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Home , Specific heat capacity

Last Time Physics 172H Modern Mechanics Einstein Model of (springs + balls)

(q+ N −1) ! # microstates Ω = N oscillators q!() N − 1 ! & q quanta

Instructor: Dr. Mark Haugan Office: PHYS 282 [email protected] TAs: Alex Kryzwda [email protected] Fundamental assumption of John Lorenz [email protected] Over time, an isolated system in a given macrostate (total energy) is equally likely to be found in any of its microstates (microscopic distribution of energy). Lecture 23: & Interactions, Section 12.6

Equilibrium = Most Probable Distribution Clicker Question

There is thermal transfer of 5000 J of energy into 1) 5000 K S ≡ k ln Ω a system. This transfer increases the of 2) 100 K the system by 50 J/K. 3) 50 K dS dS dS =1 + 2 = 0 What is the approximate of the 4) 0.01 K dq1 dq 1 dq 1 system? 5) 0.0002 K 1 ≡ dS dS dS T dEint 1= 2 dq1 dq 2 ANSWER: ∆EEEWQ= ∆ + ∆ = + 1 ≡ dS int other T dE int 1≡dS ≈ 50 J / K = 1 1 T dE5000 J 100 K Energy is exchanged until the most probable distribution is reached. int

Two Ways to Change a ’s Energy Capacity

How much thermal energy must flow into a system to change its temperature by a certain amount? compress the solid (force through a distance): the system’s SHAPE changes, so, its energy levels do a) Large amount → Large b) Small amount → Small heat capacity

This has to do with degrees of freedom -- where are HEAT (energy transfer due to temperature difference) all the microscopic places the system can store systems’ energy levels don’t change. Instead, quanta are transferred to or from the system. energy. More modes = higher heat capacity. (Energy will go into every mode it can...) Heat Capacity Heat Capacity of Solids How much thermal energy must flow into or out of a system to cause a specific temperature change? Which has the higher heat capacity, ∂E Lead (Pb) or Aluminum (Al)? C = int (Heat = energy transferred) ∂T

Water has a high heat capacity. Live near : keeps surroundings cooler in summer and warmer in Lead Aluminum winter.

Which is Bigger? Heat Capacity for Pb and Al

Initially the 1 blocks of Pb and Al are at a temperature very Each block contains 6 x 10 23 atoms. near (0 K). We will add 1 J of energy to the aluminum mAl = 27g mPb = 207g block, and 1 J of energy to the lead block, and see which block has Same crystal structure. the larger increase in temperature. Which shows the right relative sizes? ∆ Esystem ∆ ENatom atoms A) B) C) C = ≡ atom ∆T ∆T

Specific Heat Capacity ANSWER: We will step through a chain of reasoning using statistical Diameters of all atoms are roughly the same. mechanics to answer this question and, thus, determine whether The difference in is due primarily to the mass of the nuclei aluminum or lead has the higher heat capacity at low . (which in either case is just a “dot” at the center of the atom)

CLICKER QUESTION CLICKER QUESTION k = 16 N/m k = 5 N/m Al Pb We add 1 J of energy to each block. Given the fact that Al has mAl = 27g mPb = 207g (1 mole of each) the greater energy-level spacing, which block now has the larger number of quanta of energy, q?

Einstein model = independent quantum harmonic oscillators. A) Al Which is the energy level diagram for Al? Which for Pb? Answer: B) Pb C) Same for both. katom ∆E =ℏω = ℏ 4.0e-21 J matom

ANSWER:

A) Al Pb 16> 5 Al has 1 quanta Pb has ≈ 5 quanta B) Pb Al 27 207 Note: Energy spacing of levels is not shown correctly for this well! CLICKER QUESTION CLICKER QUESTION

Which block has the largest number N of quantized oscillators? The Pb block has more quanta corresponding to the 1J of thermal energy. Therefore, in which block is there a larger number of ways Ω A) N is greater in the Al of arranging the thermal energy? B) N is greater in the Pb C) N is the same for Pb and Al ( + − ) Ω = q N 1 ! # microstates − N oscillators ANSWER: q!() N 1 ! & q quanta Both blocks have 1 mole = 6 x 10 23 atoms = 18 x 10 23 oscillators. A) The number of ways Ω is greater in the Al B) The number of ways Ω is greater in the Pb C) The number of ways Ω is the same in the Pb and the Al

CLICKER QUESTION CLICKER QUESTION

Originally the temperature of the blocks was near absolute zero, with almost no thermal energy in the blocks. How many The Pb block has the larger number of ways Ω to arrange the ways are there to arrange zero energy in a block? Just 1. So energy. So which block now has the larger entropy S? what was the initial entropy in a block?

A) The entropy S is now greater in the Al A) 0 J/K B) The entropy S is now greater in the Pb B) 1 J/K C) The entropy S is the same in the Al and the Pb C) infinite

ANSWER: ANSWER: Entropy S = kln Ω is larger for Pb. (Remember 1 = x 0 ⇒ ln1=0). Entropy S = kln Ω = kln1 = 0.

CLICKER QUESTION CLICKER QUESTION

∆ We found that after adding 1 J to each block, the entropy S is now We added the same amount of energy E = 1 J to each block, and greater in the Pb block. Both blocks started with zero entropy. the entropy change ∆S was greater in the Pb block. Which block Therefore which block experienced a larger change in entropy now has the higher temperature? ∆S? A) The temperature of the Al is now higher

A) The entropy change ∆S was greater in the Al B) The temperature of the Pb is now higher B) The entropy change ∆S was greater in the Pb C) The temperature of the Al and Pb are the same ∆ C) The entropy change S was the same in the Pb and the Al ANSWER: 1 ∆S ≡ Thus, 1/T is bigger for Pb ∆ TEint CLICKER QUESTION Specific Heat and Quantization The original temperature was 0 K, and the final temperature of the Al block is higher than that of the Pb block, so the Al block has the ∆ At high temperatures, average energy per oscillator is very high. larger change in temperature, T. At low temperatures, which block Energy levels are packed so tight they look continuous. has the greater specific heat capacity , C = ( ∆E / ∆T) / 6e23? Classical (non-quantum) theory is a good approximation here!

A) The low-temperature heat capacity per atom of Al is greater B) The low-temperature heat capacity per atom of Pb is greater C) Same for both.

At low temperatures, energy quantization is really apparent: energies are nowhere near continuous. Measured heat capacities Classical theory fails badly here.

Specific Heat and Quantization Spring Model Limitations

At high temperatures, average energy per oscillator is high. The non-uniform spacing of the levels means that the spring-model is breaking down . Thus, the Einstein model starts to break down at high temperatures.

classical expectation

NOTE: The classical expectation was known to disagree with the data long ago. Quantization of energy levels solved this paradox, one of the first signs of the need for .