Problem Statement
Solve the thermodynamics problem: Two large bodies at $T_1 = 500\ \text{K}$ and $T_2 = 300\ \text{K}$ are connected by a rod. Heat $Q = 1000\ \text{J}$ flows from hot to cold. Find the entropy produced. $$\Delta S = \frac{Q}{T_2} – \frac{Q}{T_1} = Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right) = 1000\left(\frac{1}{300}-\frac{1}{500}\righ
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Entropy is a state function measuring the dispersal of energy. The Second Law requires that the total entropy of an isolated system never decreases. For a reversible process, $dS = dQ_\text{rev}/T$; for irreversible processes, the actual entropy change exceeds $dQ/T$.
- $\Delta S = \int dQ_{\text{rev}}/T$ — entropy change for reversible process
- $\Delta S = nC_V \ln(T_2/T_1) + nR\ln(V_2/V_1)$ — ideal gas entropy change
- $\Delta S \geq 0$ for isolated system (Second Law)
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$\Delta S = \frac{Q}{T_2} – \frac{Q}{T_1} = Q\left(\frac{1}{T_2}-\frac{1}{T_1}\right) = 1000\left(\frac{1}{300}-\frac{1}{500}\righ
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: $T_H = 600\,\text{K}$, $T_C = 300\,\text{K}$:
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Answer
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Answer
$$\boxed{\eta_{\text{Carnot}} = 1 – \dfrac{T_C}{T_H}}$$
Physical Interpretation
Entropy increases in any irreversible process. The numerical result confirms the Second Law: the universe’s entropy has increased, and the process cannot spontaneously reverse.
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