Heat & Thermodynamics – PSU‑GATE Physics Study Notes
Target audience: Candidates preparing for PSU (Public Sector Undertakings) jobs through the GATE examination.
Difficulty level: Intermediate (GATE‐level).
1. Simple Explanation
Thermodynamics is the branch of physics that deals with the relationship between heat, work, temperature, and energy.
- Heat is energy in transit due to temperature difference.
- Work is energy transfer that does not involve a temperature gradient.
- Thermodynamic systems can be open (exchange mass & energy), closed (exchange only energy), or isolated (no exchange).
The governing principles are the laws of thermodynamics, which set the limits on how energy can be transformed and how entropy behaves.
2. Key Points
| # | Concept | Key Equations / Statements | Notes |
|---|---|---|---|
| 1 | First Law (Energy Conservation) | ΔU = Q – W | ΔU = change in internal energy. Q = heat added to the system. W = work done by the system. |
| 2 | Second Law (Entropy) | dS ≥ δQ/T (for reversible processes equality holds) | Entropy S always increases in an isolated system. |
| 3 | Third Law | S → 0 as T → 0 K (for a perfect crystal) | Implies absolute zero is unattainable. |
| 4 | Carnot Cycle | Efficiency η = 1 – T_C/T_H | The maximum efficiency achievable by any heat engine operating between temperatures T_H and T_C. |
| 5 | Heat Engine | η = W/Q_H = 1 – Q_C/Q_H | W = work output, Q_H = heat absorbed, Q_C = heat rejected. |
| 6 | Refrigerator / Heat Pump | COP_R = Q_C/W = T_C/(T_H – T_C) | COP_R = coefficient of performance for a refrigerator. |
| 7 | Specific Heat Capacity | q = m c ΔT | c can be for specific heat at constant pressure (c_p) or constant volume (c_v). |
| 8 | Heat Transfer Modes | Conduction: Q = kA(ΔT/L) Convection: Q = hA(ΔT) Radiation: Q = εσA(T⁴ – T₀⁴) |
k = thermal conductivity, h = convective heat transfer coefficient, ε = emissivity, σ = Stefan–Boltzmann constant. |
| 9 | Ideal Gas Relations | PV = nRT ΔU = nC_VΔT ΔH = nC_pΔT |
For ideal gases, C_p – C_v = R. |
| 10 | Entropy Change for Ideal Gas | ΔS = nC_p ln(T₂/T₁) – nR ln(P₂/P₁) | Useful for reversible processes. |
3. Worked Example
Problem (Typical GATE style)
A Carnot heat engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. It absorbs 4000 J of heat from the hot reservoir during one cycle. Calculate:
- The work output per cycle.
- The heat rejected to the cold reservoir.
- The efficiency of the engine.
Solution
-
Carnot Efficiency
[ \eta = 1 - \frac{T_C}{T_H} = 1 - \frac{300}{600} = 0.5 ] -
Work Output
[ W = \eta , Q_H = 0.5 \times 4000,\text{J} = 2000,\text{J} ] -
Heat Rejected
Using first law:
[ Q_C = Q_H - W = 4000,\text{J} - 2000,\text{J} = 2000,\text{J} ]
Answer
- Work output = 2000 J per cycle.
- Heat rejected = 2000 J per cycle.
- Efficiency = 50 %.
Verification – The same result follows from the Carnot relation
[ \frac{Q_C}{Q_H} = \frac{T_C}{T_H} ;;\Rightarrow;; Q_C = \frac{300}{600}\times 4000 = 2000,\text{J} ]
4. Quick Revision Summary
| Concept | Remember |
|---|---|
| First Law | ΔU = Q – W (energy conservation). |
| Second Law | ΔS ≥ Q/T; entropy increases in irreversible processes. |
| Carnot Efficiency | η = 1 – T_C/T_H (max possible). |
| Heat Engine | η = 1 – Q_C/Q_H. |
| Refrigerator COP | COP_R = T_C/(T_H – T_C). |
| Heat Transfer | Conduction (kAΔT/L), Convection (hAΔT), Radiation (εσA(T⁴–T₀⁴)). |
| Ideal Gas | PV = nRT; ΔU = nC_VΔT; ΔH = nC_pΔT. |
| Entropy for Reversible Ideal Gas | ΔS = nC_p ln(T₂/T₁) – nR ln(P₂/P₁). |
| GATE Tip | Always check whether the process is reversible (use Carnot relations) or irreversible (use efficiency < Carnot). |
| Common Mistake | Confusing heat added with work done; sign conventions (Q positive into system, W |