Complete Theory

3

Worked Examples

2
Example 1Free adiabatic expansion — Entropy increases even without heat
Example 2Spontaneous heat flow — Why heat goes from hot to cold

Exercises with Solutions

2
Exercise 1Irreversible heating — A copper block in contact with a hot sourceHard
Problem to solve
A copper block of mass m=2m=2 kg at T1=20°CT_1=20°C is placed in contact with a source at TH=200°CT_H=200°C. Compute ΔS\Delta S of the block, the source, and the universe. (cCu=385c_{Cu}=385 J/(kg·K))
Given data
m=2 kgT_1=293 KT_H=473 Kc_Cu=385 J/(kg·K)
Exercise 2Irreversible cycle — Entropy production in a real engineVery Hard
Problem to solve
A heat engine operates between TH=500T_H=500 K and TC=200T_C=200 K. It absorbs QH=3000Q_H=3000 J and rejects QC=1800Q_C=1800 J to the cold reservoir. Compute η\eta, ηCarnot\eta_{Carnot} and ΔSuniv\Delta S_{univ}. Why is ΔSuniv>0\Delta S_{univ} > 0 even though the engine is cyclic?
Given data
T_H=500 KT_C=200 KQ_H=3000 JQ_C=1800 J

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Integrative Problems

Problems combining all chapters — exam level
Problem 1The Thermoelectric Power Plant: from Gas to Molecules to EntropyEXTREME
A thermoelectric power plant uses n=5 moln = 5\,\mathrm{mol} of a diatomic gas (γ=7/5\gamma = 7/5, Cv=5R/2C_v = 5R/2) running through the following cycle on a PV diagram:

State A: PA=1 atmP_A = 1\,\mathrm{atm}, TA=300 KT_A = 300\,\mathrm{K}. A→B: adiabatic compression to VB=VA/8V_B = V_A/8 (compression ratio r=8r = 8). B→C: isochoric, heating to TC=2400 KT_C = 2400\,\mathrm{K} (combustion). C→D: adiabatic expansion to VD=VAV_D = V_A (return to original volume). D→A: isochoric, cooling (Otto cycle).
📌 Problem data
n=5 moln = 5\,\mathrm{mol}γ=7/5=1.4,  Cv=5R/2\gamma = 7/5 = 1.4,\; C_v = 5R/2TA=300 K,  PA=1 atmT_A = 300\,\mathrm{K},\; P_A = 1\,\mathrm{atm}r=VA/VB=8r = V_A/V_B = 8TC=2400 KT_C = 2400\,\mathrm{K}
(a)Ideal Gas — Thermodynamic States(b)First Law — Work and Heat for Each Process(c)Cycles — Otto vs Carnot Efficiency(d)Kinetic Theory — Molecules in Motion(e)Entropy — Second Law and Global Balance