**Temperature** is a measure of the degree of hotness or coldness of a body. It is a relative measure and indicates the thermal state of matter. We perceive temperature through touch, but this sense is unreliable and limited in range for scientific purposes.
**Heat** is the form of energy transferred between two or more systems (or a system and its surroundings) by virtue of temperature difference. When a temperature difference exists, heat spontaneously flows from the higher temperature region to the lower temperature region until thermal equilibrium is established.
**Key Distinctions:**
**SI Units:**
**Real-Life Example:** A blacksmith heating an iron ring—the ring gains heat from the furnace, its temperature rises, allowing it to expand and fit on a wheel rim. Once removed, it loses heat to the environment and contracts.
---
Temperature is measured using a **thermometer**, an instrument based on variation of physical properties with temperature.
**Physical Property Used:** Volume change of liquids with temperature. Liquids like mercury and alcohol expand linearly with temperature over wide ranges, making them ideal thermometer liquids.
**Thermometer Calibration:**
Two fixed reference points standardize temperature scales:
These are measured under standard atmospheric pressure (1 atm = 101,325 Pa).
**Temperature Scales:**
**Celsius Scale:**
**Fahrenheit Scale:**
**Conversion Formula:**
The relationship between Celsius and Fahrenheit temperatures is linear (straight line):
$$\frac{t_F - 32}{180} = \frac{t_C}{100}$$
Or: **t_F = (9/5)t_C + 32**
Or: **t_C = (5/9)(t_F - 32)**
**Example:** Convert 25°C to Fahrenheit:
t_F = (9/5) × 25 + 32 = 45 + 32 = 77°F
**Worked Problem:** If water boils at 212°F, what is this in Celsius?
t_C = (5/9)(212 - 32) = (5/9) × 180 = 100°C ✓
---
**Boyle's Law (constant temperature):**
For a fixed mass of gas at constant temperature:
$$PV = \text{constant}$$
Or: **P₁V₁ = P₂V₂** (at constant T)
Pressure and volume are inversely proportional.
**Charles' Law (constant pressure):**
For a fixed mass of gas at constant pressure:
$$\frac{V}{T} = \text{constant}$$
Or: **V₁/T₁ = V₂/T₂** (at constant P)
Volume and absolute temperature are directly proportional.
**Combined Gas Law:**
For a fixed mass of gas:
$$\frac{PV}{T} = \text{constant}$$
**Ideal Gas Equation (General Form):**
For any quantity of any gas at low density:
$$PV = \mu RT$$
Where:
**Absolute Zero and Kelvin Scale:**
Extrapolating pressure-temperature graphs for low-density gases shows that pressure approaches zero at **T = -273.15°C**. This is **absolute zero**, the theoretical lowest possible temperature where all molecular motion ceases (in classical sense). No substance can be cooled below absolute zero.
**Kelvin (Absolute) Temperature Scale:**
**Temperature Conversion:**
$$T(K) = t(°C) + 273.15$$
For most calculations, use: **T(K) = t(°C) + 273**
**SI unit of temperature is Kelvin (K).**
**Relationship between scales:**
**Example:** Room temperature at 25°C in Kelvin:
T = 25 + 273 = 298 K
**Gas Thermometer Principle:**
A constant-volume gas thermometer uses P ∝ T relationship. By measuring pressure change at constant volume, temperature can be determined. All gases give identical readings regardless of type (at low densities)—this property validates the ideal gas law.
---
**Definition:** Thermal expansion is the increase in dimensions (length, area, or volume) of a substance due to increase in its temperature. Most substances expand on heating and contract on cooling.
