Does your cup of coffee contain heat? Does it contain temperature? Explain.© BrainMass Inc. brainmass.com December 24, 2021, 11:53 pm ad1c9bdddf
SOLUTION This solution is FREE courtesy of BrainMass!
Heat and temperature are not fundamental properties of physical objects. These quantities only have a meaning at the level of a statistical description of physical objects. It is analogous to considering the average weight of people in a group and then asking if the group contains this average. While you can calculate it from the properties of the system, it only exists at the group level as a statistical quantity.
To be able to describe the macroscopic world we can observe in a tractable way, requires one to be able to eliminate the detailed physics of microscopic world of atoms and molecules that the macroscopic objects consists of. While we may also be interested in atoms and molecules, physics would not be useful in practice if you could not describe the behavior of things like a cup of coffee without having to invoke all the details of 10^25 molecules. The problem we then face is that the laws of physics don't allow for decoupling of the macroscopic and microscopic degrees of freedom of a system.
Take e.g. a ball that can collide with other balls. We know that we can describe the physics here using conservation of momentum and conservation of energy. Suppose that we take into account that the ball is not a point mass and that it consists of a large number of molecules. Then as far as conservation of momentum is concerned, there is no problem, because the sum of the momenta of all the molecules equals the momentum of the center of mass of the ball. But the total energy of the molecules is not equal to the kinetic energy of the center of mass. We can write the total energy of N molecules with mass m as:
V(x1,x2,...x_N) + 1/2 m (v1^2 + v2^2 + ...+v_N^2)
where V is the potential energy function that depends on the positions x_1,...x_N of the N molecules and v1,..,v_N are the velocities of the molecules. If we write the velocity of the jth molecule as vi = U + wi where U is the velocity of the center of mass, then the kinetic energy becomes:
1/2 m N U^2 + 1/2 m (w1^2 + w2^2 + ...+w_N^2) = 1/2 M U^2 + 1/2 m (w1^2 + w2^2 + ...+w_N^2)
where M is the mass of the ball. Here we've used that the summation of the relative speeds w.r.t. the center of mass cancels out.
The total energy is thus equal to the center of mass kinetic energy plus the potential energy of the molecules plus the kinetic energy relative to the center of mass. The sum of the latter two quantities is the internal energy of the system.
This thus means that energy present at the macroscopic level such as the center of mass of the balls can vanish into the internal energy in the form of potential and kinetic energy of the molecules. The only way to be able to do physics without getting bogged down into having to describe what all the molecules are doing is to describe the molecules statistically and then to keep track of how much energy there is in the form of internal energy.
For any arbitrary system one defines what the so-called external parameters of the system are. These are variables such as the volume of the system that one wants to keep track of. All the other variables are going to be treated statistically. An decrease in the internal energy of the system due to a change in the external parameters is, by definition, work performed by the system. The total change in the internal energy plus the work done by the system is, by definition, the heat absorbed by the system. If you were to keep track of all the molecules in your system by adding all the positions of the molecules as your external parameters of the system, then all changes in energy would be due to work and there would be no such thing as heat. While this is, of course, totally impractical, this illustrates that the definition of heat is subjective, depending on the choice of the external parameters.
The definition of temperature is less general than heat, it can only be defined for systems in so-called thermal equilibrium. If we have an isolated system of molecules that has settled down for a long time, we can assume that it is equally likely to be present in any of the possible physical states it could conceivably be found in. Since energy is conserved all such possible states will have the same energy. According to quantum mechanics, there are only a finite number of states possible within some narrow energy interval. In our macroscopic description of the system we can only specify the internal energy with some finite accuracy delta E, so, all possible states within an interval of delta E around the specified internal energy are possible. Let's denote the possible number of states as a function of the specified internal energy E as Omega(E). Then let's consider what happens if we bring two systems each with its own Omega functions, denoted as Omega1 and Omega2 with different internal energies E1 and E2 into thermal contact.
The combined system is an isolated system with total internal energy E = E1 +E2 where E1 and E2 are the internal energies of subsystem 1 and subsystem 2. The total number of states available for the combined system is:
Omega(E) = sum over E1 of Omega1(E1) * Omega2(E - E1)
In equilibrium, all states are equally likely, the most likely partitioning of the energy E into E1 and E2 is thus that term in the summation that is the largest. So, fr what E1 is the maximum of Omega1(E1) * Omega2(E - E1)? obtained. If we take the logarithm, we get:
Log[ Omega1(E1)] + Log[Omega2(E - E1)]
Differentiating w.r.t. E1 and setting the derivative equal to zero, yields:
d Log(Omega1)/dE1 = d Log(Omega2)/dE2
where we've put E2 = E- E1 and used the chain rule. So, in equilibrium the most likely partitioning of the energy will lead to the quantities d LogOmega/dE for the subsystems to become equal. If these quantities are equal to begin with then the internal energies will stay just as they are, as no heat will flow. If not, then heat will flow between the systems until these quantities become equal.
We then define the temperature T in terms of d Log(Omega)/dE by putting:
1/(k T) = d Log(Omega)/ dE
where k is Boltzmann's constant.© BrainMass Inc. brainmass.com December 24, 2021, 11:53 pm ad1c9bdddf>