JEE Advanced problems often don't give all values directly. You need to know typical orders of magnitude — e.g., α for metals ~ 10⁻⁵ K⁻¹, thermal conductivity of metals ~ 10–400 W/mK. Without this intuition, you can't sanity-check your answers.
Interactive Thermal Calculator
🧮 Calculator
- Select the type of calculation from the dropdown
- Enter the known values in their respective units
- Click Calculate to get result with the formula shown
- For Stefan: use Kelvin, not Celsius
- For linear expansion: use exact α (e.g., 12×10⁻⁶ = 0.000012)
Always do a sanity check: Is the answer in the right order of magnitude? Linear expansion of 1m steel rod over 100°C ≈ 1.2mm. If your answer is 1.2 cm, you've made a power-of-10 error.
Typical Physical Values — Know These Cold
Specific Heat Capacities
| Substance | c (J/kg·K) | Memory Aid |
|---|---|---|
| Water | 4186 | Highest common liquid |
| Ice | 2100 | Half of water |
| Steam | 2010 | ~half of water |
| Copper | 385 | ~10× less than water |
| Iron/Steel | 460 | About 500 |
| Aluminium | 900 | ~1000 |
| Mercury | 140 | Very low |
Thermal Conductivity (K)
| Material | K (W/m·K) | Conductor? |
|---|---|---|
| Silver | 429 | Best conductor |
| Copper | 385 | Good conductor |
| Iron | 50 | Medium |
| Glass | 0.8 | Poor conductor |
| Water | 0.6 | Poor |
| Wood | 0.1 | Insulator |
| Air | 0.024 | Best insulator |
Expansion Coefficients (α for common materials)
In JEE Main, if a problem gives "a metal rod" without specifying material, use α = 10⁻⁵ K⁻¹ as the order of magnitude. For steel/iron specifically: 12 × 10⁻⁶ K⁻¹.
Order-of-Magnitude Estimation Problems
These are the thinking problems JEE Advanced tests. Build this intuition systematically.
The key to estimation: (1) Identify the dominant formula, (2) Plug in order-of-magnitude values, (3) Count powers of 10. If you get 10² and the answer choices are 10⁻², 10¹, 10², 10⁵ — pick 10². Precision comes from correctly using α, β, γ values.
Graph Interpretation for Thermal Chapter
JEE Main and NEET regularly ask graph-based questions. The key is knowing what SHAPE each thermal process produces and what the SLOPE represents.
Shape: Exponential decay curve. Starts steep, becomes flatter as T approaches T₀.
Slope: dT/dt = −k(T−T₀). Slope decreases as temperature approaches T₀.
The curve NEVER touches the T₀ line — only approaches asymptotically. Examiners check this.
Shape: Bell curve (Planck distribution). Rises, peaks at λ_max, then falls.
Key: As T increases → peak shifts LEFT (smaller λ, higher freq.) AND area under curve INCREASES (T⁴).
Wien's Law: λ_max↑ × T₂↑ = λ_max↓ × T₁↓ = constant. This is why hotter stars look blue-white.
Key Graph Patterns — Quick Reference
| Plot | X-axis | Y-axis | Shape | Slope Represents |
|---|---|---|---|---|
| Newton's Cooling | time (t) | Temperature T | Exponential decay | Rate of cooling (−k) |
| Newton's Cooling (log form) | time (t) | ln(T−T₀) | Straight line | −k (cooling constant) |
| Heating Curve | Heat added Q | Temperature T | Staircase (rises + flat) | 1/(mc) for rising; 0 for flat |
| Thermal Expansion | Temperature T | Length L | Linear | αL₀ |
| Heat Conduction (steady) | Position x | Temperature T | Linear (for uniform rod) | −(T₁−T₂)/L = temp gradient |
| Blackbody Spectrum | Wavelength λ | Intensity E | Bell curve (Planck) | Peak at λ_max = b/T |
Ready for Solved Problems?
6 problem types — direct formula, conceptual, multi-step, graph-based, A&R, case-based.