CBSE Class 12 Physics Conduction Theory and Examples

CBSE Class 12 Physics Conduction Theory and Examples Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

Heat moves from one place to another by three modes, namely conduction, convection and radiation. This Class 12 Physics chapter on conduction explains how heat flows through a solid from particle to particle, the meaning of thermal conductivity, steady state and isothermal surfaces, the temperature gradient, the law of heat transfer through conduction, thermal resistance, the comparison of heat conduction with electrical conduction, and the series, parallel and mixed combinations of rods, with solved examples.

CHAPTER - CONDUCTION

Heat can be transmitted from one place to the other by three modes: (1) Conduction, (2) Convection and (3) Radiation. Only conduction is discussed in this chapter.

1. CONDUCTION

Conduction is the phenomenon of transfer of heat through one part of a body to another, from particle to particle in the direction of fall of temperature, without any actual movement of the particles. For example, when one end of a solid rod is heated, its other end becomes hot. In conduction, particles at a higher temperature give heat to particles at a lower temperature, so molecular collisions transfer the energy from one end to the other.

2. THERMAL CONDUCTIVITY

Thermal conductivity of a solid is a measure of its ability to conduct heat. Substances differ considerably in this respect. For example, copper and silver are very good conductors of heat, whereas glass and wood are bad conductors.

3. TERMS RELATED TO THERMAL CONDUCTIVITY

(i) Variable state: As the temperature of every cross-section of the rod goes on increasing, the rod is said to be in the variable state.

(ii) Steady state: This is the state of the rod in which the temperature of each part becomes constant and there is no further absorption of heat anywhere in the rod. In the steady state, the temperature of each part of the rod is constant, but the temperature of different parts is not the same.

Note: The theory of conduction is valid only in the steady state.

(iii) Isothermal surface: In a metallic rod in which heat flows from one end to the other, after some time a steady state is reached. In this state, the temperature of every transverse section of the rod is the same over the whole area of section. This transverse section behaves as an isothermal surface.

Properties of isothermal surfaces:

(1) At an isothermal surface, the temperature of all the molecules is the same.
(2) Isothermal surfaces never intersect each other.
(3) The shape of an isothermal surface depends on the shape of the conductor and the nature of the flow of heat. The flow can be one-dimensional, two-dimensional or three-dimensional.
(4) The direction of heat flow is always normal to the isothermal surface.

(iv) Temperature gradient: The rate of change of temperature with distance along the direction of flow of heat is called the temperature gradient. If a metallic rod has temperatures θ and θ - Δθ at two isothermal surfaces which are a distance Δx apart, then the temperature gradient is Δθ / Δx = - dθ / dx.

(1) The negative sign shows that the temperature decreases with distance in the direction of heat flow.
(2) The unit of temperature gradient is °C per metre.

4. LAW OF HEAT TRANSFER THROUGH CONDUCTION

In the steady state, the rate of flow of heat dQ / dt through the cross-section of a conductor is:

(i) directly proportional to the area A of the surface, dQ / dt ∝ A.
(ii) directly proportional to the temperature gradient (- dθ / dx), that is, dQ / dt ∝ (- dθ / dx).

On combining the two, dQ / dt = - kA (dθ / dx) ... (1)

Here k is a constant called the coefficient of thermal conductivity of the material. Its unit is J / (m · s · K).

If dH is the amount of heat transferred in time interval dt, then dQ / dt = dH / dt = KAΔθ / Δx, which is the rate of heat flow.

If heat enters from one end such that the temperature decreases with distance in the direction of heat flow, then dH / dt = - KA dθ / dx.

Special point: If heat enters from one end and the temperature is measured from the other end, opposite to the direction of flow, then dH / dt = KA dθ / dx.

5. THERMAL RESISTANCE

Equation (1) can be rewritten as dQ / dt = dθ / (dx / KA). This is analogous to Ohm's law, I = V / R. The quantity dx / KA is called the thermal resistance, in analogy with electrical resistance. A greater thermal resistance means better thermal insulation and poorer thermal conduction.

