Question

A cyclic engine exchanges heat with two reservoirs maintained at $100^\circ\text{ C}$ and $300^\circ\text{ C}$ respectively. The maximum work (in J) that can be obtained from $1000\text{ J}$ of heat extracted from the hot reservoir is:

• A. $349\text{ J}$
• B. $651\text{ J}$
• C. $200\text{ J}$
• D. $500\text{ J}$

Hint:

Always convert temperatures to Kelvin when calculating thermal efficiency or entropy. Using Celsius values ($\frac{100}{300}$) will lead to incorrect results!

Answer 1

The Correct Option is A

Concept: According to the Second Law of Thermodynamics, the maximum thermal efficiency ($\eta_{\text{max}}$) achievable by any heat engine operating between two fixed temperatures is limited by the efficiency of a reversible Carnot engine. This maximum efficiency is given by: \[ \eta_{\text{max}} = 1 - \frac{T_{\text{L}}}{T_{\text{H}}} \] where $T_{\text{L}}$ and $T_{\text{H}}$ are the absolute temperatures of the cold and hot reservoirs, respectively. Efficiency is also defined as the ratio of work produced ($W$) to the heat extracted from the hot reservoir ($Q_{\text{H}}$): \[ \eta = \frac{W}{Q_{\text{H}}} \implies W_{\text{max}} = \eta_{\text{max}} \cdot Q_{\text{H}} \]

Step 1: Convert reservoir temperatures to the absolute Kelvin scale.
\[ T_{\text{H}} = 300^\circ\text{ C} + 273.15 = 573.15\text{ K} \] \[ T_{\text{L}} = 100^\circ\text{ C} + 273.15 = 373.15\text{ K} \]

Step 2: Calculate the maximum Carnot efficiency $\eta_{\text{max}}$.
Using the absolute temperatures: \[ \eta_{\text{max}} = 1 - \frac{373.15}{573.15} = \frac{573.15 - 373.15}{573.15} = \frac{200}{573.15} \approx 0.34893 \]

Step 3: Determine the maximum work output $W_{\text{max}}$.
Given that $Q_{\text{H}} = 1000\text{ J}$: \[ W_{\text{max}} = \eta_{\text{max}} \cdot Q_{\text{H}} = 0.34893 \times 1000 = 348.93\text{ J} \] Rounding to the nearest integer gives $349\text{ J}$.