Problem 2.102 — Carnot Engine: Maximum Work

Problem Statement

Solve the thermodynamics problem: An ideal gas Carnot engine uses 1 mol gas, $T_1=400\ \text{K}$, $T_2=200\ \text{K}$, absorbs $Q_1=5\ \text{kJ}$ per cycle. Find max work and heat rejected. $$\eta_{Carnot} = 1-\frac{T_2}{T_1} = 1-\frac{200}{400} = 0.5 = 50\%$$ $$W = \eta Q_1 = 0.5\times5000 = 2500\ \text{J} = 2.5\ \text{kJ}$$ $$Q_2

Given Information

• See problem statement for all given quantities.

Physical Concepts & Formulas

The Carnot engine is the idealized heat engine operating between two reservoirs at temperatures $T_H$ (hot) and $T_C$ (cold). It represents the maximum possible efficiency for any heat engine operating between those two temperatures — a fundamental result of the Second Law of Thermodynamics. No real engine can exceed Carnot efficiency.

• $\eta_{\text{Carnot}} = 1 – T_C/T_H$ — efficiency depends only on absolute temperatures
• $W = Q_H \eta$ — net work output equals heat absorbed times efficiency
• $Q_C = Q_H(1 – \eta)$ — heat rejected to cold reservoir

Step-by-Step Solution

Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.

Worked Calculation

$$\eta_{Carnot} = 1-\frac{T_2}{T_1} = 1-\frac{200}{400} = 0.5 = 50\%$$

$$W = \eta Q_1 = 0.5\times5000 = 2500\ \text{J} = 2.5\ \text{kJ}$$

$$Q_2

Given Information

• Temperatures, pressures, volumes, and process type as given
• Universal gas constant $R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}$
• $C_p$ and $C_v$ or $\gamma = C_p/C_v$ as applicable

Physical Concepts & Formulas

The First Law of Thermodynamics $\Delta U = Q – W$ is an energy balance: internal energy increases when heat flows in and decreases when the gas does work. For ideal gases, internal energy depends only on temperature: $\Delta U = nC_v \Delta T$. Different processes have different constraints: isothermal ($T = \text{const}$, $W = nRT\ln(V_f/V_i)$), adiabatic ($Q=0$, $PV^\gamma = \text{const}$), isobaric ($P = \text{const}$, $W = P\Delta V$), isochoric ($V = \text{const}$, $W = 0$). Carnot efficiency sets the upper bound for any heat engine: $\eta = 1 – T_C/T_H$.

• $\Delta U = Q – W$ — First Law
• $PV = nRT$ — Ideal Gas Law
• $W_{\text{isothermal}} = nRT\ln(V_f/V_i)$
• $PV^\gamma = \text{const}$ — adiabatic process
• $\eta_{\text{Carnot}} = 1 – T_C/T_H$ — maximum efficiency

Step-by-Step Solution

Step 1 — Identify the process (isothermal, adiabatic, isobaric, isochoric).

Step 2 — Write the appropriate work expression and compute $W$.

Step 3 — Find $\Delta U = nC_v\Delta T$.

Step 4 — Apply First Law: $Q = \Delta U + W$.

Worked Calculation

Substituting all values with units:

Carnot engine: $T_H = 600\,\text{K}$, $T_C = 300\,\text{K}$:

$$

Answer

$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$

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.