Problem Statement
An engine absorbs $Q_1=10\ \text{kJ}$ from a source at $T_1=500\ \text{K}$ and rejects $Q_2=7\ \text{kJ}$ to a sink at $T_2=300\ \text{K}$. Is this engine reversible, irreversible-allowed, or violating the second law?
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
This problem applies fundamental physics principles to the scenario described. The solution requires identifying the relevant conservation laws and equations of motion, then solving systematically with careful attention to units and sign conventions.
- See the step-by-step solution for the specific equations applied.
- All quantities are in SI units unless otherwise stated.
Step-by-Step Solution
Step 1 — Identify given quantities and set up the problem: Check Clausius inequality for a cycle:
Step 2 — Apply the relevant physical law or equation: $$\oint\frac{dQ}{T} = \frac{Q_1}{T_1} – \frac{Q_2}{T_2} = \frac{10000}{500} – \frac{7000}{300} = 20 – 23.33 = -3.33\ \text{J/K}$$
Step 3 — Solve algebraically for the unknown: Since $\oint dQ/T = -3.33 < 0$, this is an allowed irreversible engine. (If it were zero, it would be reversible; if positive, it would violate the second law.)
Step 4 — Substitute numerical values with units: Actual efficiency: $\eta = W/Q_1 = 3000/10000 = 30\%$.
Step 5 — Compute and check the result: Carnot efficiency: $\eta_C = 1-300/500 = 40\%$. Since $30\% < 40\%$, confirmed as sub-Carnot (irreversible). ✓
Worked Calculation
$$\oint\frac{dQ}{T} = \frac{Q_1}{T_1} – \frac{Q_2}{T_2} = \frac{10000}{500} – \frac{7000}{300} = 20 – 23.33 = -3.33\ \text{J/K}$$
$$\text{Numerical result} = \text{given expression substituted with values}$$
$$\boxed{\oint\frac{dQ}{T} = \frac{Q_1}{T_1} – \frac{Q_2}{T_2} = \frac{10000}{500} – \frac{7000}{300} = 20 – 23.33 = -3.33\ \text{J/K}}$$
Check Clausius inequality for a cycle:
$$\oint\frac{dQ}{T} = \frac{Q_1}{T_1} – \frac{Q_2}{T_2} = \frac{10000}{500} – \frac{7000}{300} = 20 – 23.33 = -3.33\ \text{J/K}$$
Since $\oint dQ/T = -3.33 < 0$, this is an allowed irreversible engine. (If it were zero, it would be reversible; if positive, it would violate the second law.)
Actual efficiency: $\eta = W/Q_1 = 3000/10000 = 30\%$.
Carnot efficiency: $\eta_C = 1-300/500 = 40\%$. Since $30\% < 40\%$, confirmed as sub-Carnot (irreversible). ✓
Answer
$$\boxed{\oint\frac{dQ}{T} = \frac{Q_1}{T_1} – \frac{Q_2}{T_2} = \frac{10000}{500} – \frac{7000}{300} = 20 – 23.33 = -3.33\ \text{J/K}}$$
Physical Interpretation
The Carnot efficiency sets an absolute upper bound imposed by thermodynamics — no real engine, however well engineered, can exceed it. Higher hot-reservoir temperature or lower cold-reservoir temperature both increase efficiency.
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