**Cause:** Increased thermal energy causes atoms/molecules to vibrate more vigorously, requiring more space.
**Types of Thermal Expansion:**
For a rod of length **l** experiencing temperature change **ΔT**:
$$\frac{\Delta l}{l} = \alpha_l \Delta T$$
Or: **ΔL = L₀ α_l ΔT** (absolute change)
**Final length:** **L = L₀(1 + α_l ΔT)**
Where:
**α_l depends on material type but is independent of length.**
**Typical Values (Table 10.1):**
| Material | α_l (10⁻⁵ K⁻¹) |
|----------|----------------|
| Aluminium | 2.5 |
| Brass | 1.8 |
| Copper | 1.7 |
| Iron | 1.2 |
| Silver | 1.9 |
| Glass (pyrex) | 0.32 |
| Lead | 0.29 |
**Key Observation:** Metals have larger linear expansion coefficients than glass. Copper expands ~5 times more than pyrex glass for same temperature change.
**Worked Example (Blacksmith Problem):**
Iron ring diameter at 27°C = 5.231 m
Rim diameter at 27°C = 5.243 m
Coefficient of linear expansion for iron: α_l = 1.2 × 10⁻⁵ K⁻¹
Using **L_T2 = L_T1[1 + α_l(T₂ - T₁)]**:
5.243 = 5.231[1 + 1.2 × 10⁻⁵(T₂ - 27)]
Solving: T₂ = 218°C
The ring must be heated to ~218°C to expand and fit the rim.
For a rectangular sheet of area **A**:
$$\frac{\Delta A}{A} = \alpha_A \Delta T = 2\alpha_l \Delta T$$
**Proof (Example 10.1):**
Consider rectangle with length a, breadth b. When temperature increases by ΔT:
Since α_l ≈ 10⁻⁵ K⁻¹, the term (α_l)² ab(ΔT)² ≈ 0 (negligible)
Therefore: **ΔA/A = 2α_l ΔT**
Or: **α_A = 2α_l**
**Final area:** **A = A₀(1 + 2α_l ΔT)**
For a solid or liquid experiencing temperature change **ΔT**:
$$\frac{\Delta V}{V} = \alpha_v \Delta T$$
Or: **ΔV = V₀ α_v ΔT**
**Final volume:** **V = V₀(1 + α_v ΔT)**
Where **α_v** = coefficient of volume expansion (K⁻¹)
**Relationship between α_v and α_l:**
Consider a cube of side length **l** expanding equally in all directions:
Neglecting higher order terms (Δl << l):
Therefore: **α_v = 3α_l**
**Typical Values (Table 10.2):**
| Material | α_v (K⁻¹) |
|----------|-----------|
| Aluminium | 7 × 10⁻⁵ |
| Brass | 6 × 10⁻⁵ |
| Iron | 3.55 × 10⁻⁵ |
| Mercury | 18.2 × 10⁻⁵ |
| Water | 20.7 × 10⁻⁵ |
| Alcohol | 110 × 10⁻⁵ |
| Glass (pyrex) | 1 × 10⁻⁵ |
| Invar | 2 × 10⁻⁶ |
Gases expand far more than solids and liquids at ordinary temperatures.
**For an ideal gas at constant pressure:**
From PV = μRT:
$$\Delta V = \frac{\mu R \Delta T}{P}$$
$$\frac{\Delta V}{V} = \frac{\Delta T}{T}$$
**Therefore:** **α_v = 1/T** (temperature dependent)
At 0°C (273 K): α_v = 1/273 ≈ 3.7 × 10⁻³ K⁻¹
At room temperature (~300 K): α_v ≈ 3.3 × 10⁻³ K⁻¹
This is **1000 times larger** than typical liquid or solid expansion coefficients.
---
**Water exhibits unique behavior:** It **contracts on heating** from 0°C to 4°C (opposite to normal thermal expansion).
**Characteristics:**
**Environmental Significance:**
Lakes and ponds freeze from the **top down**, not bottom up:
1. As water cools toward 4°C, surface water becomes denser and sinks
2. Warmer water rises from bottom (convection)
3. Once surface reaches below 4°C, it becomes less dense and remains at top
4. Ice forms on surface, insulating deeper water
5. Aquatic life survives in liquid water below the ice layer
**Without this anomaly:** Water would freeze from bottom up, killing fish and plant life. This unique property is crucial for freshwater ecosystem survival.
---
When thermal expansion is **prevented** by rigid constraints, the substance develops internal stress called **thermal stress**.
