Unit 3 · Chapter 01

01Thermal Energy and Heat

Distinguish thermal energy, temperature, and heat; calculate heat transfer using specific heat capacity; analyze conduction, convection, and radiation.

Thermal energy is the most common form of energy in everyday life. Understanding heat transfer is essential for engineering, climate science, and thermodynamics.

How does energy move between objects, and why does the same amount of heat raise the temperature of some materials more than others?

Lesson Overview

Thermal energy, temperature, and heat are three related but distinct concepts. Thermal energy is the total kinetic energy of all particles in an object. Temperature is the average kinetic energy per particle — a measure of how hot or cold something is. Heat is the transfer of thermal energy between objects at different temperatures, always flowing from hot to cold. In this lesson you will learn to convert between temperature scales, calculate heat absorbed or released using specific heat capacity and latent heat, analyze the three modes of heat transfer, apply thermal expansion, and solve calorimetry problems using conservation of energy.

Temperature Scales

Celsius (°C)

Based on water: 0 °C = freezing point, 100 °C = boiling point at 1 atm.

°C = (°F − 32) × 5/9

°C = K − 273.15

Fahrenheit (°F)

Common in the US: 32 °F = freezing, 212 °F = boiling at 1 atm.

°F = °C × 9/5 + 32

°F = (K − 273.15) × 9/5 + 32

Kelvin (K)

SI unit. Absolute zero (0 K) = no thermal motion. No negative values.

K = °C + 273.15

K = (°F − 32) × 5/9 + 273.15

Heat Transfer Methods

🔥 Conduction

Heat transfer through direct contact between particles. Occurs in solids (especially metals). Energy passes from particle to particle via collisions.

Q/t = kAΔT/d

k = thermal conductivity, A = area, d = thickness

🌊 Convection

Heat transfer by bulk movement of a fluid (liquid or gas). Warm fluid rises, cool fluid sinks, creating convection currents.

No single formula — governed by fluid dynamics

Examples: boiling water, ocean currents, weather

☀️ Radiation

Heat transfer via electromagnetic waves (infrared). Requires no medium — works through a vacuum. All objects emit radiation.

P = εσAT⁴

ε = emissivity, σ = 5.67×10⁻⁸ W/m²K⁴

Key Equations

Q = mcΔT

Specific heat capacity (sensible heat)

Q = mL

Latent heat (phase change)

K = °C + 273.15

Celsius → Kelvin conversion

ΔL = αLΔT

Linear thermal expansion

P = εσAT⁴

Stefan–Boltzmann radiation law

Q_lost = Q_gained

Calorimetry / conservation of energy

Worked Examples

Example 1

How much heat is required to raise the temperature of 2 kg of water from 20 °C to 80 °C? (c_water = 4,186 J/kg·°C)

Identify: m = 2 kg, c = 4,186 J/kg·°C, ΔT = 80 − 20 = 60 °C

Formula: Q = mcΔT

Substitute: Q = 2 × 4,186 × 60

Q = 2 × 251,160 = 502,320 J

Answer:Q ≈ 502,320 J ≈ 502 kJ
Example 2

How much heat is needed to completely melt 0.5 kg of ice at 0 °C? (L_f = 334,000 J/kg)

Identify: m = 0.5 kg, L_f = 334,000 J/kg

Formula: Q = mL

Substitute: Q = 0.5 × 334,000

Q = 167,000 J

Answer:Q = 167,000 J = 167 kJ
Example 3

A steel rod is 2 m long at 20 °C. The temperature rises by 100 °C. How much does it expand? (α_steel = 12 × 10⁻⁶ /°C)

Identify: α = 12 × 10⁻⁶ /°C, L = 2 m, ΔT = 100 °C

Formula: ΔL = αLΔT

Substitute: ΔL = 12 × 10⁻⁶ × 2 × 100

ΔL = 12 × 10⁻⁶ × 200 = 2,400 × 10⁻⁶ m = 0.0024 m

Answer:ΔL = 0.0024 m = 2.4 mm
Example 4

A 0.2 kg copper block (c = 385 J/kg·°C) at 150 °C is dropped into 0.5 kg of water (c = 4,186 J/kg·°C) at 20 °C. Find the final equilibrium temperature.

Set heat lost by copper = heat gained by water: m_Cu × c_Cu × (T_i,Cu − T_f) = m_w × c_w × (T_f − T_i,w)

0.2 × 385 × (150 − T_f) = 0.5 × 4,186 × (T_f − 20)

77 × (150 − T_f) = 2,093 × (T_f − 20)

11,550 − 77T_f = 2,093T_f − 41,860

11,550 + 41,860 = 2,093T_f + 77T_f

53,410 = 2,170 T_f

T_f = 53,410 ÷ 2,170 ≈ 24.6 °C

Answer:T_f ≈ 24.6 °C
Example 5

Convert normal body temperature 98.6 °F to Celsius and Kelvin.

Celsius: °C = (°F − 32) × 5/9 = (98.6 − 32) × 5/9 = 66.6 × 5/9

°C = 333/9 = 37.0 °C

Kelvin: K = °C + 273.15 = 37.0 + 273.15

Answer:37.0 °C and 310.15 K

Guided Problems

Guided Problem 1

How much heat is released when 3 kg of aluminum (c = 900 J/kg·°C) cools from 200 °C to 25 °C?

Hint: Use Q = mcΔT. ΔT = 200 − 25 = 175 °C. Multiply m × c × ΔT.

