K is above the Debye temperature, a good approximation for C is





C  3R 3 8.31J.mol  1 .K  1 24.9 J.mol  1 .K  1

and, converting this to specific heat





c  24. 9J.mol  1 .K  1  1mol / 63.55g1000g / kg 392J.kg  1 .K

7) The constant A in the equation for the heat capacity, Cv = AT3, is

1

12  4 R 3 5 D

where R

is the gas constant and θD is the Debye temperature (K). Estimate θ D for aluminum, given that the specific heat is 4.60 J/kg-K at 15 K. For aluminum, we want to compute the Debye temperature, θD, given the expression for A above and the heat capacity at 15 K. First of all, let us determine the magnitude of A, as A





C 4.60J.mol 1K  1 1kg / 1000g  26.98g / mol    3.6810  5 J.mol-1K - 4 T3 15K 3

Then using the expression for A 12 4 R A 5 3D

and solving for θD D

 12 4 R     5 A  

13





 12 4 8.31J.mol 1 .K 1   5 1 4   5 3.6810 J.mol .K 





13

 375K

8) A 0.4 m rod of a metal elongates by 0.48 mm on heating from 20 to 100C. Determine the value of the linear coefficient of thermal expansion for this material. The linear coefficient of thermal expansion for this material may be determined using  

0.4810  3 m L L 15.0 10  6 K -1   L 0 T L0  Tf  Ti   0.4m 100  20  C

9) Compute the density for iron at 1000 K given that its room-temperature (300K) density is 7870 kg/m3. Assume that the volume coefficient of thermal expansion α V, is equal to 3αL. αL of iron = 11.810-6 K-1. Let us use as the basis for this determination 1 m3 of material at 20°C, which has a mass of 7870 kg, it is assumed that this mass will remain constant upon heating to 1000 K. Let us compute the volume expansion of this cubic centimeter of iron as it is heated to 1000 K.

V v    V  v V0 T V 0 T Also, αv = 3αL, Therefore, the volume, V, of this specimen of Fe at 1000 K is just V V 0  V V 0 1   v T  V 0 1 3  l T 









 1m3 1  3 11 .8 10 6 K  1 1000K  300K 



3

1.02478 m

thus the density is just the 7870 kg divided by the new volume 

7870kg 1.02478m

3

 7679.69kg.m  3

10) Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature averages 4C. If a joint space of 5.4 mm is allowed between the standard 11.9-m long rails, what is the hottest possible temperature that can be tolerated without the introduction of thermal stresses? The coefficient of linear thermal expansion of 1025 steel is 12.0 x 10-6 °C-1. For these railroad tracks, each end is allowed to expand one-half of the joint space distance, or the track may expand a total of this distance (5.4 mm). T 

L  L 0

 Tf Ti 

L 5.4 10  3 m 4 C     41.8 C  L 0 12.0 10 6  C 1 11 .9m 





...