**Physical Situation:**
A rod fixed at both ends experiences temperature increase but cannot expand. External forces must provide compressive strain to prevent expansion.
**Calculation of Thermal Stress:**
If a rod of length l, cross-sectional area A, coefficient of linear expansion α_l is prevented from expanding when temperature increases by ΔT:
**Compressive strain:** ε = (Δl/l) = α_l ΔT
**Thermal stress:** σ = Y × ε = **Y α_l ΔT**
Where **Y** = Young's modulus (Pa)
**Force developed:** **F = A × σ = AY α_l ΔT**
**Worked Example:**
Steel rail: length L₀ = 5 m, area A = 40 cm² = 40 × 10⁻⁴ m²
Temperature increase: ΔT = 10°C
α_l (steel) = 1.2 × 10⁻⁵ K⁻¹
Y (steel) = 2 × 10¹¹ N/m²
Strain: ε = 1.2 × 10⁻⁵ × 10 = 1.2 × 10⁻⁴
Stress: σ = 2 × 10¹¹ × 1.2 × 10⁻⁴ = 2.4 × 10⁷ N/m²
Force: F = 2.4 × 10⁷ × 40 × 10⁻⁴ = 9.6 × 10⁵ N ≈ 10⁵ N
This enormous force can **bend or fracture rails** if not accommodated by expansion joints.
**Practical Applications:**
---
**Definition:** Specific heat capacity of a substance is the amount of heat required to raise the temperature of unit mass of the substance by one unit temperature without any change in its state.
**Mathematical Definition:**
$$s = \frac{1}{m} \frac{\Delta Q}{\Delta T}$$
Or: **ΔQ = m × s × ΔT**
Where:
**SI Unit:** J kg⁻¹ K⁻¹ or J kg⁻¹ °C⁻¹
**Key Points:**
**Physical Meaning:**
**Worked Example:**
How much heat is needed to raise the temperature of 2 kg of water from 20°C to 80°C?
s(water) = 4186 J kg⁻¹ K⁻¹
ΔQ = m × s × ΔT = 2 × 4186 × (80 - 20) = 2 × 4186 × 60 = 502,320 J ≈ 5.02 × 10⁵ J
---
When heat capacity is expressed per mole instead of per unit mass:
$$C = \frac{1}{\mu} \frac{\Delta Q}{\Delta T}$$
Or: **ΔQ = μ × C × ΔT**
Where:
**Relation between s and C:**
$$C = M \times s$$
Where **M** = molar mass (kg/mol)
Or: **s = C/M**
**For Gases—Two Important Cases:**
When gas is allowed to expand freely at constant pressure during heating:
When volume is held constant during heating (rigid container):
**Mayer's Relation:** C_p - C_v = R (8.31 J mol⁻¹ K⁻¹)
Detailed treatment covered in Chapter 11 (Thermodynamics).
---
**Definition:** Heat capacity of a body is the amount of heat required to raise its temperature by one unit without change of state.
$$S = \frac{\Delta Q}{\Delta T}$$
**SI Unit:** J K⁻¹ or J °C⁻¹
**Relationship to Specific Heat Capacity:**
$$S = m \times s$$
Heat capacity is **extensive** (depends on mass), while specific heat capacity is **intensive** (independent of mass).
**Thermal Inertia:**
Heat capacity measures the **thermal inertia** of a body—resistance to temperature change. Large heat capacity means large heat input needed for temperature change.
**Experimental Observations (from heating experiments):**
1. **Varying ΔT, constant mass, same substance:**
2. **Varying mass, constant ΔT, same substance:**
3. **Different substances, same m and ΔT:**
These observations validate: **ΔQ = m s ΔT**
---
**Definition:** Calorimetry is the branch of physics dealing with measurement of heat energy and thermal properties using a calorimeter.
