Heat & Temperature: Scales, Transfer, Expansion & Calorimetry

TAKEAWAYS:

• Temperature measures how hot/cold an object is (in °C, °F, or K), while heat is energy transferred between objects at different temperatures.
• Thermal expansion makes solids/liquids/gases expand when heated—critical for bridges, railway tracks, and mercury thermometers.
• Heat transfer happens via conduction (solids), convection (fluids), and radiation (all media), with applications in refrigerators, ovens, and solar panels.
• Calorimetry uses the principle of heat exchange (Q = mcΔT) to solve mixture problems, like cooling coffee with milk.
• Specific heat capacity tells how much heat a substance absorbs per °C (e.g., water has a high value, so it cools slowly).
• NEB loves numerical problems on mixtures, expansion, and heat transfer—always check units (J, cal, °C) and sign conventions.

1. Temperature and Its Measurement

What is temperature? Temperature is a measure of how hot or cold an object is. It tells us the average kinetic energy of the particles in a substance. The higher the temperature, the faster the particles move.

Temperature scales: Three common scales are used:

1. Celsius (°C): Water freezes at 0°C and boils at 100°C (used in most countries).
2. Fahrenheit (°F): Water freezes at 32°F and boils at 212°F (used in the USA).
3. Kelvin (K): Absolute scale where 0 K = −273.15°C (no negative temperatures; used in science).

Conversion formulas:

°C = \frac{5}{9}(°F - 32)
°F = \frac{9}{5}°C + 32
K = °C + 273.15

How is temperature measured? Common thermometers use:

• Liquid expansion (mercury or alcohol in glass tubes).
• Bimetallic strips (used in ovens and fire alarms).
• Thermocouples (for high temperatures in industries).
• Digital sensors (thermistors in modern devices).

Example 1: Convert 98.6°F to Celsius Use the formula: $$°C=\frac{5}{9}(98.6-32)=\frac{5}{9}(66.6)=37°C$$ Answer: 98.6°F = 37°C (normal human body temperature).

2. Thermal Expansion

When objects are heated, their particles gain energy and move faster, causing the object to expand. This happens in solids, liquids, and gases, but the effect is most noticeable in gases.

Types of Expansion:

1. Linear Expansion (Solids):

   • Expansion in one dimension (e.g., railway tracks, bridges).
   • Formula: $$\Delta L=L_0\alpha \Delta T$$ where:
     • $\Delta L$ = change in length,
     • $L_0$ = original length,
     • $\alpha$ = coefficient of linear expansion (varies by material),
     • $\Delta T$ = change in temperature.

2. Area Expansion (Solids):

   • Expansion in two dimensions (e.g., metal sheets).
   • Formula: $$\Delta A=A_0\beta \Delta T$$ where $\beta =2\alpha$ (for isotropic materials).

3. Volume Expansion (Liquids & Gases):

   • Expansion in three dimensions (e.g., mercury in thermometers, alcohol in car radiators).
   • Formula: $$\Delta V=V_0\gamma \Delta T$$ where $\gamma =3\alpha$ (for solids) or is given for liquids/gases.

Example 2: A steel rod expands from 1 m to 1.001 m when heated from 20°C to 120°C. Find its coefficient of linear expansion ($\alpha$). Given:

• $L_0=1\text{m}$,
• $\Delta L=1.001-1=0.001\text{m}$,
• $\Delta T=120-20=100°C$.

Using $\Delta L=L_0\alpha \Delta T$: $$\begin{gathered} 0.001=1\times \alpha \times 100 \\ \alpha =\frac{0.001}{100}=1.0\times 10^{-5}{\text{°C}}^{-1} \end{gathered}$$ Answer: $\alpha =1.0\times 10^{-5}{\text{°C}}^{-1}$.

Why is thermal expansion important?

• Pros: Used in thermostats, bimetallic strips, and expansion joints in bridges/roads.
• Cons: Can cause cracks in buildings, burst pipes, or loose-fitting lids if not accounted for.

3. Heat and Its Transfer

What is heat? Heat is the transfer of energy from a hotter object to a colder one. It is measured in joules (J) or calories (cal) (1 cal = 4.18 J).

Modes of Heat Transfer:

Example 3: Why does a metal spoon feel hotter than a plastic one when dipped in hot soup?

• Metal conducts heat fast (high thermal conductivity), so it transfers heat to your hand quickly.
• Plastic conducts heat slowly (low thermal conductivity), so it feels cooler.

flowchart TD
    A["Hot Object\n(e.g., stove)"] -->|"Heat Transfer"| B["Conduction\n(Solids)"]
    A -->|"Heat Transfer"| C["Convection\n(Fluids)"]
    A -->|"Heat Transfer"| D["Radiation\n(All)"]
    B --> E["Metal spoon\n(feels hot fast)"]
    C --> F["Boiling water\n(circulation)"]
    D --> G["Sunlight\n(heats Earth)"]

4. Calorimetry and Heat Exchange

Principle of Calorimetry: When two objects at different temperatures are mixed, heat lost = heat gained (assuming no heat is lost to the surroundings).

Formula: $$Q=mc\Delta T$$ where:

• $Q$ = heat energy (J or cal),
• $m$ = mass (kg or g),
• $c$ = specific heat capacity (J/kg·°C or cal/g·°C),
• $\Delta T$ = temperature change (°C or K).

