Temperature and Thermal Expansion

Temperature scales, thermal equilibrium, and thermal expansion of materials

🌡️ 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: TK=TC+273.15T_K = T_C + 273.15 TC=TK−273.15T_C = T_K - 273.15

Celsius ↔ Fahrenheit: TF=95TC+32T_F = \frac{9}{5}T_C + 32 TC=59(TF−32)T_C = \frac{5}{9}(T_F - 32)

Temperature Changes:

  • ΔTK=ΔTC\Delta T_K = \Delta T_C (same size degree)
  • ΔTF=95ΔTC\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:

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

where:

  • ΔL\Delta L = change in length
  • α\alpha = coefficient of linear expansion (1/°C or 1/K)
  • L0L_0 = original length
  • ΔT\Delta T = temperature change

Final length: L=L0(1+αΔT)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:

ΔA=2αA0ΔT\Delta A = 2\alpha A_0 \Delta T

or using area expansion coefficient β=2α\beta = 2\alpha:

ΔA=βA0ΔT\Delta A = \beta A_0 \Delta T

Volume Expansion

For a 3D object:

ΔV=3αV0ΔT\Delta V = 3\alpha V_0 \Delta T

or using volume expansion coefficient γ=3α\gamma = 3\alpha:

ΔV=γV0ΔT\Delta V = \gamma V_0 \Delta T

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

Water: βV≈210×10−6\beta_V \approx 210 \times 10^{-6} /°C Mercury: βV≈180×10−6\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 α\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: ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T
    • Area: ΔA=2αA0ΔT\Delta A = 2\alpha A_0 \Delta T
    • Volume: ΔV=3αV0ΔT\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 ΔTK=ΔTC\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 TK=TC+273.15=25+273.15=298.15 KT_K = T_C + 273.15 = 25 + 273.15 = 298.15 \text{ K}

Part (b): 300 K to Celsius TC=TK−273.15=300−273.15=26.85°CT_C = T_K - 273.15 = 300 - 273.15 = 26.85°\text{C}

Part (c): 98.6°F to Celsius TC=59(TF−32)=59(98.6−32)T_C = \frac{5}{9}(T_F - 32) = \frac{5}{9}(98.6 - 32) TC=59(66.6)=37.0°CT_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 TK=TC+273.15=25+273.15=298.15 KT_K = T_C + 273.15 = 25 + 273.15 = 298.15 \text{ K}

Part (b): 300 K to Celsius TC=TK−273.15=300−273.15=26.85°CT_C = T_K - 273.15 = 300 - 273.15 = 26.85°\text{C}

Part (c): 98.6°F to Celsius TC=59(TF−32)=59(98.6−32)T_C = \frac{5}{9}(T_F - 32) = \frac{5}{9}(98.6 - 32) TC=59(66.6)=37.0°CT_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: L0=1000L_0 = 1000 m
  • Initial temp: T1=20°CT_1 = 20°\text{C}
  • Final temp: T2=40°CT_2 = 40°\text{C}
  • Linear expansion: α=11×10−6\alpha = 11 \times 10^{-6} /°C

Find: Change in length ΔL\Delta L

Solution:

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

Step 2: Apply linear expansion formula. ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T ΔL=(11×10−6)(1000)(20)\Delta L = (11 \times 10^{-6})(1000)(20) ΔL=0.22 m=22 cm\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: L0=1000L_0 = 1000 m
  • Initial temp: T1=20°CT_1 = 20°\text{C}
  • Final temp: T2=40°CT_2 = 40°\text{C}
  • Linear expansion: α=11×10−6\alpha = 11 \times 10^{-6} /°C

Find: Change in length ΔL\Delta L

Solution:

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

Step 2: Apply linear expansion formula. ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T ΔL=(11×10−6)(1000)(20)\Delta L = (11 \times 10^{-6})(1000)(20) ΔL=0.22 m=22 cm\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: r0=10.0r_0 = 10.0 cm =0.100= 0.100 m
  • ΔT=100−20=80°C\Delta T = 100 - 20 = 80°\text{C}
  • α=24×10−6\alpha = 24 \times 10^{-6} /°C

Solution:

Part (a): New radius

Step 1: Find change in radius. Δr=αr0ΔT=(24×10−6)(0.100)(80)\Delta r = \alpha r_0 \Delta T = (24 \times 10^{-6})(0.100)(80) Δr=1.92×10−4 m=0.0192 cm\Delta r = 1.92 \times 10^{-4} \text{ m} = 0.0192 \text{ cm}

Step 2: Find new radius. r=r0+Δr=10.0+0.0192=10.0192 cmr = r_0 + \Delta r = 10.0 + 0.0192 = 10.0192 \text{ cm}

Part (b): Change in volume

Method 1: Using volume expansion. ΔV=3αV0ΔT\Delta V = 3\alpha V_0 \Delta T

Initial volume: V0=43πr03=43π(0.100)3=4.19×10−3 m3V_0 = \frac{4}{3}\pi r_0^3 = \frac{4}{3}\pi (0.100)^3 = 4.19 \times 10^{-3} \text{ m}^3

Change: ΔV=3(24×10−6)(4.19×10−3)(80)\Delta V = 3(24 \times 10^{-6})(4.19 \times 10^{-3})(80) ΔV=2.41×10−5 m3=24.1 cm3\Delta V = 2.41 \times 10^{-5} \text{ m}^3 = 24.1 \text{ cm}^3

Method 2: Calculate volumes directly. Vf=43πrf3=43π(0.10019)3=4.214×10−3 m3V_f = \frac{4}{3}\pi r_f^3 = \frac{4}{3}\pi (0.10019)^3 = 4.214 \times 10^{-3} \text{ m}^3 ΔV=Vf−V0=2.4×10−5 m3\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: r0=10.0r_0 = 10.0 cm =0.100= 0.100 m
  • ΔT=100−20=80°C\Delta T = 100 - 20 = 80°\text{C}
  • α=24×10−6\alpha = 24 \times 10^{-6} /°C

Solution:

Part (a): New radius

Step 1: Find change in radius. Δr=αr0ΔT=(24×10−6)(0.100)(80)\Delta r = \alpha r_0 \Delta T = (24 \times 10^{-6})(0.100)(80) Δr=1.92×10−4 m=0.0192 cm\Delta r = 1.92 \times 10^{-4} \text{ m} = 0.0192 \text{ cm}

Step 2: Find new radius. r=r0+Δr=10.0+0.0192=10.0192 cmr = r_0 + \Delta r = 10.0 + 0.0192 = 10.0192 \text{ cm}

Part (b): Change in volume

Method 1: Using volume expansion. ΔV=3αV0ΔT\Delta V = 3\alpha V_0 \Delta T

Initial volume: V0=43πr03=43π(0.100)3=4.19×10−3 m3V_0 = \frac{4}{3}\pi r_0^3 = \frac{4}{3}\pi (0.100)^3 = 4.19 \times 10^{-3} \text{ m}^3

Change: ΔV=3(24×10−6)(4.19×10−3)(80)\Delta V = 3(24 \times 10^{-6})(4.19 \times 10^{-3})(80) ΔV=2.41×10−5 m3=24.1 cm3\Delta V = 2.41 \times 10^{-5} \text{ m}^3 = 24.1 \text{ cm}^3

Method 2: Calculate volumes directly. Vf=43πrf3=43π(0.10019)3=4.214×10−3 m3V_f = \frac{4}{3}\pi r_f^3 = \frac{4}{3}\pi (0.10019)^3 = 4.214 \times 10^{-3} \text{ m}^3 ΔV=Vf−V0=2.4×10−5 m3\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
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