WBCHSE Class 11 Physics Notes For Thermal Conduction

Transmission of Heat Thermal Conduction

Conduction can be explained easily using the laws of motion for particles. When a part of a solid is heated, the molecules of that region get energized and start vibrating with large amplitudes about their equilibrium positions.

As a result, die number of collisions with the neighbouring molecules increases, and a part of their kinetic energy (due to vibration) is transferred to the neighbouring molecules. These molecules in their turn transfer some energy to their adjoining molecules.

Thus thermal motion occurs or heat travels from one molecule to another, or from one layer to another layer without any net displacement the molecules. Clearly, a material medium is necessary for the transfer of heat by conduction.

Read and Learn More: Class 11 Physics Notes

• Thermal Conductivity
• Coefficient of Thermal Conductivity

Thermal conductivity: The capacity of conduction of heat is called the thermal conductivity of a matter. Different substances have different thermal conductivities. Generally, metallic substances are better conductors of heat than non-metallic and other substances.

• A piece of wood, with one end kept in fire, can be held in our hands for a long time but we cannot hold a piece of iron, with one end kept in fire for too long. Iron can conduct heat more easily than wood. So the thermal conductivity of iron is more than that of wood. Materials which can conduct heat easily are called thermal conductors.
• Almost all metals are conductors silver is the best thermal conductor. Copper and aluminium come next in line. In metals there are many free electrons that move around haphazardly in the metal-like molecules of an ideal gas.
• These free electrons can carry energy from hotter to a colder region within a body more quickly and thus take part in conduction of heat along with the molecules of a metal making the metal a good conductor.
• Materials that cannot conduct heat easily are called bad conductors of heat or thermal insulators. Nonmetals do not have mobile free electrons to carry heat. In this case, heat is solely transmitted by atomic vibrations. Thus they are insulators or bad conductors.

Polystyrene, cotton, fiberglass, paper, cork, water, wood, rubber, asbestos, etc. are bad conductors of heat. All liquids except mercury, are generally bad conductors. All gasses are bad conductors. Vacuum cannot conduct heat and therefore is an ideal insulator.

WBBSE Class 11 Thermal Conduction Notes

Coefficient of thermal conductivity: Thermal conductivity is the property of a material that indicates its ability to conduct heat. The physical quantity used to denote it is called the coefficient of thermal conductivity.

Let a rectangular plate of cross-sectional area A and thickness d maintain temperatures θ₂ and θ₁ on its two opposite faces, where θ₂ > θ₁.

In this condition, heat passes perpendicularly through the slab from the hotter surface to the colder surface. If Q amount of heat is transferred perpendicularly across the plate in time t, then it is observed experimentally that,

1. Q is directly proportional to A, i.e., Q ∝ A
2. Q is directly proportional to (θ₂-θ₁), i.e., Q (θ₂-θ₁)
3. Q is directly proportional to t, i.e., Q ∝ t and
4. Q is inversely proportional to d, i.e., Q \(xs\propto \frac{1}{d}\)

Hence, \(Q \propto \frac{A\left(\theta_2-\theta_1\right) t}{d} or, Q=\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)…(1)

or, \(\frac{Q}{t}=\frac{k A\left(\theta_2-\theta_1\right)}{d}\)…(2)

Understanding Heat Conduction in Physics

Here k is a physical constant and it depends on the material of the plate, k is known as the coefficient of thermal conductivity or in brief, thermal conductivity of the material.

∴ \(\frac{\theta_2-\theta_1}{d}\) is the space rate of change of temperature in the direction of heat flow and is called the temperature gradient. Q/t is the time rate of heat flow and is called the heat current.

In case, the temperature gradient is non-uniform, equation (2) can be generalised as, \(\frac{d Q}{d t}=k A \frac{d \theta}{d x}\)

[where dθ is the infinitesimal change in temperature across an infinitesimal thickness dx of the material]

From equation (1), Q = k when A = 1, θ₂-θ₁ = 1, d = 1 and t = 1

Coefficient of thermal conductivity Definition: Heat conducted perpendicularly across the opposite faces of a cuboid of unit cross-sectional area and unit width in unit time, when the temperature difference of the two faces is unity, is called the coefficient of thermal conductivity of the material of the cuboid.