**Principle:** Based on **conservation of thermal energy** (Law of Thermal Exchanges):
When two bodies at different temperatures are brought into thermal contact and isolated from surroundings:
**Calorimeter Setup:**
**Basic Calorimetry Equation:**
When hot and cold substances mix:
$$m_1 s_1 (T_1 - T_f) = m_2 s_2 (T_f - T_2)$$
Where:
**Or in general form:** **Σ(ΔQ) = 0** for isolated system
**Worked Example:**
100 g of aluminum at 80°C is mixed with 200 g of water at 20°C.
s(Al) = 900 J kg⁻¹ K⁻¹, s(water) = 4186 J kg⁻¹ K⁻¹
Heat lost by aluminum: Q₁ = 0.1 × 900 × (80 - T_f)
Heat gained by water: Q₂ = 0.2 × 4186 × (T_f - 20)
At equilibrium: Q₁ = Q₂
0.1 × 900 × (80 - T_f) = 0.2 × 4186 × (T_f - 20)
90(80 - T_f) = 837.2(T_f - 20)
7200 - 90T_f = 837.2T_f - 16,744
23,944 = 927.2T_f
T_f ≈ 25.8°C
**Applications:**
**Limitations:**
**Corrections:**
---
**Definition:** Change of state refers to transformation between different phases of matter (solid ↔ liquid ↔ gas) at constant temperature during heat absorption/release.
**Key Feature:** During phase change, temperature remains constant despite continuous heat flow.
**Melting:** Process of converting solid to liquid by heating at constant temperature (melting point).
**Example:** Ice melting at 0°C under standard pressure
**Latent Heat of Fusion (L_f):**
$$Q_f = m L_f$$
Where:
**SI Unit:** J kg⁻¹
**Typical Values:**
**Freezing:** Reverse of melting (liquid → solid)
**Worked Example:**
How much heat is needed to melt 250 g of ice at 0°C?
L_f(ice) = 3.34 × 10⁵ J/kg
Q = m × L_f = 0.25 × 3.34 × 10⁵ = 8.35 × 10⁴ J
**Boiling:** Conversion of liquid to gas by heating at constant temperature (boiling point).
**Example:** Water boiling at 100°C at standard pressure
**Latent Heat of Vaporization (L_v):**
$$Q_v = m L_v$$
Where:
**SI Unit:** J kg⁻¹
**Typical Values:**
**Note:** L_v >> L_f for same substance (water: 2.26 × 10⁶ vs 3.34 × 10⁵)
Much more heat needed to vaporize than to melt.
**Condensation:** Reverse of vaporization (gas → liquid)
**Worked Example:**
How much heat is released when 1 kg of steam at 100°C condenses to water at 100°C?
L_v(water) = 2.26 × 10⁶ J/kg
Q = m × L_v = 1 × 2.26 × 10⁶ = 2.26 × 10⁶ J
**Definition:** Direct conversion of solid to gas without passing through liquid phase at specific temperature and pressure (sublimation point).
**Example:** Dry ice (solid CO₂) converts to CO₂ gas below -78°C at 1 atm
Naphthalene (mothballs) sublimes at room temperature
Frost on grass converts directly to water vapor
**Latent Heat of Sublimation (L_s):**
$$Q_s = m L_s$$
**Relationship:** L_s = L_f + L_v (approximately)
Sublimation energy equals fusion energy plus vaporization energy.
**Deposition:** Reverse of sublimation (gas → solid)
---
When substances undergo phase changes during heat exchange:
**Total Heat Calculation:**
$$\Delta Q_{total} = \Delta Q_{sensible} + \Delta Q_{latent}$$
Sensible heat: raises/lowers temperature (Q = mcΔT)
Latent heat: changes state at constant temperature (Q = mL)
**Worked Example:**
Convert 100 g ice at -10°C to steam at 110°C
s_ice = 2100 J kg⁻¹ K⁻¹, s_water = 4186 J kg⁻¹ K⁻¹, s_steam = 2000 J kg⁻¹ K⁻¹
L_f = 3.34 × 10⁵ J/kg, L_v = 2.26 × 10⁶ J/kg
**Step 1:** Heat ice from -10°C to 0°C
Q₁ = 0.1 × 2100 × 10 = 2,100 J
**Step 2:** Melt ice at 0°C
Q₂ = 0.1 × 3.34 × 10⁵ = 3.34 × 10⁴ J
**Step 3:** Heat water from 0°C to 100°C
Q₃ = 0.1 × 4186 × 100 = 4.186 × 10⁴ J
**Step 4:** Vaporize water at 100°C
Q₄ = 0.1 × 2.26 × 10⁶ = 2.26 × 10⁵ J
**Step 5:** Heat steam from 100°C to 110°C
Q₅ = 0.1 × 2000 × 10 = 2,000 J
**Total Heat:** Q_total = 2.1 × 10³ + 3.34 × 10⁴ + 4.186 × 10⁴ + 2.26 × 10⁵ + 2 × 10³
= 300,500 J ≈ 3 × 10⁵ J
Vaporization dominates heat requirement.
