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Can a "miserly" system, with a concave-up entropy-energy graph, ever be in stable thermal equilibrium with another system? Explain.

Short Answer

Expert verified

No, a "miserly" system, with a concave-up entropy-energy graph, cannot be in stable thermal equilibrium with another system.

Step by step solution

01

Given Introduction

If two objects are in thermal equilibrium, their temperatures are said to be the same. It can be expressed in terms of the system's entropy and energy, the measure of randomness being the entropy. It is determined by the amount of energy that is not available for work.

02

Explanation

Mathematically, temperature can be defined as:

1T=∂S∂U

Where,

∂Sis the change in entropy and ∂Uis the change in the internal energy of the body.

Systems in thermal equilibrium have the same temperature because their entropy-versus-energy graphs have the same slopes. The systems are coupled by gravity in a concave-up graph between entropy-energy graphs, and the temperature lowers when energy is supplied because the energy is stored as potential energy and the average kinetic energy is reduced. To put it another way, the heat capacity is negative. The entropy energy plot will thus be concave up in that situation.

Assume that there are two miserly systems, A and B. The temperature for both systems is the same. When energy is transmitted from system B to system A, system B becomes hotter, and a runway effect is observed because more energy is transferred spontaneously from system B to system A. As a result, the temperature of system B rises significantly above the temperature of system A.

As a result, two miserly systems can exist in thermal equilibrium with one another. These systems, however, are not stable.

03

Final answer

Hence, a "miserly" system, with a concave-up entropy-energy graph, cannot be in stable thermal equilibrium with another system.

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Most popular questions from this chapter

Use the definition of temperature to prove the zeroth law of thermodynamics, which says that if system A is in thermal equilibrium with system B, and system B is in thermal equilibrium with system C, then system A is in thermal equilibrium with system C. (If this exercise seems totally pointless to you, you're in good company: Everyone considered this "law" to be completely obvious until 1931, when Ralph Fowler pointed out that it was an unstated assumption of classical thermodynamics.)

Consider an ideal two-state electronic paramagnet such as DPPH, with μ=μB. In the experiment described above, the magnetic field strength was 2.06T and the minimum temperature was 2.2K. Calculate the energy, magnetization, and entropy of this system, expressing each quantity as a fraction of its maximum possible value. What would the experimenters have had to do to attain99% of the maximum possible magnetization?

Experimental measurements of the heat capacity of aluminum at low temperatures (below about 50K) can be fit to the formula

CV=aT+bT3

where CVis the heat capacity of one mole of aluminum, and the constants aand bare approximately a=0.00135J/K2and b=2.48×10-5J/K4. From this data, find a formula for the entropy of a mole of aluminum as a function of temperature. Evaluate your formula at T=1Kand at T=10K, expressing your answers both in conventional units (J/K)and as unitless numbers (dividing by Boltzmann's constant).

In order to take a nice warm bath, you mix 50 liters of hot water at 55°C with 25 liters of cold water at 10°C. How much new entropy have you created by mixing the water?

Use a computer to reproduce Table 3.2 and the associated graphs of entropy, temperature, heat capacity, and magnetization. (The graphs in this section are actually drawn from the analytic formulas derived below, so your numerical graphs won't be quite as smooth.)

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