The coefficient of thermal conductivity of copper is 0.92 CGS unit means that 0.92 cal of heat will flow per second normally from the hotter surface to the colder surface of a copper cuboid of 1 cm² cross-sectional area and 1 cm width, with 1 °C temperature difference between the two surfaces.

Note that for an ideal conductor, k is infinity, and for an ideal insulator k is zero. But in practice, no substance is found for which k → ∞ or k = 0.

The value of k for a substance changes slightly with the change in temperature. With the increase in temperature, the values of k for solids and liquids decrease, but that for gases increases.

Units of k: From equation (1), k = \(\frac{Q \times d}{A \times\left(\theta_2-\theta_1\right) \times t}\)

Dimension of k: From equation (1),

⇒ \([k]=\frac{[Q][d]}{[A]\left[\theta_2-\theta_1\right][t]}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~L}}{\mathrm{~L}^2 \Theta \mathrm{T}}=\mathrm{MLT}^{-3} \Theta^{-1}\)

The coefficient of thermal conductivity (k) of a few substances in CGS system

Thermal Resistance and Thermal Resistivity: Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d} \text { or, } \frac{Q}{t}=\frac{\theta_2-\theta_1}{\frac{1}{k} \cdot \frac{d}{A}}\)….(1)

where \(\frac{Q}{t}\) denotes the rate of flow of heat through a conductor of thickness d, area of cross-section A, and coefficient of thermal conductivity k.

An equation similar to (1) is obtained for the rate of flow of charge or current through an electrical conductor of length d, area of cross-section A, resistance R, and resistivity ρ having a potential difference (V₂-V₁) across its length, this equation is

⇒ \(\frac{q}{t}=I \text { (current) }=\frac{V_2-V_1}{R}=\frac{V_2-v_1}{\rho \frac{d}{A}}\)….(2)

Comparing equations (1) and (2) it is apparent that

1. V₂ – V₁, the potential difference is similar to the temperature difference. θ₂-θ₁ in this case.
2. \(\frac{d}{k A}\) is similar to the electrical resistance R. Hence, \(\frac{d}{k A}\) is termed as the thermal resistance of the conductor.
3. \(\frac{1}{k}\) is similar to ρ. Hence, \(\frac{1}{k}\) is termed as thermal resistivity of the conductor.

• Pre-steady state
• Steady State
• Thermal Diffusivity

When a metal rod is heated by putting one of its ends into a source of heat, it is observed that heat is conducted along the length of the rod. As a result, the temperatures of the transverse layers situated perpendicular to the length of the rod increase gradually.

Heat received by any layer from its previous layer is spent in three ways:

1. By absorption so that the layer’s temperature increases,
2. By radiation from its outer surface, and
3. By transmission to the next layer. Same happens for each of the consecutive transverse layers of the rod.

• Heat absorbed by each layer increases its temperature and it is seen that the temperature of a layer decreases with the distance of the layer from the heat source.
• This type of absorption of heat and rise in temperature goes on for sometime. This condition of the metal rod is called the pre-steady state of heating.
• After a reasonable interval of time, the temperature of each of the transverse layers of the rod is seen to have reached a steady value. This means that the temperature does not rise anymore and that the layers are not absorbing heat anymore.

• Some of the heat received from the previous layer is radiated and the rest is conducted to the next layer. This state is called steady state of heating. In this state the temperature of the layers also decreases with the distance of the layers from the heat source.
• Conduction of heat through any part of the rod depends on its coefficient of thermal conductivity. In the steady state of heating, a part of the heat received by a layer from its previous one is absorbed to increase its own temperature.

The rise in temperature of a layer depends on the specific heat of the material of the rod. Hence, in a pre-steady state, heat conduction through a rod depends on both

1. The coefficient of thermal conductivity, and
2. Specific heat of the material of the rod.

On the other hand, since there is no absorption of heat by any layer in the steady state, heat conduction along a rod depends only on the coefficient of thermal conductivity of its material.

Thermal diffusivity or thermometric conductivity: if a substance has a coefficient of thermal conductivity k, specific heat s, and density ρ, it can be proved that during the pre-steady state of heating the rate of increase in temperature for any section of the rod does not depend only on the value of k.