---
**Definition:** Heat transfer is the spontaneous flow of thermal energy from regions of higher temperature to regions of lower temperature.
**Driving Force:** Temperature difference (ΔT) between regions
**Three Mechanisms:**
**Definition:** Heat transfer through a material without bulk motion of the material. Heat flows from molecule to molecule by direct contact.
**Mechanism:**
**Examples:**
**Thermal Conductivity:**
Heat conducted through material depends on:
1. Temperature difference (ΔT)
2. Material's thermal conductivity (λ or k)
3. Cross-sectional area (A)
4. Length/thickness (d)
**Fourier's Law of Heat Conduction:**
Heat flow rate (power):
$$H = -\lambda A \frac{dT}{dx}$$
For steady-state with uniform temperature gradient:
$$H = \lambda A \frac{\Delta T}{d}$$
Or: **Q = λ A ΔT t / d**
Where:
**SI Unit of λ:** W m⁻¹ K⁻¹ (or J s⁻¹ m⁻¹ K⁻¹)
**Thermal Conductivity Values (approx):**
| Material | λ (W m⁻¹ K⁻¹) |
|----------|-----------------|
| Silver | 429 |
| Copper | 385 |
| Aluminum | 205 |
| Steel | 50 |
| Brick | 0.9 |
| Concrete | 1.4 |
| Glass | 0.8 |
| Water | 0.6 |
| Air | 0.026 |
| Wood | 0.1-0.2 |
**Key Observations:**
**Thermal Resistance (R-value):**
$$R = \frac{d}{\lambda}$$
Analogous to electrical resistance. Higher R means poorer conductor.
**Worked Example:**
A copper rod 10 cm long, 2 cm² cross-section, with ends at 100°C and 0°C. How much heat flows in 1 minute?
λ
Q1. What is the SI unit of heat?
Answer: A — Joule is the SI unit of energy, including heat energy; other units are either non-SI or measure different quantities (W measures power, K measures temperature).
Q2. The ice point and steam point of water on the Celsius scale are:
Answer: B — By definition, pure water freezes at 0°C and boils at 100°C under standard atmospheric pressure on the Celsius scale.
Q3. Using the relation (tF − 32)/180 = tC/100, what is 32°F in Celsius?
Answer: A — Substituting tF = 32: (32 − 32)/180 = tC/100 gives 0/180 = tC/100, so tC = 0°C (ice point).
Q4. If a gas at constant volume has pressure P₁ at temperature T₁, what is its pressure at temperature 2T₁?
Answer: C — From ideal gas law at constant volume: P/T = constant, so P₁/T₁ = P₂/2T₁, giving P₂ = 2P₁.
Q5. The value of the universal gas constant R in SI units is:
Answer: B — R = 8.31 J mol⁻¹ K⁻¹ is the SI value used in the ideal gas equation PV = µRT; other options use non-SI units.
Q6. A gas thermometer reads the same for all low-density gases because:
Answer: B — All low-density gases exhibit identical expansion behaviour following PV/T = constant, making gas thermometers universal; other factors are not related to thermometric consistency.
Q7. Which statement about absolute zero is NOT correct?
