Question

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

Short answer

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

Step-by-step solution

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.

Explanation

Mathematically, temperature can be defined as:

$\frac{1}{T}=\left(\frac{\partial S}{\partial U}\right)$

Where,

$\partial S$is the change in entropy and $\partial U$is 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.

Final answer

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