Instead, this rate depends on the ratio \(\frac{k}{\rho s}\). This ratio \(\frac{k}{\rho s}\) is called the thermal diffusivity or thermometric conductivity of the substance.

∴ Thermal diffusivity, h = \(\frac{k}{\rho s}\)

= \(\frac{\text { coefficient of thermal conductivity }}{\text { density } \times \text { specific heat }}\)

= \(\frac{\text { coefficient of thermal conductivity }}{\text { thermal capacity per unit volume of the substance }}\)

[ρ = mass per unit volume of a substance]

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Thermal diffusivity or thermometric conductivity Definition: The ratio of the coefficient of thermal conductivity and the thermal capacity per unit volume of a substance is called the thermal diffusivity of the substance.

• Even if the coefficient of thermal conductivity of a substance is less than that of another substance, its thermal diffusivity may be relatively more. For example, lead has a coefficient of thermal conductivity less than that of iron but the thermal diffusivity of lead is more.
• Due to this reason, when two similar wax-coated rods of lead and iron are kept in the same source of heat, the different parts of the lead rod get heated faster than those of the iron rod, during the pre-steady state.
• In steady state let the temperature of a part on the iron rod situated l distance away from the source of heat be t₁°C and the temperature of a part on the lead rod situated l distance away from the source of heat be t₁°C. Then we can see that t₁°C > t₁°C.

So, in a pre-steady state, the wax coating on the lead rod melts along its length at a faster rate than that on the iron rod. But, finally, in the steady state, it is seen that the wax on iron rod has melted for a greater distance than the wax on lead rod.

Practical Illustrations of Conduction of Heat

1. Cooking utensils are usually made of copper or aluminum because of their high conductivity. Heat is conducted quickly through them. For the same reason, boiler tubes are generally made of copper.

2. Davy’s safety lamp: In 1815, Sir Humphry Davy invented a lamp for the safety of coal miners using the property of good conductivity of copper wire gauze. In this lamp, an ordinary oil lamp is kept inside a fence of copper wire mesh with very fine pores.

• The inflammable gases if present in the coal mine penetrate through the mesh and enter the lamp. So, the flame begins to burn brightly and changes its colour. As the pores in the wire mesh are very fine and made of copper, the flame cannot propagate through them to light the inflammable gases outside the mesh.
• Copper being a good conductor, rapidly distributes heat from the flame throughout the entire surface of the copper mesh. Thus immediately outside the mesh, the temperature is not high enough to sustain the burning reaction. So no explosion can take place. This kind of wire mesh is known as a flame arrestor.
• In modern times, this lamp has been replaced by other types of lamps.

3. Woolen clothes are used during winter. They keep us warm. So they are also known as warm clothes. Woolen yarn traps a lot of air, which is a bad conductor of heat. Thus body heat cannot easily escape through these clothes. So we keep warm. Ordinary cotton yarn does not trap as much air. So they do not provide enough warmth.

• If we wear two relatively finer clothes instead of a thick cloth, we feel warmer. This happens because the two clothes trap a layer of air between them. Air is a thermal insulator so it prevents the heat from our body from escaping. So we feel warm.
• For the same reason, in colder countries, the windows have double panes of glass. Air is trapped between the two layers of glass. As air is an insulator, heat cannot escape from the room This keeps the room warm.

4. The kettles are made of aluminum as it is a good con ductor. So the water in the kettle gets heated fast but at the same time, the handle also gets heated. So an insulator is needed.

Aluminum kettles have wooden handles or cane handles to provide insulation, as wood or cane are bad conductors and heat cannot pass from the hot kettle to the hand.

5. Sawdust is used to cover ice to prevent it from melting. Sawdust is a bad conductor and this also traps a lot of air (also a bad conductor). Thus, heat from outside cannot reach the ice block and this prevents it from melting.

6. Villagers use the property of conductivity of different substances to build houses. Thatched roofs of mud houses, which comprise hay and air spaces (both insulators), prevent heat from entering the house during the scorching summer, while in the winter the heat generated within the house does not escape to the outside.

7. A wooden chair feels warmer on touch in winter than an iron chair in the same room at the same temperature. During winter, the temperature of the chairs is generally less than that of our bodies. Iron, being a good conductor, conducts heat away from our body rapidly.

