Problem Statement
Explain the Clausius statement of the Second Law: Heat cannot by itself flow from a cold body to a hot body. Show how this is related to entropy.
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
Explain the Clausius statement of the Second Law: Heat cannot by itself flow from a cold body to a hot body. Show how this is related to entropy.
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
Explain the Clausius statement of the Second Law: “Heat cannot by itself flow from a cold body to a hot body.” Show how this is related to entropy.
Given Information
- Heat $Q$ flows from cold body at $T_C$ to hot body at $T_H$ spontaneously (hypothetical violation)
Physical Concepts & Formulas
If heat $Q$ flowed from cold to hot spontaneously:
$$\Delta S_{cold} = -\frac{Q}{T_C}, \quad \Delta S_{hot} = +\frac{Q}{T_H}$$
$$\Delta S_{univ} = -\frac{Q}{T_C} + \frac{Q}{T_H} = Q\left(\frac{1}{T_H} – \frac{1}{T_C}\right)$$
Since $T_H > T_C$: $\frac{1}{T_H} < \frac{1}{T_C}$, so $\Delta S_{univ} < 0$.
Step-by-Step Solution
Step 1: Test entropy change if heat flows cold→hot.
$$\Delta S_{univ} = Q\left(\frac{1}{T_H} – \frac{1}{T_C}\right) < 0 \quad \text{(for } T_H > T_C \text{)}$$
Step 2: Second Law of Thermodynamics (entropy statement).
$\Delta S_{univ} \geq 0$ always (equals zero only for reversible processes).
Since the hypothetical process gives $\Delta S_{univ} < 0$, it violates the Second Law.
Step 3: Equivalence.
The Clausius statement (no spontaneous cold→hot heat flow) and Kelvin statement (no perfect heat engine) are both equivalent forms of $\Delta S_{univ} \geq 0$.
Worked Calculation
For $Q = 100$ J, $T_C = 300$ K, $T_H = 400$ K (hypothetical reverse flow):
$$\Delta S_{univ} = 100\left(\frac{1}{400} – \frac{1}{300}\right) = 100 \times \left(-\frac{1}{1200}\right) = -0.083 \text{ J/K} < 0$$
Answer
Spontaneous cold→hot heat flow would decrease universe entropy (violating $\Delta S_{univ} \geq 0$) — hence it never occurs naturally.
Physical Interpretation
The Second Law determines the direction of natural processes. Heat, diffusion, and chemical reactions all proceed in the direction that increases total entropy. A refrigerator does move heat from cold to hot, but it requires work input — and the work input generates entropy elsewhere, ensuring $\Delta S_{univ} > 0$ overall. There is no violation; you just have to account for all entropy contributions.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\Delta S_{univ} = 100\left(\frac{1}{400} – \frac{1}{300}\right) = 100 \times \left(-\frac{1}{1200}\right) = -0.083 \text{ J/K} < 0}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\boxed{\Delta S_{univ} = 100\left(\frac{1}{400} – \frac{1}{300}\right) = 100 \times \left(-\frac{1}{1200}\right) = -0.083 \text{ J/K} < 0}}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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