◈ Termodinamica
Gas perfetti, trasformazioni termodinamiche, I e II principio, macchine termiche, cicli di Carnot e Otto, entropia.
Complete Theory
Worked Examples
Example 1Free adiabatic expansion — Entropy increases even without heat
Example 2Spontaneous heat flow — Why heat goes from hot to cold
Exercises with Solutions
Exercise 1Irreversible heating — A copper block in contact with a hot sourceHard
📋 Problem to solve
A copper block of mass kg at is placed in contact with a source at . Compute of the block, the source, and the universe. ( 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 K and K. It absorbs J and rejects J to the cold reservoir. Compute , and . Why is even though the engine is cyclic?
📌 Given data
T_H=500 KT_C=200 KQ_H=3000 JQ_C=1800 J
Recommended Books
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Integrative Problems
Problems combining all chapters — exam levelProblem 1The Thermoelectric Power Plant: from Gas to Molecules to EntropyEXTREME
A thermoelectric power plant uses of a diatomic gas (, ) running through the following cycle on a PV diagram:
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
State A: , . A→B: adiabatic compression to (compression ratio ). B→C: isochoric, heating to (combustion). C→D: adiabatic expansion to (return to original volume). D→A: isochoric, cooling (Otto cycle).
📌 Problem data
n = 5\,\mathrm{mol}\gamma = 7/5 = 1.4,\; C_v = 5R/2T_A = 300\,\mathrm{K},\; P_A = 1\,\mathrm{atm}r = V_A/V_B = 8T_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