• It then quickly spreads the heat to other parts of the chair. So, we feel colder on touching the iron chair. Wood is a bad conductor of heat, and so the heat from our bodies is not spread to the other parts of the chair. So a wooden chair does not feel that cold.
• For the same reason, an iron chair feels warmer on touch than a wooden chair in summer. During summer, the temperature of the chair is generally more than that of our bodies.
• Now iron, being a good conductor, conducts heat from other parts of the chair to our hands rapidly. On the other hand, wood being a bad conductor, conducts heat to our hands only from the part that we touch. So a wooden chair feels relatively less hot.

8. During winter, the birds fluff up their feathers to keep themselves warm. The feathers trap air in them. Air, being an insulator, does not allow heat from the body to escape. So the birds keep warm.

9. New quilts make us feel warmer than old quilts do. Cotton in old quilts contract. So the amount of air trapped inside them decreases. But new quilts trap more air. Air being an insulator stops heat from our bodies from escaping. So we feel warmer.

10. During winters, walking barefeet on a cemented floor makes us feel very cold. Cement is a good conductor. Heat from our bodies gets conducted to the floor and spreads to other parts rapidly. So we feel cold. Carpeted floors feel warmer as carpets are insulators. So the heat from our bodies is not conducted and spread through the carpet.

11. A dip in a pond in summer afternoon is comfortable. Sun heats up the top layer of water. But, water being a bad conductor, does not conduct heat down to the depths of the pond. Hence, the lower water layers are comfortably cool for a dip.

Transmission of Heat Thermal Conductivity Numerical Examples

Short Answer Questions on Heat Conduction

Example 1. A copper piece of thickness 5 cm and area of cross-section 100 cm² is so heated that the temperature difference between the two opposite surfaces is 5°C. If the conductivity of copper is 0.9 CGS unit, find the quantity of heat that would flow in 5 min across the surfaces.
Solution:

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

[Given, k=0.9 CGS unit, \(A=100 \mathrm{~cm}^2, \theta_2-\theta_1=5^{\circ} \mathrm{C}, t=5 \times 60=300 \mathrm{~s}, d=5 \mathrm{~cm}\)]

∴ Q = \(\frac{0.9 \times 100 \times 5 \times 300}{5}=27000 \mathrm{cal}\) .

Example 2. Thickness of an iron plate is 4 mm and area 150 cm². Its surfaces are at 100°C and 30°C and 3940 cal of heat is conducted in 1 s across the sur¬faces. Calculate the thermal conductivity of iron.
Solution:

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d} \text { or, } k=\frac{Q d}{A\left(\theta_2-\theta_1\right) t}\)

[In this case, Q = 3940 cal, d = 0.4 cm, A = 150 cm², θ₂-θ₁ = 100-30 = 70°C, t= 1 s]

∴ k = \(\frac{3940 \times 0.4}{150 \times 70 \times 1}=0.15 \text { CGS unit }\)

Example 3. There are five glass windows in a room. Area of each glass window is 2 m2 and the thickness is 2 mm. Temperatures inside and outside the room are 20°C and -5°C respectively. Find the amount of heat conducted outside per minute through the glass windows. k for glass = 0.002 CGS unit
Solution:

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

[Given, k = 0.002 CGS unit, A = 5 x 2 x 10⁴ cm², θ₂ – θ₁ = 20 – (-5) = 25°C, t = 60 s, d = 0.2 cm]

∴ Q = \(\frac{0.002 \times 5 \times 2 \times 10^4 \times 25 \times 60}{0.2}=15 \times 10^5 \mathrm{cal} .\)

Example 4. The surface area of an iron cube is 4 cm². Two opposite surfaces of the cube are in contact with ice and steam respectively. What mass of ice will melt in 10 min? The conductivity of the material of the cube = 0.2 CGS unit and latent heat of fusion of ice = 80 cal · g^-1.
Solution:

Area of each surface of the cube A = 4 cm²

Each side of the cube = √4= 2 cm = thickness of the cube = d

θ₂ = 100°C (steam), θ₂ = 0°C (ice), t = 10x 60 = 600 s, k = 0.2 CGS unit

∴ Heat conducted in 10 min = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}=\frac{0.2 \times 4 \times 100 \times 600}{2}=24000 \mathrm{cal}\)

Latent heat of fusion of ice = 80 cal · g^-1

∴ Mass of ice melted = \(\frac{24000}{80}\) = 300 g.

