WBCHSE Class 11 Physics Notes For Transmission of Heat 

Transmission of Heat

Different Mechanisms Of Heat Transfer: When two bodies having different temperatures are brought in contact with each other, heat flows from the hot¬ter body to the colder one.

Again, when there is a temperature difference between different points of a body, heat flows in die body from the point of higher temperature to the point of lower temperature.

The flow of heat from one place to another is called transmission of heat: Heat is transmitted in three different processes:

1. Conduction,
2. Convection and
3. Radiation.

 

1. Conduction: It is the process in which heat energy is transmitted from a hotter region to a colder region of a material without any displacement of molecules. Conduction usually takes place in solids. When one end of an iron rod is placed in fire, the other end becomes so hot that it becomes difficult to hold that end with our hand. In this case heat gets transmitted through the rod from the end held in fire to the other end. This is known as heat conduction.
2. Convection: It is the process in which heat is transmitted from a hotter region to a colder region of a material by the actual movement of heated molecules. This happens only in fluids i.e., in liquids and gases because here molecules can move freely. If we take some liquid in a container and heat it from below, the top part of the liquid gets heated mainly through convection.
3. Radiation: This is the process in which heat is transferred from one place to another in the form of electro-magnetic radiation in the absence of any material medium or without heating a material medium if it is present between the two places. Heat from the sun reaches the earth by radiation.

Comparison among the three modes of transmission of heat

 

Transmission Of Heat Numerical Examples

Example 1. Four metal pieces of the same surface area and of thickness 1 cm, 2 cm, 3cm, and 4cm respectively are connected serially with each other; Coefficients of thermal conductivity of the respective pieces are 0.2 CGS, 0.3 CGS, 0.1 CGS, and 0.4 CGS units. Find the equivalent conductivity of the system.
Solution:

It is known that, \(\frac{x_1+x_2+x_3+x_4}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}+\frac{x_3}{k_3}+\frac{x_4}{k_4}\)

or, \(\frac{1+2+3+4}{k}=\frac{1}{0.2}+\frac{2}{0.3}+\frac{3}{0.1}+\frac{4}{0.4} \)

or, \(\frac{10}{k}=51.67 \text { or, } k=0.1935 \text { CGS. }\)

Example 2. A thick composite plate is formed by two plates of equal thickness kept one over the other. If the conductivities of the material of the constituent plates are k₁ and k₂, show that the equivalent conductivity of the thick plate, k = \(\frac{2 k_1 k_2}{k_1+k_2}\)
Solution:

Is is known, \(\frac{x_1+x_2}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}\)

Here, \(x_1=x_2=x\) (say)

∴ \(\frac{x+x}{k}=\frac{x}{k_1}+\frac{x}{k_2} or, \frac{2}{k}=\frac{1}{k_1}+\frac{1}{k_2} or, k=\frac{2 k_1 k_2}{k_1+k_2}\).

Example 3. A 75 cm long copper rod and a 125 cm long steel rod are joined face to face. Each rod is of a circular cross-section of diameter 2 cm. Temperatures at the two ends of the composite rod are 100 °C and 0°C and the outer surface of the rod is insulated. Find the temperature at the junction of the two rods What is the rate of conduction of heat through the junction?

1. k for copper = 9.2 x 10^-2 kcal · m^-1 · °C^-1 · s^-1 and
2. k for steel = 1.1 x 10^-2  kcal · m^-1 · °C^-1 · s^-1

Solution: Let the temperature of the junction = θ Conductivity of copper,

⇒ \(k_1=\frac{9.2 \times 10^{-2} \times 10^3}{10^2}=0.92 \text { CGS unit }\)

and that of steel, \(k_2=\frac{1.1 \times 10^{-2} \times 10^3}{10^2}=0.11 \text { CGS unit }\)

Area of cross-section of each rod, \(A=\pi(1)^2=\pi \mathrm{cm}^2\)

∴ Rate of flow of heat through the junction, \(\frac{Q}{t}=\frac{k_1 A\left(\theta_2-\theta\right)}{l_1}=\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}\)

or, \(k_1 l_2(100-\theta)=k_2 l_1 \theta\)

or, \(0.92 \times 125 \times(100-\theta)=0.11 \times 75 \times \theta \text { or, } \theta=93.3^{\circ} \mathrm{C}\)

∴ Rate of conduction, \(\frac{Q}{t}=\frac{k_2 A\left(\theta-\theta_1\right)}{l_2}=\frac{0.11 \times \pi \times 93.3}{125}=0.258 \mathrm{cal} \cdot \mathrm{s}^{-1}\)

Example 4. The thickness of each metal in a composite bar is 0. 01 m and the temperature of the two external surfaces are 100 °C and 30 °C. If the conductivities of the metals be 0.2 CGS unit and 0.3 CGS unit respectively, find the temperature on the interface.
Solution:

Let the temperature on the interface = θ.

At steady state, rate of flow of heat will be the same through both the plates.

