3.1 Heat Transmission & Expansion

1. Modes of Transmission of Heat

Heat always flows from a body at a higher temperature to a body at a lower temperature. This transfer of thermal energy can take place through three distinct modes:

Concept 1. Conduction: The process of heat transfer from one particle of a body to the adjacent particle without the actual mass movement of the particles. It primarily takes place in solids.

Convection

The process of heat transfer where the actual movement of the heated particles takes place from one place to another. This is the primary mode of heat transfer in fluids (liquids and gases).

Example: Boiling water in a pot; the hot water at the bottom rises while the cold water at the top sinks.

Radiation

The process of heat transfer which does not require any material medium. Heat travels directly in the form of electromagnetic waves at the speed of light.

Example: Heat reaching the Earth from the Sun; warmth felt sitting near a campfire.

2. Good and Bad Conductors of Heat

Fact Good Conductors: Substances which allow heat to pass through them easily are called good conductors. They have high thermal conductivity.
Examples: Silver (best conductor), Copper, Aluminum, Brass, Iron.

Bad Conductors (Insulators): Substances which do not allow heat to pass through them easily are called bad conductors. They have very low thermal conductivity.

Examples: Wood, Glass, Rubber, Air, Wool, Asbestos, and Plastics.

3. Law of Thermal Conductivity

Consider a solid block of thickness x and cross-sectional area A. Let the two opposite faces be maintained at steady temperatures T&sub1; and T&sub2; (where T&sub1; > T&sub2;). The quantity of heat (Q) flowing perpendicularly between the faces is:

Directly proportional to the Area of cross-section (A)
Directly proportional to the temperature difference (T&sub1; - T&sub2;)
Directly proportional to the time (t) of heat flow
Inversely proportional to the thickness or distance (x) between the faces

Combining these, the Law of Thermal Conductivity is given by: Q = K · A · (T&sub1; - T&sub2;) · t / x

Important Coefficient of Thermal Conductivity (K): It is defined as the quantity of heat that flows in one second through a unit cube of the material when its opposite faces are maintained at a temperature difference of 1 degree.

S.I. Unit: Watt per meter-kelvin (W / (m·K)) or Joules per second-meter-kelvin (J / (s·m·K)).

4. Thermal Expansion of Solids

When a solid is heated, its dimensions usually increase. This phenomenon is called thermal expansion. Depending on the shape of the solid, expansion can be of three types:

1. Linear Expansion (α)

The increase in the length of a solid rod on heating is called linear expansion.

Coefficient of Linear Expansion (α): The increase in length per unit original length per degree rise in temperature.
Formula: ΔL = L&sub0; · α · ΔT

2. Areal or Superficial Expansion (β)

The increase in the surface area of a solid on heating is called aerial expansion.

Coefficient of Areal Expansion (β): The increase in area per unit original area per degree rise in temperature.
Formula: ΔA = A&sub0; · β · ΔT

3. Cubical or Volume Expansion (γ)

The increase in the volume of a solid body on heating is called cubical expansion.

Coefficient of Cubical Expansion (γ): The increase in volume per unit original volume per degree rise in temperature.
Formula: ΔV = V&sub0; · γ · ΔT

Concept Relation Between α, β, and γ:
For isotropic solids (materials having uniform properties in all directions), the coefficients of expansion are related as follows:
β = 2α and γ = 3α
Therefore, the ratio is: α : β : γ = 1 : 2 : 3

5. Solved Numericals

Numerical 1: Thermal Conductivity

Question: A glass window pane is 2 m high, 1.5 m wide, and 4 mm thick. The temperature inside the room is 25°C and outside is 5°C. If the coefficient of thermal conductivity of glass is 0.8 W/(m·K), calculate the amount of heat lost per second through the window.

Solution:
Given:
Area (A) = 2 m × 1.5 m = 3 m²
Thickness (x) = 4 mm = 0.004 m
Temp Diff (T&sub1; - T&sub2;) = 25 - 5 = 20°C (or 20 K)
Time (t) = 1 second
Thermal conductivity (K) = 0.8 W/(m·K)

Using the formula: Heat per second (Q/t) = K · A · (T&sub1; - T&sub2;) / x
Q/t = (0.8 × 3 × 20) / 0.004
Q/t = 48 / 0.004 = 12,000 Watts (or 12 kW)

Numerical 2: Linear Expansion

Question: An iron rod has a length of 5 meters at 20°C. Find its length when heated to 100°C. (Coefficient of linear expansion of iron, α = 1.2 × 10&supmin;&sup5; /°C).

Solution:
Given:
Original length (L&sub0;) = 5 m
Change in temp (ΔT) = 100 - 20 = 80°C
α = 1.2 × 10&supmin;&sup5; /°C

Increase in length (ΔL) = L&sub0; · α · ΔT
ΔL = 5 × (1.2 × 10&supmin;&sup5;) × 80
ΔL = 480 × 10&supmin;&sup5; m = 0.0048 m

New length = Original length + ΔL
New length = 5 + 0.0048 = 5.0048 meters

Numerical 3: Coefficients of Expansion

Question: The coefficient of linear expansion of copper is 1.7 × 10&supmin;&sup5; /°C. What will be its coefficient of cubical expansion and coefficient of areal expansion?

Solution:
Given: α = 1.7 × 10&supmin;&sup5; /°C

1. Coefficient of areal expansion (β):
β = 2α
β = 2 × (1.7 × 10&supmin;&sup5;)
β = 3.4 × 10&supmin;&sup5; /°C

2. Coefficient of cubical expansion (γ):
γ = 3α
γ = 3 × (1.7 × 10&supmin;&sup5;)
γ = 5.1 × 10&supmin;&sup5; /°C