Example 5. The length and the diameter of a metal rod are 31.41 cm and 4 cm respectively. One end of the rod is kept in steam at 100°C and the other end Is in ice at 0°C. Conductivity k of the material of the rod = 0.9 CGS unit Find the rate of melting ice per minute.
Solution:

Here, A = π(2)² = 4×3.141 cm², d = 31.41 cm, θ₂ – θ₁ = 100 – 0 = 100°C, k = 0.9 CGS unit and f = 60 s

∴ Heat conducted per minute = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}=\frac{0.9 \times 4 \times 3.141 \times 100 \times 60}{31.41}=2160 \mathrm{cal}\)

Latent heat of fusion of ice = 80 cal • g^-1

∴ Ice melted per minute = \(\frac{2160}{80}=27 \mathrm{~g}\)

Example 6. External diameter of a thin hollow sphere is 10 cm. This is filled with water at 100°C. The sphere is then immersed in melting ice. Find the rate of supply of heat to the sphere so that the temperature of water in the sphere is maintained. The thickness of the sphere is 2 mm and the coefficient of thermal conductivity of its material is 0.002 CGS unit
Solution:

Rate of conduction of heat to the ice through the wall of the hollow sphere,

⇒ \(\underset{t}{Q}=\frac{k A\left(\theta_2-\theta_1\right)}{d} \mathrm{cal} \cdot \mathrm{s}^{-1}\)

Here, \(k=0.002 CGS unit,\)

A = \(4 \pi(5)^2=100 \pi=100 \times 3.14=314 \mathrm{~cm}^2\)

⇒ \(\theta_2-\theta_1=100-0=100^{\circ} \mathrm{C} \text { and } d=2 \mathrm{~mm}=0.2 \mathrm{~cm}\)

∴ Q = \(\frac{0.002 \times 314 \times 100}{0.2}=314 \mathrm{cal} \cdot \mathrm{s}^{-1}\)

So, heat has to be supplied at the same rate i.e., 314 cal · s^-1 to keep the temperature of water in the sphere constant.

Example 7. Steam at 100°C is passed through a copper pipe of length 2 m, thickness 2 mm, and circumference 20 cm. Through the outlet 25000 g of water at 100°C is coming out per min. What is the temperature at the outer surface of the pipe? The coefficient of thermal conductivity of copper =0.9 CGS unit and latent heat of vaporization of water = 540 cal • g^-1.
Solution:

Let the temperature of the outer surface of the pipe = θ₁ temperature of inner surface, θ₂ = 100°C, the thickness of the pipe, d = 2 mm = 0.2 cm.

Total surface area of the pipe, A = circumference x length = 20 x 200 = 4000 cm²

Time, t = 1 min = 60 s

Hence, heat conducted per minute through the surface,

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}=\frac{0.9 \times 4000 \times\left(100-\theta_1\right) \times 60}{0.2} \mathrm{cal}\)

Latent heat of vaporisation of water = 540 cal • g^-1

Mass of steam condensed to water in 1 min

= \(\frac{0.9 \times 4000 \times\left(100-\theta_1\right) \times 60}{0.2 \times 540} \mathrm{~g}\)

∴ \(\frac{0.9 \times 4000 \times\left(100-\theta_1\right) \times 60}{0.2 \times 540}=25000\)

or, \(100-\theta_1=12.5 \text { or, } \theta_1=87.5^{\circ} \mathrm{C}\) .

