What this chapter is about
This chapter explores how matter responds to changes in temperature. You will study the concepts of heat and temperature, learn how to measure temperature using different scales, and understand the various ways heat transfers from one body to another. The chapter builds on your earlier understanding of energy and introduces thermal expansion, specific heat capacity, latent heat, and the three modes of heat transfer: conduction, convection and radiation.
A Class 11 student meets this topic because thermal physics forms the foundation for thermodynamics, which you will study next. Understanding how materials expand when heated explains everyday phenomena like gaps left in railway tracks, the working of thermometers, and why lakes freeze from the top. These ideas connect physics to engineering, meteorology and biology.
After studying this chapter, you should be able to convert temperatures between Celsius, Fahrenheit and Kelvin scales, calculate heat absorbed or released during temperature changes and phase transitions, apply the principle of calorimetry, and explain heat transfer in practical situations such as cooking, insulation and climate.
Key ideas
- Temperature is a measure of the average kinetic energy of molecules in a substance; it determines the direction of heat flow between two bodies in contact.
- Heat is energy transferred between bodies due to a temperature difference; SI unit is joule (J), though calorie (cal) is sometimes used (1 cal = 4.186 J).
- Thermal equilibrium occurs when two bodies in contact reach the same temperature and no net heat flows between them (zeroth law of thermodynamics).
- Linear, area and volume expansion: when temperature rises by ΔT, a solid's length changes by ΔL = αL₀ΔT, area by ΔA = βA₀ΔT (β ≈ 2α), and volume by ΔV = γV₀ΔT (γ ≈ 3α), where α, β, γ are expansion coefficients.
- Specific heat capacity (c): heat needed to raise 1 kg of a substance by 1 K; Q = mcΔT.
- Latent heat (L): heat needed to change the phase of 1 kg of a substance without temperature change; Q = mL.
- Conduction transfers heat through a material without bulk motion; rate depends on thermal conductivity, area, temperature gradient and thickness.
- Convection transfers heat by bulk movement of fluid; radiation transfers heat through electromagnetic waves and needs no medium.
Formulas and facts to remember
- Formula: T(K) = T(°C) + 273.15 · Meaning: Conversion from Celsius to Kelvin
- Formula: T(°F) = (9/5)T(°C) + 32 · Meaning: Conversion from Celsius to Fahrenheit
- Formula: ΔL = αL₀ΔT · Meaning: Linear expansion of a solid
- Formula: ΔV = γV₀ΔT, γ ≈ 3α · Meaning: Volume expansion; γ is about three times α
- Formula: Q = mcΔT · Meaning: Heat absorbed or released during temperature change
- Formula: Q = mL · Meaning: Heat absorbed or released during phase change
- Formula: H = kA(T₁ − T₂)/d · Meaning: Rate of heat conduction through a slab (Fourier's law)
- Formula: P = σAT⁴ · Meaning: Power radiated by a black body (Stefan–Boltzmann law); σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴
- Water has an unusually high specific heat capacity (about 4186 J kg⁻¹ K⁻¹), which moderates climate near coasts.
- Anomalous expansion of water: water is densest at 4 °C, so ice floats and aquatic life survives in winter.
- Good conductors (metals) have high k; insulators (wood, air, wool) have low k.
Worked examples
Example 1: Temperature conversion
A weather report says the temperature in Shimla is 5 °C. Express this in Kelvin and Fahrenheit.
Solution
Kelvin: T(K) = 5 + 273.15 = 278.15 K ≈ 278 K
Fahrenheit: T(°F) = (9/5) × 5 + 32 = 9 + 32 = 41 °F
Example 2: Thermal expansion of a steel rod
A steel rod is 2.000 m long at 20 °C. Its coefficient of linear expansion α = 1.2 × 10⁻⁵ K⁻¹. Find its length when heated to 70 °C.
Solution
Temperature rise ΔT = 70 − 20 = 50 K
Change in length ΔL = αL₀ΔT = 1.2 × 10⁻⁵ × 2.000 × 50 = 1.2 × 10⁻³ m = 1.2 mm
New length = 2.000 + 0.0012 = 2.0012 m
Example 3: Calorimetry (heat exchange)
A 200 g iron block at 100 °C is dropped into 500 g of water at 25 °C in an insulated container. Specific heat of iron = 450 J kg⁻¹ K⁻¹; of water = 4186 J kg⁻¹ K⁻¹. Find the final temperature.
Solution
Let final temperature = T.
Heat lost by iron = heat gained by water (no heat escapes).
m₁c₁(100 − T) = m₂c₂(T − 25)
0.200 × 450 × (100 − T) = 0.500 × 4186 × (T − 25)
90(100 − T) = 2093(T − 25)
9000 − 90T = 2093T − 52325
9000 + 52325 = 2093T + 90T
61325 = 2183T
T = 61325 / 2183 ≈ 28.1 °C
The final equilibrium temperature is about 28 °C. Notice how the large specific heat of water keeps the temperature rise small despite the hot iron.
Common mistakes
- Using Celsius instead of Kelvin in formulas like Stefan–Boltzmann law → Always convert to Kelvin when the formula involves absolute temperature or T⁴.
- Forgetting that γ ≈ 3α for volume expansion of solids → Remember the relation; do not use α directly for volume problems.
- Confusing heat and temperature, treating them as the same quantity → Heat is energy transferred; temperature is a state variable that determines the direction of transfer.
- Ignoring sign conventions in calorimetry (adding heat lost and gained directly) → Set heat lost by hot body equal to heat gained by cold body, keeping magnitudes positive.
- Assuming latent heat causes a temperature change → During phase change, temperature stays constant even though heat is absorbed or released.
Quick revision
- Temperature scales: K = °C + 273; °F = (9/5)°C + 32.
- Linear expansion: ΔL = αL₀ΔT; for volume, use γ ≈ 3α.
- Heat exchange: Q = mcΔT (temperature change) or Q = mL (phase change).
- Conduction rate: H = kAΔT/d; metals conduct well, air and wool insulate.
- Stefan–Boltzmann law for radiation: P = σAT⁴ (T in kelvin).
- Water is densest at 4 °C; this anomalous expansion lets aquatic life survive freezing winters.