6. COMPARISON OF HEAT CONDUCTION WITH ELECTRICAL CONDUCTION

Electrical conduction

(1) Electrical charge flows from higher potential to lower potential.
(2) The rate of flow of charge is called electric current, I = dq / dt.
(3) By Ohm's law, I = (v1 - v2) / R.
(4) The electrical resistance is R = ρ ℓ / A = ℓ / (σA), where ρ is resistivity and σ is conductivity.

Thermal conduction

(1) Heat flows from higher temperature to lower temperature.
(2) The rate of flow of heat is called heat current, I = dQ / dt.
(3) The heat current is given by I = (T1 - T2) / R_Th.
(4) The thermal resistance is R_Th = ℓ / (KA), where K is the thermal conductivity of the conductor.

7. COMBINATION OF STRAIGHT RODS

7.1 Series combination

Take two rods of the same cross-sectional area A, with lengths ℓ1 and ℓ2 and thermal conductivities K1 and K2. The ends of the composite rod are kept at temperatures θ1 and θ2, and θ is the common temperature of the junction.

In the steady state, the same heat current passes through both rods. So:

Heat current in the first rod = K1A(θ1 - θ) / ℓ1 ... (1)
Heat current in the second rod = K2A(θ - θ2) / ℓ2 ... (2)

Equating (1) and (2) gives the junction temperature θ = (K1ℓ2θ1 + K2ℓ1θ2) / (K1ℓ2 + K2ℓ1).

Putting this value of θ in (1): \( \frac{dH}{dt} = \frac{K_1A}{\ell_1}\left[\theta_1 - \frac{K_1\ell_2\theta_1+K_2\ell_1\theta_2}{K_1\ell_2+K_2\ell_1}\right] = \frac{A(\theta_1-\theta_2)}{\frac{\ell_2}{K_2}+\frac{\ell_1}{K_1}} = \frac{\theta_1-\theta_2}{\frac{\ell_2}{AK_2}+\frac{\ell_1}{AK_1}} \) ... (3)

(i) In terms of thermal resistance, if R_eq is the net resistance of the combination, then \( \frac{dH}{dt} = \frac{\theta_1-\theta_2}{R_{eq}} \) ... (4). Equating (3) and (4): \( R_{eq} = \frac{\ell_2}{AK_2}+\frac{\ell_1}{AK_1} = R_1 + R_2 \) ... (5)

(ii) If K_eq is the equivalent coefficient of thermal conductivity of the combination, the thermal resistance is \( R = \frac{\ell_1+\ell_2}{AK_{eq}} \) ... (6). Comparing (5) and (6): \( \frac{\ell_2}{AK_2}+\frac{\ell_1}{AK_1} = \frac{\ell_1+\ell_2}{AK_{eq}} \), so \( K_{eq} = \frac{\ell_1+\ell_2}{\frac{\ell_1}{K_1}+\frac{\ell_2}{K_2}} \).

Using the thermal resistance, the net resistance of the series combination is R_eq = R1 + R2, which gives K_eq = (ℓ1 + ℓ2) / (ℓ1 / K1 + ℓ2 / K2).

7.2 Parallel combination

Take two rods of the same length ℓ and cross-sectional areas A1 and A2, with thermal conductivities K1 and K2. The ends of the rods are kept at temperatures θ1 and θ2, and the temperature difference is the same between the ends of each rod.

The total heat current is dH / dt = dH1 / dt + dH2 / dt = (K1A1 / ℓ + K2A2 / ℓ)(θ1 - θ2).

(i) If R_eq is the equivalent thermal resistance, then 1 / R_eq = 1 / R1 + 1 / R2.
(ii) If K_eq is the equivalent thermal conductivity of the parallel combination, then R_eq = ℓ / (K_eq(A1 + A2)), which gives K_eq = (K1A1 + K2A2) / (A1 + A2).

