NEET · Physics

NEET Thermal Properties: Expansion, Calorimetry, Latent Heat and Heat Transfer

Own the thermal segment explicitly embedded in official Unit 7: heat, temperature, thermal expansion, specific heat, calorimetry, change of state, latent heat, and heat transfer by conduction, convection and radiation.

Subject: Physics
Official syllabus unit: Unit 7: Properties of Solids and Liquids

In short

Thermal Properties owns the heat-and-temperature portion of official Unit 7. Use it for thermal expansion, specific heat, calorimetry, phase change, latent heat and heat transfer. Thermodynamics begins only when the question is about equilibrium, state variables, heat/work/internal energy and thermodynamic laws/processes.

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Ashwini M.

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Also in Physics: Kumar Gandharva Saxena

Syllabus mapping

Syllabus mapping
Unit	Topics
Focused part of a larger current official unit	Temperature is not heat, Expansion, Calorimetry and phase change, Heat transfer

Unit
Focused part of a larger current official unit
Topics
Temperature is not heat, Expansion, Calorimetry and phase change, Heat transfer

Before this chapter

Concepts in this chapter

1. Temperature is not heat

Temperature characterizes thermal state; heat is energy transferred because of temperature difference. Avoid language that treats an object as 'containing heat' in the thermodynamic sense used by the syllabus.

2. Expansion

For small temperature changes over which expansion coefficients are approximately constant, linear expansion is proportional to original length and temperature change. Area and volume approximations for isotropic solids can be related to the linear coefficient under small-strain assumptions.

3. Calorimetry and phase change

Heating within one phase uses Q=mcΔT when specific heat is treated as constant over the interval. During an ideal phase change at constant transition temperature, latent heat uses Q=mL. Calorimetry problems should be solved by energy balance with a clearly defined insulated/closed model when heat loss is neglected.

4. Heat transfer

Conduction transfers energy through a material temperature gradient; convection involves bulk fluid motion; radiation transfers energy electromagnetically and does not require a material medium. Formula use must match the model and geometry.

5. Decision aids for this chapter

Heat vs temperature distinction card

Thermal process selector: temperature change vs phase change vs transfer

Unit 7 boundary diagram

Formula sheet

Linear expansion: ΔL = α L_0 ΔT. m. α K^-1. Small expansion with approximately constant coefficient over ΔT.
Linear expansion
Use when — m. α K^-1. Small expansion with approximately constant coefficient over ΔT.
Volume expansion isotropic solid: ΔV ≈ βV_0ΔT, with β≈3α. β K^-1. Approximate small isotropic expansion relation.
Volume expansion isotropic solid
Use when — β K^-1. Approximate small isotropic expansion relation.
Sensible heat: Q = mcΔT. J. m kg, c J kg^-1 K^-1. Use when no phase change occurs and c is treated constant.
Sensible heat
Use when — J. m kg, c J kg^-1 K^-1. Use when no phase change occurs and c is treated constant.
Latent heat: Q = mL. J. L J kg^-1. Ideal phase change; temperature remains at transition temperature under the simple model.
Latent heat
Use when — J. L J kg^-1. Ideal phase change; temperature remains at transition temperature under the simple model.
Conduction rate slab: Q/t = kA(T_hot-T_cold)/L. W. Steady one-dimensional conduction through uniform slab, constant k, area A, thickness L.
Conduction rate slab
Use when — W. Steady one-dimensional conduction through uniform slab, constant k, area A, thickness L.
Net thermal radiation: P_net = εσA(T^4-T_s^4). W. Grey-body approximation with emissivity ε; absolute temperatures in kelvin; surroundings represented by T_s.
Net thermal radiation
Use when — W. Grey-body approximation with emissivity ε; absolute temperatures in kelvin; surroundings represented by T_s.

Worked examples

A 2.0 m rod has linear expansion coefficient 2.0×10^-5 K^-1. It is heated by 50 K. Estimate the length increase.

Concept being tested: Thermal expansion

Prompt: A 2.0 m rod has linear expansion coefficient 2.0×10^-5 K^-1. It is heated by 50 K. Estimate the length increase.

Decisive condition: Small expansion and constant coefficient are assumed, so use linear expansion.

Method choice: ΔL=αLΔT.

Reasoning/calculation: ΔL=2.0×10^-5 × 2.0 × 50 = 2.0×10^-3 m = 2.0 mm.

Unit/sign/condition check: Kelvin difference equals Celsius-degree difference numerically for ΔT; final length change has unit m/mm.

PI if answered incorrectly: Execution Error if ΔT is omitted or units are mishandled; Decision / Selection Error if the coefficient is not approximately constant.

Common mistakes and what they actually indicate

Using Q=mcΔT during a phase change
Decision / selection error
Why it happens
Preparation Intelligence classification: Decision / Selection Error.
How it is corrected
Use latent heat for the ideal constant-temperature phase-change segment.
Using Celsius values in T^4 radiation expressions
Execution error
Why it happens
Preparation Intelligence classification: Execution Error.
How it is corrected
Use absolute temperature in kelvin.
Calling heat a state variable stored in the body
Knowledge gap
Why it happens
Preparation Intelligence classification: Knowledge Gap.
How it is corrected
Heat is energy transfer associated with temperature difference.
Using β=3α without small isotropic-expansion assumptions
Needs review
Why it happens
Preparation Intelligence classification: Needs Review.
How it is corrected
State the approximation and material model.

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