MixedJEE Physics · Original learning card5 original chapter questions

Heat Engines and Efficiency

For a cyclic heat engine absorbing magnitude Q_H from a hot reservoir and rejecting magnitude Q_C to a cold reservoir, W = Q_H − Q_C and η = W/Q_H = 1 − Q_C/Q_H, with 0 ≤ η < 1.

Why this shows up in the exam

Finding work from heat flows · Efficiency of a specified gas cycle · Comparing actual engines with reversible limits

Learn the idea

A heat engine converts only part of absorbed heat into cyclic work and rejects the rest. Because the working substance returns to its initial state each cycle, its net work is the difference between heat received and heat rejected.

🧠 Memory hook: Input heat splits into work plus rejected heat.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • W = Q_H - Q_C — cycle energy balance using positive heat magnitudes
  • η = W/Q_H = 1 - Q_C/Q_H — thermal efficiency of any heat engine

How to approach it

  1. 1Draw heat-in, work-out, and heat-out arrows
  2. 2Use the cycle energy balance
  3. 3Compare the result with the Carnot upper bound when temperatures are given

Common slip-ups that cost marks

  • •Dividing work by rejected heat
  • •Treating signed Q_C as a positive term without changing formula
  • •Claiming complete conversion of cyclic heat into work

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 5

A gas absorbs 500 J of heat and does 200 J of work. What is the change in its internal energy?

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