Question

A heat engine receives 1000 J of heat from a reservoir at 1000 K and rejects 600 J of heat to a sink at 300 K. This engine is

• A. Reversible
• B. Irreversible
• C. Impossible
• D. Perfectly efficient

Hint:

Alternatively, apply the Clausius inequality directly: \[ \oint \frac{\delta Q}{T} = \frac{Q_H}{T_H} - \frac{Q_C}{T_C} = \frac{1000}{1000} - \frac{600}{300} = 1 - 2 = -1 \] Since \(\oint \frac{\delta Q}{T} < 0\), the cycle is completely possible but fundamentally irreversible.

Answer 1

The Correct Option is B

Concept: The thermodynamic behavior and viability of any heat engine cycle are evaluated using the Second Law of Thermodynamics, specifically via Clausius' Theorem and Carnot's efficiency limits.
• A cycle is reversible if its efficiency matches the Carnot efficiency, or equivalently if the cyclic integral of entropy change is zero (\(\oint \frac{\delta Q}{T} = 0\)).
• A cycle is irreversible if its performance falls below the Carnot limit but still satisfies basic energy conservation laws (\(\oint \frac{\delta Q}{T} < 0\)).
• A cycle is impossible if its calculated thermal efficiency exceeds the Carnot efficiency, which violates the Second Law (\(\oint \frac{\delta Q}{T} > 0\)).

Step 1: Calculate the actual thermal efficiency (\(\eta_{\text{actual}}\))
The heat engine receives energy \(Q_H = 1000\text{ J}\) and rejects energy \(Q_C = 600\text{ J}\). The thermal efficiency of any power cycle is given by: \[ \eta_{\text{actual}} = 1 - \frac{Q_C}{Q_H} \] Substituting the provided energy values: \[ \eta_{\text{actual}} = 1 - \frac{600}{1000} = 1 - 0.60 = 0.40 \quad \implies \quad 40% \]

Step 2: Calculate the maximum reversible Carnot efficiency (\(\eta_{\text{Carnot}}\))
The operating temperatures of the hot reservoir and cold sink are \(T_H = 1000\text{ K}\) and \(T_C = 300\text{ K}\), respectively. The maximum theoretical efficiency achievable between these thermal limits is defined as: \[ \eta_{\text{Carnot}} = 1 - \frac{T_C}{T_H} \] Substituting the absolute temperature values: \[ \eta_{\text{Carnot}} = 1 - \frac{300}{1000} = 1 - 0.30 = 0.70 \quad \implies \quad 70% \]

Step 3: Compare efficiency terms to determine the engine's classification
Let us compare our results: \[ \eta_{\text{actual}} = 40% \] \[ \eta_{\text{Carnot}} = 70% \] We observe that: \[ 0.40 < 0.70 \quad \implies \quad \eta_{\text{actual}} < \eta_{\text{Carnot}} \] Because the engine's actual efficiency is strictly less than the maximum theoretical Carnot efficiency limit, the engine does not violate the Second Law of Thermodynamics. However, because it falls short of the ideal limit, the cycle contains real-world thermodynamic losses, classifying it as an irreversible engine. This corresponds to Option (2).