Skip to content
III. Long Answer Questions · Q23

Q.Explain the second law of thermodynamics in terms of entropy.

Tamil Nadu DgeTextbookSubjectiveImportance★★★★★
60% · 76/126 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Step 1. Since QH/TH=QL/TLQ_H/T_H=Q_L/T_L for a reversible (Carnot) engine, the quantity Q/TQ/T is defined as entropy, a state variable; QH/THQ_H/T_H is the entropy received from the hot reservoir, and QL/TLQ_L/T_L the entropy given out to the cold reservoir -- equal for a reversible engine, so its net entropy change over one cycle is zero.

Step 2. For real, irreversible engines, instead QL/TL>QH/THQ_L/T_L>Q_H/T_H -- entropy given out exceeds entropy received, so total entropy increases. This motivates the general entropy statement: for every process that actually occurs in nature (all of which are irreversible), the total entropy always increases; only for an idealised reversible process does entropy stay unchanged.

Step 3. This directly explains why heat only ever flows spontaneously from hot to cold: the entropy gained by the cooler object (+Q/Tcold+Q/T_{cold}) exceeds in magnitude the entropy lost by the hotter object (−Q/Thot-Q/T_{hot}, since Tcold<ThotT_{cold}<T_{hot}), so this direction increases total entropy. The reverse flow (cold to hot) would DECREASE total entropy, which the second law forbids. …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.