Example 8. A cubical box of side 20 cm and wall thickness 0.2 cm, is filled with ice at 0 °C and then immersed in water at 100°C. The conductivity of the material of the box is 0.02 CGS unit. How long will it take to melt the whole ice in the box? Latent heat of fusion of ice = 80 cal • g^-1 density of ice at 0 °C = 0.9 g • cm^-3.
Solution:

Area of the six surfaces of the cubical box, A = 6 x (20)² = 2400 cm², thickness of the wall, d = 0.2 cm, θ₂ = 100°C, θ₁ = 0°C, k = 0.02 CGS unit

Neglecting thickness of the wall, volume of the box = volume of ice = (20)³ cm³ = 8 x 103 cm3

∴ Mass of ice = 8 x 103 x 0.9 = 7200 g

Heat required to melt the whole ice = 7200 x 80 = 576000 cal

Heat conducted to the box from outside in 1s

= \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}=\frac{0.02 \times 2400 \times 100 \times 1}{0.2}=24 \times 10^3 \mathrm{cal}\)

∴ Time required to melt the ice, t = \(\frac{576000}{24000}\) = 24 S

Example 9. A hollow metallic cube of side 10 cm and thickness lcm, is filled with ice and is kept immersed in water at 100 °C. Find the rate of melting of ice per minute if the conductivity of the material of the cube be 0.5 CGS unit and the latent heat of fusion of ice = 80 cal · g^-1.
Solution:

If m mass of ice melts in 1 min or 60 s, heat conducted to the cube from outside = 80 m cal.

Length of the inner side of the hollow cube =10-1-1 = 8 cm.

Hence, average length of the side = \(\frac{10+8}{2}=9 \mathrm{~cm}\)

∴ Area of 6 surfaces, A = 6 x 9 x 9 cm²

Now, \(Q=\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

or, \(80 \mathrm{~m}=\frac{0.5 \times 6 \times 9 \times 9 \times(100-0) \times 60}{1}\)

or, \(m=\frac{0.5 \times 6 \times 9 \times 9 \times 100 \times 60}{80}\)

= 18225 g

Real-Life Examples of Thermal Conductivity

Example 10. One end of a metal rod is attached to a heat source at 100°C. At a steady state of conduction, the temperature of a point at a distance of 10 cm from the source is 60°C. What will be the temperature of a point, 20 cm away from the source? Ignore heat lost by radiation.
Solution:

Let the required temperature be θ

As no heat is lost by radiation, the rate of heat transfer is constant along the rod. According to the problem,

⇒ \(\frac{Q}{t}=\frac{k A(100-60)}{10}=\frac{k A(100-\theta)}{20}\)

or, \(\frac{40}{10}=\frac{100-\theta}{20} \text { or, } 80=100-\theta\)

or, θ = 100-80 = 20°C.

Example 11. A metal rod of length 10 cm and area of cross-section 2 cm² connects a piece of ice at 0°C to a steam chamber. Find the rate of melting of ice per second. Latent heat of melting of ice =80 cal · g^-1, coefficient of thermal conductivity of the material of the rod = 0.25 CGS unit.
Solution:

If the rate of melting is m g · s^-1, heat conducted per second = 80 m = \(\frac{Q}{t}\)

Now, \(\frac{Q}{t}=\frac{k A\left(\theta_2-\theta_1\right)}{d} \quad or, 80 \mathrm{~m}=\frac{0.25 \times 2 \times(100-0)}{10}\)

or, \(m=\frac{0.25 \times 2 \times 100}{10 \times 80}=0.0625 \mathrm{~g}\).

Example 12. A 5 cm thick ice layer covers the water surface of a pond. Air temperature above the ice is -10 °C. At what rate would heat be conducted per cm³ of the ice surface? k for ice = 5 x 10^-3 CGS unit.
Solution:

Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

Here, k = 5 x 10^-3 CGS unit, θ₂-θ₁ = 0-(-10) = 10°C, d = 5 cm

∴ \(\frac{Q}{A t}=\frac{k\left(\theta_2-\theta_1\right)}{d}=\frac{5 \times 10^{-3} \times 10}{5}=0.01 \mathrm{cal} \cdot \mathrm{s}^{-1} \cdot \mathrm{cm}^{-2}\)

Example 13. Two ends of a cylindrical metal rod of length 31.4 cm and radius 2 cm are kept in touch with ice at 0°C and water at 100°C respectively. At what rate will ice melt every minute? Latent heat of melting of ice = 80 cal • g^-1, coefficient of conductivity of the material of the rod, k = 0.25 CGS unit.
Solution:

Here, A = π(2)² = 4×3.14 cm², d = 31.4 cm, θ₂ – θ₁ = 100 – 0 = 100°C, k = 0.25 CGS unit and t = 60 s

