Heat & Thermodynamics – MHT CET Physics Study Notes
Target exam: MHT CET
Subject: Physics – Thermodynamics & Heat
Time‑spent: ~1 h of focused revision
1. Simple Explanation
Heat is the transfer of energy due to a temperature difference; temperature is a measure of average molecular kinetic energy.
Thermodynamics is the branch of physics that studies the relation between heat, work and energy in macroscopic systems.
A thermodynamic system is the part of the universe we focus on. The surroundings are everything else.
The state of a system is described by variables like (P) (pressure), (V) (volume), (T) (temperature), (S) (entropy), (U) (internal energy).
2. Key Points
| Topic | What to remember |
|---|---|
| First Law | ( \Delta U = Q - W ) (sign convention: (Q>0) into system, (W>0) done by system) |
| Work in common processes | • Isothermal (ideal gas): ( W = nRT \ln(V_f/V_i) ) • Adiabatic (ideal gas): ( W = \frac{P_iV_i - P_fV_f}{\gamma-1}) |
| Specific Heat | ( Q = mc\Delta T ) where (c) is specific heat capacity |
| Calorimetry | ( m_1c_1\Delta T_1 = m_2c_2\Delta T_2 ) (energy conservation in mixing) |
| Second Law | Heat flows spontaneously from hot to cold; no engine can be 100 % efficient. |
| Entropy | ( \Delta S = \int \frac{dQ_{\text{rev}}}{T} ) |
| Carnot Efficiency | ( \eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h} ) (max efficiency for any engine between (T_h) and (T_c)) |
| Thermodynamic Processes | Isobaric, isochoric, isothermal, adiabatic – each has a distinct (PV) relation. |
| Heat Transfer | Conduction: ( \dot Q = kA \frac{\Delta T}{L} ) Convection: ( \dot Q = hA \Delta T ) Radiation: ( \dot Q = \epsilon\sigma A (T^4_{\text{hot}}-T^4_{\text{cold}}) ) |
3. Worked Example
Example 1 – Heat required to heat water
Problem:
Find the amount of heat ((Q)) needed to raise the temperature of (2;\text{kg}) of water from (25^\circ\text{C}) to (75^\circ\text{C}).
Solution
- Use the specific heat formula
[ Q = m,c,\Delta T ] - Insert values
[ m = 2;\text{kg},\qquad c_{\text{water}} = 4.18\times10^3;\text{J,kg}^{-1}\text{K}^{-1},\qquad \Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50;\text{K} ] - Compute
[ Q = (2)(4.18\times10^3)(50) = 4.18\times10^2 \times 50 \times 2 ] [ Q = 2 \times 4.18\times10^3 \times 50 = 4.18\times10^5 ;\text{J} ] - Answer
[ \boxed{Q = 4.18\times10^5;\text{J}} ]
Example 2 – Carnot Engine Efficiency
Problem:
A Carnot engine operates between a hot reservoir at (900;\text{K}) and a cold reservoir at (300;\text{K}). What is its maximum efficiency?
Solution
- Use Carnot formula
[ \eta_{\text{Carnot}} = 1 - \frac{T_c}{T_h} ] - Insert temperatures
[ \eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3} ] - Convert to percentage
[ \eta = \frac{2}{3}\times 100% \approx 66.7% ] - Answer
[ \boxed{\eta_{\text{max}}\approx 66.7%}