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It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• $\frac{d}{dE}(\Omega_A \Omega_B) = 0$ for an infinitesimal transfer of internal energy from A to B
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C. At room temperature, it is about 4 % per meV, for any system.

It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• $\frac{d}{dE}(\Omega_A \Omega_B) = 0$ for an infinitesimal transfer of internal energy from A to B
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) must be equal for both systems (if $\Omega_A$ goes down by 1 ‰, $\Omega_B$ goes up by 1 ‰).
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C. At room temperature, it is about 4 % per meV, for any system.

It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• $\frac{d}{dE}(\Omega_A \Omega_B) = 0$
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C. At room temperature, it is about 4 % per meV, for any system.

It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• $\frac{d}{dE}(\Omega_A \Omega_B) = 0$ for an infinitesimal transfer of internal energy from A to B
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C. At room temperature, it is about 4 % per meV, for any system.

It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C.

It depends on what temperature is. A simple statistical argument could go like this.

• the multiplicity of two systems is the product of their multiplicities
• in thermodynamic equilibrium the multiplicity is at its maximum
• $\frac{d}{dE}(\Omega_A \Omega_B) = 0$
• the fractional change of multiplicity with internal energies ($\beta = \frac{1}{\Omega} \frac{d\Omega}{dE}$) is equal for both systems.
• identify $\beta = 1/kT$

If this is the case for systems A and B and for B and C, the fractional change of multiplicity with energy ($\beta$) should also be the same for A and C. At room temperature, it is about 4 % per meV, for any system.