Answer: C — Absolute zero cannot be practically achieved due to thermodynamic laws; it is a theoretical limit, not an experimentally reachable temperature.
Q8. If pressure P of a fixed mass of gas at constant volume is plotted against Celsius temperature, the line extrapolates to meet the temperature axis at −273.15°C because:
Answer: B — From PV = µRT (constant V), P ∝ T absolutely; the line must pass through origin in P vs T(K) graph, which corresponds to −273.15°C on Celsius scale where P = 0.
Q9. Which pair correctly matches a temperature scale property with its value?
Answer: C — Kelvin scale has ice point = 273.15 K and steam point = 373.15 K with 100 equal intervals, matching Celsius; Fahrenheit has 180 intervals; only Kelvin has true absolute zero at 0 K.
Q10. A constant-pressure gas thermometer measures temperature by reading the volume change of gas. If volume V₁ at temperature T₁ (in Kelvin) increases to V₂ at temperature T₂, derive the relation between V₁, V₂, T₁, and T₂ using Charles' law, and calculate T₂ if V₁ = 100 cm³ at T₁ = 273 K and V₂ = 150 cm³.
Answer: B — Charles' law at constant pressure: V/T = constant, so V₁/T₁ = V₂/T₂; substituting values: T₂ = T₁(V₂/V₁) = 273 × (150/100) = 273 × 1.5 = 409.5 K.
What is the SI unit of heat energy?
Joule (J) is the SI unit of heat energy.
Define temperature in physics.
Temperature is a relative measure of the hotness or coldness of a body, indicating the average kinetic energy of its particles.
State Boyle's law for gases.
At constant temperature, the product of pressure and volume of a fixed mass of gas is constant: PV = constant.
What is the relationship between Celsius and Kelvin scales?
T(K) = tC + 273.15, where both scales have the same unit size but different zero points.
Define heat in thermodynamics.
Heat is the form of energy transferred between two systems or a system and surroundings due to temperature difference.
What are the two fixed reference points for temperature scales?
Ice point (freezing point of water at 0°C) and steam point (boiling point of water at 100°C) under standard atmospheric pressure.
State Charles' law for gases.
At constant pressure, the volume of a fixed mass of gas is directly proportional to absolute temperature: V/T = constant.
What is the ideal gas equation?
PV = µRT, where P is pressure, V is volume, µ is number of moles, R is universal gas constant (8.31 J mol⁻¹ K⁻¹), and T is absolute temperature.
Convert 98.6°F to Celsius.
Using (tF - 32)/180 = tC/100: tC = (98.6 - 32) × 100/180 = 37°C.
What is absolute zero and why is it important?
Absolute zero (-273.15°C or 0 K) is the lowest possible temperature where all molecular motion theoretically stops, defining the Kelvin scale's zero point.
Define heat and temperature. Explain why heat flows from a body at higher temperature to one at lower temperature. [2 marks]
State that temperature measures hotness (average kinetic energy of particles) and heat is energy transferred due to temperature difference; explain that systems spontaneously move toward thermal equilibrium.
Derive the ideal gas equation starting from Boyle's law and Charles' law. Show all steps and explain what each law represents physically. [5 marks]
Begin with PV = const (Boyle's law at constant T) and V/T = const (Charles' law at constant P); combine these relations to show PV/T = constant for a given mass; introduce moles and universal gas constant R to obtain PV = µRT; explain that Boyle's law represents inverse pressure-volume relationship and Charles' law represents direct volume-temperature proportionality.
A blacksmith heats an iron ring before fitting it onto the wooden rim of a cart wheel. Explain this technique using the concept of thermal expansion and derive the linear expansion formula. Why cannot the ring be forced on at room temperature? [6 marks]
Explain that heating increases atomic spacing, causing the ring's diameter to expand; derive ΔL = αL₀ΔT by considering that fractional expansion is proportional to temperature change and initial length; show that at room temperature the ring's diameter is smaller than the wheel rim, making fitting impossible without expansion; at elevated temperature, the expanded ring diameter matches the rim, allowing easy fitting and subsequent tight grip upon cooling.
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