Guided Problem 2

How much heat is required to vaporize 0.25 kg of water at 100 °C? (L_v = 2,260,000 J/kg)

Hint: Use Q = mL_v. Multiply mass by the latent heat of vaporization.

Guided Problem 3

A brass rod (α = 19 × 10⁻⁶ /°C) is 1.5 m long at 0 °C. What is its length at 200 °C?

Hint: Find ΔL = αLΔT first, then add to original length: L_final = L + ΔL.

Guided Problem 4

Convert −40 °F to Celsius. What do you notice about the result?

Hint: Use °C = (°F − 32) × 5/9. Notice that −40 °F = −40 °C — the two scales intersect at this point!

Guided Problem 5

A 0.1 kg iron block (c = 450 J/kg·°C) at 300 °C is placed in 0.4 kg of water at 15 °C. Estimate the final temperature.

Hint: Set Q_lost = Q_gained: m_Fe × c_Fe × (300 − T_f) = m_w × c_w × (T_f − 15). Solve for T_f.

Key Vocabulary

Thermal Energy

The total kinetic energy of all particles in an object. Depends on both temperature and the number of particles.

Example: A large pot of warm water has more thermal energy than a small cup at the same temperature.

Temperature

The average kinetic energy per particle in a substance. Measured in °C, °F, or K.

Example: Water boils at 100 °C (373.15 K) at standard atmospheric pressure.

Heat

The transfer of thermal energy between objects due to a temperature difference. Flows from hot to cold.

Example: Placing a cold spoon in hot soup transfers heat from the soup to the spoon.

Specific Heat Capacity

The amount of heat required to raise 1 kg of a substance by 1 °C. Symbol: c, units: J/kg·°C.

Example: Water (c = 4,186 J/kg·°C) heats up more slowly than iron (c = 450 J/kg·°C).

Latent Heat

The heat absorbed or released during a phase change at constant temperature. Q = mL.

Example: Ice melts at 0 °C absorbing 334,000 J/kg without changing temperature.

Conduction

Heat transfer through direct particle-to-particle contact, primarily in solids.

Example: A metal spoon left in hot soup becomes hot at the handle end via conduction.

Convection

Heat transfer by the bulk movement of a fluid (liquid or gas) due to density differences.

Example: Hot air rises near a heater, cool air sinks — creating a convection current in a room.

Radiation

Heat transfer via electromagnetic waves (infrared). Requires no medium; can travel through a vacuum.

Example: The Sun heats Earth through 150 million km of empty space by radiation.

Workbook Check

Interactive Practice — 5 Questions

1

How much heat is needed to raise 1 kg of water (c = 4,186 J/kg·°C) from 10 °C to 60 °C?

2

What is 0 °C expressed in Kelvin?

3

Which heat transfer method does NOT require a medium (can travel through a vacuum)?

4

A steel rod (α = 12 × 10⁻⁶ /°C) is 1 m long. It is heated by 50 °C. What is ΔL?

5

In a calorimetry experiment, the principle used to find the final temperature is:

Independent Practice

1

Convert the following temperatures: (a) 37°C to Kelvin, (b) 300 K to Celsius, (c) 98.6°F to Celsius.

2

A steel rod is 2.000 m long at 20°C. How much does it expand when heated to 120°C? (α_steel = 12 × 10⁻⁶ /°C)

3

How much heat is needed to raise the temperature of 0.50 kg of water from 20°C to 80°C? (c_water = 4,186 J/kg·°C)

4

A 200 g aluminum block at 150°C is dropped into 500 g of water at 20°C. Find the final equilibrium temperature. (c_Al = 900 J/kg·°C, c_water = 4,186 J/kg·°C)

5

★ A 0.30 kg copper block (c = 385 J/kg·°C) at 250°C is placed in a 0.80 kg aluminum calorimeter (c = 900 J/kg·°C) containing 0.50 kg of water at 15°C. (a) Write the heat balance equation. (b) Find the final temperature. (c) How much heat did the water absorb? (d) What assumption did you make about heat loss to the surroundings?

Challenge
⚠️

Common Mistakes

Confusing heat (Q) and temperature (T) — treating them as the same thing

Temperature measures average KE of particles; heat is energy transferred due to a temperature difference. A large cold object can have more thermal energy than a small hot one

Forgetting to convert Celsius to Kelvin when using gas laws or thermal energy equations

Always use Kelvin (K = °C + 273.15) in any equation involving absolute temperature (gas laws, radiation, etc.)

Setting Q_lost = Q_gained without using the correct sign convention

In calorimetry: Q_lost by hot = Q_gained by cold. Write m₁c₁(T₁−T_f) = m₂c₂(T_f−T₂) and solve for T_f

Ignoring the latent heat term when a substance changes phase

During a phase change, temperature stays constant. Use Q = mL (not Q = mcΔT) for melting/freezing or vaporization/condensation

💡

Math Tips

🌡️

Temperature conversions: K = °C + 273.15; °F = (9/5)°C + 32; °C = (°F − 32) × 5/9. For physics problems, always work in Kelvin

🔢

Specific heat of water: c = 4186 J/(kg·°C) ≈ 4200 J/(kg·°C). Water has an unusually high specific heat — it resists temperature change

📐

Linear thermal expansion: ΔL = αL₀ΔT. Area expansion: ΔA ≈ 2αA₀ΔT. Volume expansion: ΔV = βV₀ΔT where β ≈ 3α for isotropic solids

Heat transfer rate by conduction: P = kA(ΔT/d). Thicker material (larger d) or lower conductivity (smaller k) reduces heat flow rate