Temperature and Thermal Expansion

Temperature scales, thermal equilibrium, and thermal expansion of materials

Written and reviewed by the Study Mondo Education TeamLast updated February 14, 2026

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🌡️ Temperature and Thermal Expansion

What is Temperature?

Temperature is a measure of the average kinetic energy of particles in a substance. Higher temperature means particles move faster on average.

💡 Microscopic view: Temperature reflects the random thermal motion of atoms and molecules.

Temperature Scales

Celsius (°C)

Water freezes: 0°C
Water boils: 100°C (at 1 atm)
Used in most of the world

Fahrenheit (°F)

Water freezes: 32°F
Water boils: 212°F
Used primarily in the US

Kelvin (K)

Absolute temperature scale
Zero Kelvin = absolute zero (no molecular motion)
Water freezes: 273.15 K
Water boils: 373.15 K
No degree symbol! (Just "K", not "°K")

Conversions:

Celsius ↔ Kelvin: $T_K = T_C + 273.15$ $T_C = T_K - 273.15$

Celsius ↔ Fahrenheit: $T_F = \frac{9}{5}T_C + 32$ $T_C = \frac{5}{9}(T_F - 32)$

Temperature Changes:

$\Delta T_K = \Delta T_C$ (same size degree)
$\Delta T_F = \frac{9}{5}\Delta T_C$

Thermal Equilibrium

Zeroth Law of Thermodynamics: If objects A and B are each in thermal equilibrium with object C, then A and B are in thermal equilibrium with each other.

This seems obvious but is fundamental—it allows us to use thermometers!

When in thermal equilibrium:

No net heat flow between objects
Objects have the same temperature
System reaches stable state

Thermal Expansion

Most materials expand when heated and contract when cooled. Molecules vibrate more at higher temperatures, increasing average spacing.

Linear Expansion

For a solid rod or beam:

$\Delta L = \alpha L_0 \Delta T$

where:

$\Delta L$ = change in length
$\alpha$ = coefficient of linear expansion (1/°C or 1/K)
$L_0$ = original length
$\Delta T$ = temperature change

Final length: $L = L_0(1 + \alpha \Delta T)$

Common Linear Expansion Coefficients:

Material	α (×10⁻⁶ /°C)
Steel	11
Aluminum	24
Copper	17
Glass	9
Concrete	12

Area Expansion

For a flat surface:

$\Delta A = 2\alpha A_0 \Delta T$

or using area expansion coefficient $\beta = 2\alpha$ :

$\Delta A = \beta A_0 \Delta T$

Volume Expansion

For a 3D object:

$\Delta V = 3\alpha V_0 \Delta T$

or using volume expansion coefficient $\gamma = 3\alpha$ :

$\Delta V = \gamma V_0 \Delta T$

For liquids, we typically use $\beta_V$ directly (not related to linear expansion):

Water: $\beta_V \approx 210 \times 10^{-6}$ /°C Mercury: $\beta_V \approx 180 \times 10^{-6}$ /°C

Special Case: Water

Water is unusual! It has maximum density at 4°C:

Below 4°C: water expands as it cools
At 0°C: ice is less dense than water (ice floats)
This property is crucial for aquatic life in winter

Without this property:

Lakes would freeze from bottom up
All aquatic life would die in winter
Earth's climate would be very different

Applications

Engineering Considerations

Expansion joints in bridges, buildings, railroads
Gaps allow for thermal expansion without buckling
Power lines sag more in summer (expansion)

Bimetallic Strips

Two metals with different $\alpha$ bonded together
Bend when heated (one expands more than other)
Used in thermostats, circuit breakers

Railway Gaps

Older railroad tracks had gaps between sections
Modern continuous welded rail uses different techniques
Still must account for thermal stress

Problem-Solving Strategy

Identify the type of expansion: linear, area, or volume
Choose reference state: usually room temperature
Apply appropriate formula:
- Linear: $\Delta L = \alpha L_0 \Delta T$
- Area: $\Delta A = 2\alpha A_0 \Delta T$
- Volume: $\Delta V = 3\alpha V_0 \Delta T$
Watch temperature scale: Use Celsius or Kelvin (difference is same)
Check sign: Heating → expansion (positive), cooling → contraction (negative)

Common Mistakes

❌ Using Fahrenheit in thermal expansion (must use Celsius or Kelvin) ❌ Forgetting that $\Delta T_K = \Delta T_C$ (changes are equal) ❌ Using wrong coefficient (linear vs. volume) ❌ Not accounting for expansion in all dimensions ❌ Assuming water behaves normally below 4°C

📚 Practice Problems

1Problem 1easy

❓ Question:

Convert the following temperatures: (a) 25°C to Kelvin, (b) 300 K to Celsius, (c) 98.6°F (body temperature) to Celsius.

