Thermal equilibrium of A and B, microstates and quanta

[karnten07]

Homework Statement

System A consisting of 10^23 oscillators for which hw=10^-20 J, is in thermal contact with system B, consisting of 2x10^23 similar oscillators. The joint system has internal energy 3x10^3 J. Calculate the number of quanta in A and B, and the temperature, when A and B are in thermal equilibrium.

Homework Equations

The Attempt at a Solution

So i have this: W(N,Q) = (N+Q-1)!/[(N-1)!Q!] N is number of oscillators and Q is number of quanta.

For a generic system and W(total) for the system A and B = $$\Sigma$$W1(E1)W2(U-E1)

Where W is the number of microstates.

How do i find the number of quanta in A and B?

 

[Jhero]

To find the number of quanta in A and B, you can use the formula Q = U/hw, where Q is the number of quanta, U is the internal energy, and hw is the energy of each quantum. From the given information, we know that U = 3x10^3 J and hw = 10^-20 J. Therefore, for system A, we have Q = (3x10^3 J)/(10^-20 J) = 3x10^23 quanta. Similarly, for system B, we have Q = (3x10^3 J)/(10^-20 J) = 6x10^23 quanta.

To find the temperature, we can use the formula U = Nhwf(T), where U is the internal energy, N is the number of oscillators, hw is the energy of each quantum, and f(T) is the average number of quanta per oscillator at temperature T. Since the systems are in thermal equilibrium, the temperature for both systems will be the same. Therefore, we can equate the two equations for U and solve for T.

U(A) = N(A)hwf(T) = (10^23)(10^-20 J)(f(T))

U(B) = N(B)hwf(T) = (2x10^23)(10^-20 J)(f(T))

Equating the two equations, we get:

(10^23)(10^-20 J)(f(T)) = (2x10^23)(10^-20 J)(f(T))

Solving for f(T), we get f(T) = 1/2. This means that at thermal equilibrium, each oscillator will have an average of 1/2 quantum. Substituting this value for f(T) in the equation for U, we get:

U = (10^23)(10^-20 J)(1/2) = 3x10^3 J

Solving for T, we get T = U/(Nhw) = (3x10^3 J)/[(10^23)(10^-20 J)] = 3x10^3 K.

Therefore, at thermal equilibrium, the number of quanta in A and B will be 3x10^23 and 6x10^23 respectively, and the temperature will be 3x10^3 K.