∴ Heat conducted, Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

= \(\frac{0.25 \times 4 \times 3.14 \times 100 \times 60}{31.4}\)

= \(6 \times 10^2 \mathrm{cal}\)

∴ Amount of molten ice = \(\frac{6 \times 10^2}{80}=7.5 \mathrm{~g}\)

∴ Ice will melt at a rate of 7.5 g • min^-1

Example 14. Water is kept at standard atmospheric pressure inside a 1.25 cm thick iron container. The bottom surface of the container is kept at 120°C and its area is 2.5 m². The coefficient of thermal conductivity of iron is 0.2 CGS unit. After water starts boiling, how much of the water will turn into vapor within an hour? Latent heat of vaporization of water = 540 cal- g^-1.
Solution:

Amount of heat conducted, Q = \(\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

Here, k = 0.2 CGS unit, A = 2.5 x 10⁴ cm², θ₂-θ₁ = 120 – 100 = 20°C, t = 60 x 60 s, d = 1.25 cm

Q = \(\frac{0.2 \times 2.5 \times 10^4 \times 20 \times 60 \times 60}{1.25}=288 \times 10^6 \mathrm{cal}\)

∴ Amount of water transformed into vapour

= \(\frac{288 \times 10^6}{540}=533300 \mathrm{~g}=533.3 \mathrm{~kg} .\)

Example 15. Water is being boiled in a flat-bottomed steel kettle stove. The base of the kettle measures 300 cm² and is of a thickness 2 mm. If the amount of steam produced is lg per min., calculate the difference temperature between the inner and outer surace of the base. Given, the thermal conductivity of steel =0.5 cal · cm^-1 °C s^-1. Latent heat of steam = 540 cal · g^-1
Solution:

Thickness of the base of the kettle, d = 2 mm = 0.2 cm area of the base, A = 300 cm²; time, t = 1 min = 60 s

Thermal conductivity of steel, k = 0.5 cal • cm^-1 • °C^-1 • s^-1

Now, \(Q=\frac{k A\left(\theta_2-\theta_1\right) t}{d}\)

∴ \(\left(\theta_2-\theta_1\right)=\frac{Q d}{k A t}=\frac{540 \times 0.2}{0.5 \times 300 \times 60}=0.012^{\circ} \mathrm{C}\)

Conduction of Heat through a Composite Slab

1. Series combination: Two rectangular slabs, S₁ and S₂, of different materials are in contact. The cross-sectional area of each slab is A their breaths are x₁ and x₂ and the coefficients of thermal conductivity are k₁ and k₂ respectively.

The composite slab is heated from the left of S₁ and heat is conducted through the junction to S₂. Heat will be released from the right surface of S₂.

Let us assume that at steady state, the temperatures on the left-hand surface and on the right-hand surface of the composite slab are θ₂ and θ₁ respectively. The temperature at the junction is θ such that θ₂> θ> θ₁.

Hence, heat conducted through S₁ in time t,

⇒ \(Q_1=\frac{k_1 A\left(\theta_2-\theta\right) t}{x_1} \text { or, } \frac{Q_1}{t}=\frac{k_1 A\left(\theta_2-\theta\right)}{x_1}=\frac{A\left(\theta_2-\theta\right)}{\frac{x_1}{k_1}}\)

Similarly, heat conducted through S₂ in time t,

⇒ \(Q_2=\frac{k_2 A\left(\theta-\theta_1\right) t}{x_2} \text { or, } \frac{Q_2}{t}=\frac{k_2 A\left(\theta-\theta_1\right)}{x_2}=\frac{A\left(\theta-\theta_1\right)}{\frac{x_2}{k_2}}\)

In steady state, Q₁ = Q₂ = Q (say)

∴ Q/t = \(\frac{A\left(\theta_2-\theta\right)}{\frac{x_1}{k_1}}=\frac{A\left(\theta-\theta_1\right)}{\frac{x_2}{k_2}}=\frac{A\left(\theta_2-\theta_1\right)}{\frac{x_1}{k_1}+\frac{x_2}{k_2}}\)…(1)

[because \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)}

Now, let us consider a single slab of thickness (x₁+ x₂) and area of cross-section A, such that if a temperature difference of (θ₂ – θ₁) is maintained between its two opposite surfaces, then the same amount of heat will be conducted through the slab per second. In this case the coefficient of thermal conductivity is called the equivalent thermal conductivity of the composite slab.