Specific Heat Capacity (c): This tells how much heat is needed to raise 1 kg (or 1 g) of a substance by 1°C.

• Water: $c=4186\text{J/kg⋅°C}$ (highest for common liquids—why oceans moderate climate!).
• Iron: $c=450\text{J/kg⋅°C}$.
• Alcohol: $c=2400\text{J/kg⋅°C}$.

Example 4: How much heat is needed to raise 2 kg of water from 20°C to 80°C? Given:

• $m=2\text{kg}$,
• $c=4186\text{J/kg⋅°C}$,
• $\Delta T=80-20=60°C$.

Using $Q=mc\Delta T$: $$Q=2\times 4186\times 60=502,320\text{J}$$ Answer: 502.32 kJ of heat is required.

Example 5: A 50 g copper block (c = 390 J/kg·°C) at 100°C is dropped into 200 g of water at 20°C. Find the final temperature. Assumptions:

• No heat is lost to the surroundings.
• Final temperature = $T_f$.

Heat lost by copper = Heat gained by water: $$m_c{c}_c(100-T_f)=m_w{c}_w(T_f-20)$$ $$0.05\times 390\times (100-T_f)=0.2\times 4186\times (T_f-20)$$ $$19.5(100-T_f)=837.2(T_f-20)$$ $$1950-19.5T_f=837.2T_f-16,744$$ $$1950+16,744=837.2T_f+19.5T_f$$ $$\begin{gathered} 18,694=856.7T_f \\ {T}_f=\frac{18,694}{856.7}\approx 21.8°C \end{gathered}$$ Answer: Final temperature ≈ 21.8°C.

5. Anomalous Expansion of Water

Most substances expand when heated, but water behaves differently between 0°C and 4°C:

• It contracts when heated from 0°C to 4°C.
• It expands when heated above 4°C or cooled below 0°C (why ice floats!).

Why does this matter?

• Aquatic life survives in winter because ice (less dense) floats on water.
• Pipes burst in freezing weather because water expands as it turns to ice.

Exam Tips for NEB Physics (Unit 9)

1. Units matter!

   • Always convert masses to kg and temperatures to °C (or K) before calculations.
   • Remember: $1\text{cal}=4.18\text{J}$.

2. Sign conventions in calorimetry:

   • Heat lost by a hot object is negative ($Q=-mc\Delta T$).
   • Heat gained by a cold object is positive ($Q=+mc\Delta T$).

3. Common mistakes to avoid:

   • Forgetting to add 273 when converting °C to K.
   • Mixing up linear ($\alpha$) and volume ($\gamma$) expansion coefficients.
   • Ignoring the anomalous expansion of water in problems involving ice or freezing.

4. NEB-style questions to practice:

   • Short answer:
     • "Why are mercury thermometers used for measuring high temperatures?"
     • "Define specific heat capacity. Why does water have a high specific heat capacity?"
   • Numerical problems:
     • Calculate the final temperature of a mixture (like Example 5).
     • Find the coefficient of linear expansion given initial/final lengths and temperature change.
     • Determine the heat required to raise the temperature of a substance.
   • Conceptual questions:
     • "Explain why a bimetallic strip bends when heated."
     • "How does the anomalous expansion of water help aquatic life in winter?"

5. Diagrams in exams:

   • Always label axes in graphs (e.g., temperature vs. time).
   • Draw clear expansion diagrams (e.g., a rod expanding with $\Delta L$ marked).
   • For calorimetry, sketch a system with heat flow arrows.

NEB Board-Style Questions (Practice)

Question 1 (Short Answer): Define the term "thermal expansion." Why are gaps left between railway tracks?

Answer: Thermal expansion is the increase in dimensions (length, area, or volume) of a substance when its temperature rises. Gaps are left in railway tracks to accommodate expansion during hot weather, preventing buckling.

Question 2 (Numerical): A 0.5 kg aluminum pan (specific heat capacity = 900 J/kg·°C) contains 2 kg of water at 20°C. How much heat is required to raise the temperature of the pan and water to 80°C? (Assume no heat loss.)

Solution: Heat for water ($Q_w$): $$Q_w=m_w{c}_w\Delta T=2\times 4186\times (80-20)=2\times 4186\times 60=502,320\text{J}$$

Heat for aluminum pan ($Q_a$): $$Q_a=m_a{c}_a\Delta T=0.5\times 900\times 60=27,000\text{J}$$

Total heat ($Q_{\text{total}}$): $$Q_{\text{total}}=Q_w+Q_a=502,320+27,000=529,320\text{J}$$

Answer: 529.32 kJ of heat is required.

Question 3 (Conceptual): Explain why a steel bridge has expansion joints, but a concrete bridge does not need them. Which material has a higher coefficient of linear expansion?

Answer:

• Steel expands more than concrete when heated, so expansion joints are needed to prevent stress.
• Concrete expands very little, so joints are unnecessary.
• Steel has a higher $\alpha$ (12 × 10⁻⁶/°C) than concrete (10 × 10⁻⁶/°C).

Final Summary Table

Remember:

• Heat is energy in transit (J or cal).
• Temperature is a measure of average kinetic energy (°C, K, °F).
• Expansion is bigger in gases > liquids > solids.
• Calorimetry problems always use $Q=mc\Delta T$ with heat lost = heat gained.

Good luck with your NEB exams! 🚀