7.3 Mixed combination

In a mixed combination of rods, the concepts of electrical series and parallel combinations are valid.

Examples based on heat conduction

Ex.1 Transmission of heat by molecular collisions is:
(A) Conduction, (B) Convection, (C) Radiation, (D) Scattering.

Answer: (A) Conduction. In conduction, the molecules remain at their place, and the transmission of energy occurs by molecular collisions.

Ex.2 The coefficient of thermal conductivity depends on:
(A) Nature of material, (B) Heat produced, (C) Difference in temperature, (D) Atmospheric pressure.

Answer: (A) Nature of material. In the formula Q = KA(θ1 - θ2) t / d, K is a constant whose value depends on the nature of the material.

Ex.3 The state when there is no more absorption of heat by the bar to raise the temperature of any part of it is called the:
(A) Variable state, (B) Steady state, (C) Both the above, (D) None of the above.

Answer: (B) Steady state. In the steady state the temperature remains constant, so neither absorption nor radiation of heat takes place.

Ex.4 A copper rod and a steel rod of equal length and equal cross-sectional area are joined in series. The free end of the copper rod is kept at 0°C and the free end of the steel rod is kept at 100°C. Find the temperature of the junction. Conductivity of copper = 390 W/m-°C and of steel = 46 W/m-°C.

Answer:
1. Heat current through the copper rod = 390 × A(0 - θ) / ℓ.
2. Heat current through the steel rod = 46 × A(θ - 100) / ℓ.
3. In series, the heat current is the same, so 390 × A(0 - θ) / ℓ = 46 × A(θ - 100) / ℓ.
4. -390θ = 46θ - 4600, so 436θ = 4600.
5. θ = 10.6°C.

Ex.5 An aluminium rod and a copper rod of equal length 1 m and cross-sectional area 1 cm² are welded together. One end is kept at 20°C and the other at 60°C. Calculate the heat taken out per second from the hot end. Thermal conductivity of aluminium is 200 W/m°C and of copper is 390 W/m°C.

Answer:
1. Heat current through the aluminium rod = 200 × (1 × 10⁻⁴) × (60 - 20) / 1 = 200 × 10⁻⁴ × 40 W.
2. Heat current through the copper rod = 390 × (1 × 10⁻⁴) × (60 - 20) / 1 = 390 × 10⁻⁴ × 40 W.
3. Total heat per second = (200 + 390) × 10⁻⁴ × 40 = 590 × 40 × 10⁻⁴ = 2.36 J.

Ex.6 Three identical rods, aluminium, copper and aluminium, are joined end to end as in arrangement (1), with the outer ends at 0°C and 100°C. Heat flows at the rate of 40 W. Find the rate of heat flow when the rods are joined as in arrangement (2), where the left aluminium rod is joined to a parallel combination of a copper rod and an aluminium rod. Thermal conductivity of aluminium is 200 W/m°C and of copper is 400 W/m°C.

Answer:
1. In arrangement (1) the rods are in series: 40 = 100 / (R1 + R2 + R3), where R = ℓ / (KA).
2. So (ℓ / A)(2 / 200 + 1 / 400) = 100 / 40, which gives ℓ / A = 200 per m.
3. In arrangement (2), the first aluminium rod is in series with the parallel combination of the other two rods, so dQ/dt = (θ1 - θ2) / R, where R = R1 + 1 / (1/R1 + 1/R2), with R1 = ℓ / (K1A) and R2 = ℓ / (K2A).
4. The parallel part has resistance (ℓ / A) / (K1 + K2), so R = (ℓ / A)[1/K1 + 1/(K1 + K2)] = 200 [1/200 + 1/600].
5. Hence dQ/dt = 100 / (200 [1/200 + 1/600]) = 600 × 100 / (200 × 4).
6. Here R = 200 × 4/600 = 4/3 in these units.
7. So the rate of heat flow = 75 W.