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Ch. 3 Solutions

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Basic

Medium

Basic

B-1. Calculating Carnot Efficiency

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Problem:

Calculate the maximum efficiency of a Carnot cycle operating between a hot reservoir at \(T_{\text{hot}} = 500\) K and a cold reservoir at \(T_{\text{cold}} = 300\) K. Also, calculate the efficiency when the cold reservoir is lowered to \(T_{\text{cold}} = 200\) K, and compare the two results.

Solution:

\(\eta_{\max} = 1 - \frac{T_{\text{cold}}}{T_{\text{hot}}}\)

For \(T_{\text{hot}} = 500\) K, \(T_{\text{cold}} = 300\) K: \(\eta = 1 - \frac{300}{500} = 1 - 0.6 = \boxed{0.4 = 40\%}\)

If \(T_{\text{cold}}\) is lowered to \(200\) K: \(\eta = 1 - \frac{200}{500} = 1 - 0.4 = \boxed{0.6 = 60\%}\)

Key point: Lowering the cold reservoir temperature increases efficiency. However, unless \(T_{\text{cold}} = 0\) K (absolute zero), the efficiency never reaches 100%. This is not a technical limitation but a fundamental constraint imposed by the second law of thermodynamics.

Medium

M-1. Entropy of Coins

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Problem:

When \(N\) coins are tossed, the number of microstates \(\Omega\) corresponding to "exactly half showing heads" is given by the binomial coefficient \(\binom{N}{N/2}\). Calculate \(\Omega\) for the cases \(N = 4\) and \(N = 100\), and verify how overwhelmingly large the number of states with "exactly half heads" becomes as \(N\) increases.

Solution:

For \(N = 4\): \(\Omega = \binom{4}{2} = \frac{4!}{2! \cdot 2!} = 6\)

The total number of states is \(2^4 = 16\), so the probability of "exactly half heads" is \(6/16 = 37.5\%\).

For \(N = 100\), using Stirling's approximation \(\ln N! \approx N\ln N - N\):

\(\ln \binom{100}{50} = \ln 100! - 2\ln 50!\) \(\approx (100\ln 100 - 100) - 2(50\ln 50 - 50)\) \(= 100\ln 100 - 100\ln 50 - 100 + 100\) \(= 100(\ln 100 - \ln 50) = 100\ln 2 \approx 69.3\)

\(\boxed{\Omega \approx e^{69.3} \approx 10^{30}}\)

The total number of states is \(2^{100} \approx 10^{30.1}\). This means the "exactly half heads" states account for a significant fraction of the total. As \(N\) becomes large, states with "approximately half heads" become overwhelmingly dominant.

M-2. Statistical Mechanical Definition of Temperature

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Problem:

When two systems (with energies \(E_1\), \(E_2\), and total energy \(E = E_1 + E_2\) held constant) are in thermal equilibrium, derive from the condition that the total entropy \(S_{\text{total}} = S_1(E_1) + S_2(E_2)\) is maximized that \(\frac{\partial S_1}{\partial E_1} = \frac{\partial S_2}{\partial E_2}\), i.e., \(T_1 = T_2\).

Solution:

Maximize the total entropy \(S_{\text{total}} = S_1(E_1) + S_2(E - E_1)\) with respect to \(E_1\).

\(\frac{dS_{\text{total}}}{dE_1} = \frac{\partial S_1}{\partial E_1} + \frac{\partial S_2}{\partial E_2}\frac{dE_2}{dE_1} = 0\)

Since \(E_2 = E - E_1\), we have \(dE_2/dE_1 = -1\):

\(\frac{\partial S_1}{\partial E_1} = \frac{\partial S_2}{\partial E_2}\)

Using the definition of temperature \(1/T = \partial S / \partial E\):

\(\boxed{\frac{1}{T_1} = \frac{1}{T_2} \quad \Rightarrow \quad T_1 = T_2}\)

Key point: The condition for thermal equilibrium is "total entropy is maximized." This is equivalent to "temperatures are equal." The statistical mechanical definition of temperature is consistent with the everyday experience that "temperatures are equal in thermal equilibrium."