💡 Show Solution

Solution:

Part (a): 25°C to Kelvin $T_K = T_C + 273.15 = 25 + 273.15 = 298.15 \text{ K}$

Part (b): 300 K to Celsius $T_C = T_K - 273.15 = 300 - 273.15 = 26.85°\text{C}$

Part (c): 98.6°F to Celsius $T_C = \frac{5}{9}(T_F - 32) = \frac{5}{9}(98.6 - 32)$ $T_C = \frac{5}{9}(66.6) = 37.0°\text{C}$

Answer:

(a) 298.15 K (approximately 298 K)
(b) 26.85°C (room temperature)
(c) 37.0°C (normal body temperature)

2Problem 2easy

❓ Question:

Convert the following temperatures: (a) 25°C to Kelvin, (b) 300 K to Celsius, (c) 98.6°F (body temperature) to Celsius.

💡 Show Solution

Solution:

Part (a): 25°C to Kelvin $T_K = T_C + 273.15 = 25 + 273.15 = 298.15 \text{ K}$

Part (b): 300 K to Celsius $T_C = T_K - 273.15 = 300 - 273.15 = 26.85°\text{C}$

Part (c): 98.6°F to Celsius $T_C = \frac{5}{9}(T_F - 32) = \frac{5}{9}(98.6 - 32)$ $T_C = \frac{5}{9}(66.6) = 37.0°\text{C}$

Answer:

(a) 298.15 K (approximately 298 K)
(b) 26.85°C (room temperature)
(c) 37.0°C (normal body temperature)

3Problem 3medium

❓ Question:

A steel bridge is 1000 m long at 20°C. How much does it expand when the temperature rises to 40°C? (α_steel = 11 × 10⁻⁶ /°C)

💡 Show Solution

Given:

Original length: $L_0 = 1000$ m
Initial temp: $T_1 = 20°\text{C}$
Final temp: $T_2 = 40°\text{C}$
Linear expansion: $\alpha = 11 \times 10^{-6}$ /°C

Find: Change in length $\Delta L$

Solution:

Step 1: Calculate temperature change. $\Delta T = 40 - 20 = 20°\text{C}$

Step 2: Apply linear expansion formula. $\Delta L = \alpha L_0 \Delta T$ $\Delta L = (11 \times 10^{-6})(1000)(20)$ $\Delta L = 0.22 \text{ m} = 22 \text{ cm}$

Answer: The bridge expands by 0.22 m or 22 cm

This is why bridges need expansion joints!

4Problem 4medium

❓ Question:

A steel bridge is 1000 m long at 20°C. How much does it expand when the temperature rises to 40°C? (α_steel = 11 × 10⁻⁶ /°C)

💡 Show Solution

Given:

Original length: $L_0 = 1000$ m
Initial temp: $T_1 = 20°\text{C}$
Final temp: $T_2 = 40°\text{C}$
Linear expansion: $\alpha = 11 \times 10^{-6}$ /°C

Find: Change in length $\Delta L$

Solution:

Step 1: Calculate temperature change. $\Delta T = 40 - 20 = 20°\text{C}$

Step 2: Apply linear expansion formula. $\Delta L = \alpha L_0 \Delta T$ $\Delta L = (11 \times 10^{-6})(1000)(20)$ $\Delta L = 0.22 \text{ m} = 22 \text{ cm}$

Answer: The bridge expands by 0.22 m or 22 cm

This is why bridges need expansion joints!