Let the equivalent thermal conductivity of the composite slab be k. Then,

⇒ \(\frac{Q}{t}=\frac{k A\left(\theta_2-\theta_1\right)}{x_1+x_2}=\frac{A\left(\theta_2-\theta_1\right)}{\frac{x_1+x_2}{k}}\)….(2)

Comparing equations (1) and (2) we have, \(\frac{x_1+x_2}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}\)…(3)

If the composite slab is made up of n number of slabs with thickness x₁, x₂, •••x_n, xn and coefficients of thermal conductivity k₁, k₂, •••, k_n respectively, then,

∴ \(\frac{x_1+x_2+\cdots+x_n}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}+\cdots+\frac{x_n}{k_n}\)….(4)

Thermal Conductivity and Its Importance

Temperature of the junction: From equation (1),

⇒ \(\frac{\theta_2-\theta}{\frac{x_1}{k_1}}=\frac{\theta-\theta_1}{\frac{x_2}{k_2}} \text { or, } \frac{\theta}{\frac{x_2}{k_2}}+\frac{\theta}{x_1}=\frac{\theta_2}{\frac{x_1}{k_1}}+\frac{\theta_1}{\frac{x_2}{k_2}}\)

or, \(\theta\left(\frac{k_2}{x_2}+\frac{k_1}{x_1}\right)=\frac{k_1 \theta_2}{x_1}+\frac{k_2 \theta_1}{x_2}\)

or, \(\theta\left(\frac{k_2 x_1+k_1 x_2}{x_1 x_2}\right)=\frac{k_1 \theta_2 x_2+k_2 \theta_1 x_1}{x_1 x_2}\)

Hence, the junction temperature, \(\theta=\frac{k_1 \theta_2 x_2+k_2 \theta_1 x_1}{k_2 x_1+k_1 x_2}\)…(5)

2. Parallel combination: In this case, two rectangular slabs S’₁ and S’₂ of same length x but of different cross-sectional area A₁ and A₂ respectively are joined at their ends so that the left ends of both slabs are kept at a temperature θ_2, and the right ends are kept at temperature θ₁.

Let k₁ and k₂ be the coefficients of thermal conductivity of S’₁. and S’₂ respectively, then heat conducted through slab S’₁ in time t,

⇒ \(Q_1 =\frac{k_1 A_1\left(\theta_2-\theta_1\right) t}{x}\)

and heat conducted through slab S’₂ in time t,

⇒ \(Q_2=\frac{k_2 A_2\left(\theta_2-\theta_1\right) t}{x}\)

The total heat passing through the composite slab in time t,

Q = \(Q_1+Q_2=\frac{\left(\theta_2-\theta_1\right) t}{x}\left(k_1 A_1+k_2 A_2\right)\)

\(\frac{Q}{t}=\frac{\theta_2-\theta_1}{x}\left(k_1 A_1+k_2 A_2\right)\)….(6)

Now, let us consider a single slab of length x, cross-sectional area A such that if a temperature difference (θ₂– θ₁) is maintained between its two opposite surfaces, the same amount of heat will be conducted through the slab per second.
Let, k be the equivalent thermal conductivity of the composite slab. Then,

⇒ \(\frac{Q}{t}=\frac{k A\left(\theta_2-\theta_1\right)}{x}\)…(7)

Comparing equations (6) and (7) we get,

∴ \(k_1 A_1+k_2 A_2=k A=k\left(A_1+A_2\right) \quad\left[because A=A_1+A_2\right]\)

k = \(\frac{k_1 A_1+k_2 A_2}{\left(A_1+A_2\right)}\)…(8)

If the composite slab is made up of n number of slabs with cross-sectional areas A₁, A₂, ……, A_n and coefficients of thermal conductivity k₁, k₂, k_n respectively, then,

⇒ \(k_1 A_1+k_2 A_2+\cdots+k_n A_n=k\left(A_1+A_2+\cdots+A_n\right)\)

k = \(\frac{k_1 A_1+k_2 A_2+\cdots+k_n A_n}{\left(A_1+A_2+\cdots+A_n\right)}\)…(9)