💡 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:

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

1. Identify the type of expansion: linear, area, or volume
2. Choose reference state: usually room temperature
3. Apply appropriate formula:
   • Linear: $\Delta L = \alpha L_0 \Delta T$
   • Area: $\Delta A = 2\alpha A_0 \Delta T$
   • Volume: $\Delta V = 3\alpha V_0 \Delta T$
4. Watch temperature scale: Use Celsius or Kelvin (difference is same)
5. 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

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)

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!

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!

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

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