5Problem 5hard

❓ Question:

An aluminum sphere has a radius of 10.0 cm at 20°C. (a) What is its radius at 100°C? (b) What is the change in volume? (α_Al = 24 × 10⁻⁶ /°C)

💡 Show Solution

Given:

Initial radius: $r_0 = 10.0$ cm $= 0.100$ m
$\Delta T = 100 - 20 = 80°\text{C}$
$\alpha = 24 \times 10^{-6}$ /°C

Solution:

Part (a): New radius

Step 1: Find change in radius. $\Delta r = \alpha r_0 \Delta T = (24 \times 10^{-6})(0.100)(80)$ $\Delta r = 1.92 \times 10^{-4} \text{ m} = 0.0192 \text{ cm}$

Step 2: Find new radius. $r = r_0 + \Delta r = 10.0 + 0.0192 = 10.0192 \text{ cm}$

Part (b): Change in volume

Method 1: Using volume expansion. $\Delta V = 3\alpha V_0 \Delta T$

Initial volume: $V_0 = \frac{4}{3}\pi r_0^3 = \frac{4}{3}\pi (0.100)^3 = 4.19 \times 10^{-3} \text{ m}^3$

Change: $\Delta V = 3(24 \times 10^{-6})(4.19 \times 10^{-3})(80)$ $\Delta V = 2.41 \times 10^{-5} \text{ m}^3 = 24.1 \text{ cm}^3$

Method 2: Calculate volumes directly. $V_f = \frac{4}{3}\pi r_f^3 = \frac{4}{3}\pi (0.10019)^3 = 4.214 \times 10^{-3} \text{ m}^3$ $\Delta V = V_f - V_0 = 2.4 \times 10^{-5} \text{ m}^3$ ✓

Answer:

(a) New radius: 10.02 cm (tiny change!)
(b) Volume change: 24 cm³

6Problem 6hard

❓ Question:

An aluminum sphere has a radius of 10.0 cm at 20°C. (a) What is its radius at 100°C? (b) What is the change in volume? (α_Al = 24 × 10⁻⁶ /°C)

💡 Show Solution

Given:

Initial radius: $r_0 = 10.0$ cm $= 0.100$ m
$\Delta T = 100 - 20 = 80°\text{C}$
$\alpha = 24 \times 10^{-6}$ /°C

Solution:

Part (a): New radius

Step 1: Find change in radius. $\Delta r = \alpha r_0 \Delta T = (24 \times 10^{-6})(0.100)(80)$ $\Delta r = 1.92 \times 10^{-4} \text{ m} = 0.0192 \text{ cm}$

Step 2: Find new radius. $r = r_0 + \Delta r = 10.0 + 0.0192 = 10.0192 \text{ cm}$

Part (b): Change in volume

Method 1: Using volume expansion. $\Delta V = 3\alpha V_0 \Delta T$

Initial volume: $V_0 = \frac{4}{3}\pi r_0^3 = \frac{4}{3}\pi (0.100)^3 = 4.19 \times 10^{-3} \text{ m}^3$

Change: $\Delta V = 3(24 \times 10^{-6})(4.19 \times 10^{-3})(80)$ $\Delta V = 2.41 \times 10^{-5} \text{ m}^3 = 24.1 \text{ cm}^3$

Method 2: Calculate volumes directly. $V_f = \frac{4}{3}\pi r_f^3 = \frac{4}{3}\pi (0.10019)^3 = 4.214 \times 10^{-3} \text{ m}^3$ $\Delta V = V_f - V_0 = 2.4 \times 10^{-5} \text{ m}^3$ ✓

Answer:

(a) New radius: 10.02 cm (tiny change!)
(b) Volume change: 24 cm³

Explain using:

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Heat and Specific Heat

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Temperature scales, thermal equilibrium, and thermal expansion of materials

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Temperature and Thermal Expansion

Try the Interactive Version!

🌡️ Temperature and Thermal Expansion

What is Temperature?

Temperature Scales

Celsius (°C)

Fahrenheit (°F)

Kelvin (K)

Conversions:

Thermal Equilibrium

Thermal Expansion

Linear Expansion

Common Linear Expansion Coefficients:

Area Expansion

Volume Expansion

Special Case: Water

Applications

Engineering Considerations

Bimetallic Strips

Railway Gaps

Problem-Solving Strategy

Common Mistakes

📚 Practice Problems

1Problem 1easy

❓ Question:

2Problem 2easy

❓ Question:

3Problem 3medium

❓ Question:

4Problem 4medium

❓ Question:

5Problem 5hard

❓ Question:

6Problem 6hard

❓ Question:

Practice Lab

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