∴ \(\frac{Q}{t}=\frac{k_1 A(100-\theta)}{x_1}=\frac{k_2 A(\theta-30)}{x_2}\)

or, \(\frac{0.2(100-\theta)}{1}=\frac{0.3(\theta-30)}{1} \text { or, } 200-2 \theta=3 \theta-90\)

or, \(5 \theta=290 \text { or, } \theta=58^{\circ} \mathrm{C} .\)

Example 5. Three metal rods of the same length and area of cross-section are attached in series. Conductivity of the three metals are k, 2k, and 3k. Free end of the first rod is kept at 200°C while the other end of the combination is kept of 100 °C. Find the temperatures of the two junctions at steady state. Assume that no heat loss occurs due to radiation.
Solution:

Let the required temperatures be θ₁  and θ₂.

At steady state, \(\frac{Q}{t}=\frac{k A\left(200-\theta_1\right)}{l}=\frac{2 k A\left(\theta_1-\theta_2\right)}{l}=\frac{3 k A\left(\theta_2-100\right)}{l}\)

∴ \(200-\theta_1=2\left(\theta_1-\theta_2\right)=3\left(\theta_2-100\right)\)

As \(200-\theta_1=2\left(\theta_1-\theta_2\right)\)

∴ \(3 \theta_1-2 \theta_2=200\)…..(1)

Also, \(200-\theta_1=3\left(\theta_2-100\right)\)

or, \(\theta_1+3 \theta_2=500\)…..(2)

Solving (1) and (2), \(\theta_1=145.45^{\circ} \mathrm{C} \text { and } \theta_2=118.18^{\circ} \mathrm{C} \text {. }\)

Example 6. A composite block is constructed with three plates of equal thickness and of equal cross-sectional area. The coefficients of conductivity of the three plates are k₁, k₂, and k₃ respectively. If the coefficient of conductivity of the composite block is k, then prove that k = \(=\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)
Solution:

Let the thickness of each plate be d and cross-sectional area be A.

Q amount of heat flows through each plate in time t, The temperatures of the two ends of the composite slab are T₁ and T₂ respectively.

Also, the temperatures of the consecutive junctions

So, \(Q=\frac{k_1 A\left(T_1-T_1{ }^{\prime}\right) t}{d}\)

= \(\frac{k_2 A\left(T_1{ }^{\prime}-T_2{ }^{\prime}\right) t}{d}=\frac{k_3 A\left(T_2{ }^{\prime}-T_2\right) t}{d}\) or, \(\frac{Q d}{A t}=\frac{T_1-T_1{ }^{\prime}}{\frac{1}{k_1}}=\frac{T_1{ }^{\prime}-T_2{ }^{\prime}}{\frac{1}{k_2}}=\frac{T_2{ }^{\prime}-T_2}{\frac{1}{k_3}}\)

or, \(\frac{Q d}{A t}=\frac{T_1-T_2}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)…(1)

[by componendo and dividendo process]

Again, since thickness of the composite slab is 3d,

Q = \(\frac{k A\left(T_1-T_2\right) t}{3 d} \text { or, } \frac{Q d}{A t}=\frac{T_1-T_2}{\frac{3}{k}}\)….(2)

From (1) and (2) we get, \(\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}=\frac{3}{k} \text { or, } k=\frac{3}{\frac{1}{k_1}+\frac{1}{k_2}+\frac{1}{k_3}}\)

Example 7. The temperature of a room is kept fixed at 20 °C with the help of an electric heater of resistance 20Ω. The heater is connected to a 200 V main. Temperature is same everywhere in the room. The area of the window through which heat is conducted outside from the room is 1 m² and the thickness of the glass is 0.2 cm. Then calculate the temperature outside. [Given, coefficient of thermal conductivity of glass = 0.002 cal • m^-1 • ° C^-1 • s^-1 and / = 4.2 J/s
Solution:

Heat generated by the heater in 1s,

⇒ \(H_1=\frac{V^2}{R J}=\frac{(200)^2}{20 \times 4.2}=\frac{200 \times 2000}{20 \times 42}=476.19 \mathrm{cal}\)

Heat conducted through the window in 1s,

⇒ \(H_2=\frac{k A\left(\theta_2-\theta_1\right)}{d}=\frac{0.002 \times 10^4 \times\left(\theta_2-\theta_1\right)}{0.2} \mathrm{cal}\)

Here, \(H_2=H_1 \quad \text { or, } \frac{0.002 \times 10^4 \times\left(\theta_2-\theta_1\right)}{0.2}=476.19\)

or, \(\left(\theta_2-\theta_1\right)=4.7619\)

∴ \(\theta_1=\theta_2-4.76=20-4.76=15.24^{\circ} \mathrm{C}\)

Example 8. The temperature of a room is kept fixed at 20°C. With the help of an electric heater when the temperature outside is -10°C. The total area of the walls in the room is 137 m2. The walls have three layers—the innermost layer is made of wood and 2.5 cm thick, the middle layer is made of cement and 1 cm thick, the outermost layer is made of bricks and 25 cm thick. Then calculate the power of the heater. [Given, coefficients of thermal conductivity of wood, cement, and brick are 0.125 W · m^-1 • °C^-1, 1.5 W · m^-1 • °C^-1 and 1.0 W • m^-1 • °C^-1 respectively]
Solution:

If k is the equivalent thermal conductivity of the wall made of three layers of different materials then,

k = \(\frac{x_1+x_2+x_3}{\frac{x_1}{k_1}+\frac{x_2}{k_2}+\frac{x_3}{k_3}}\)

Here, \(x_1=2.5 \mathrm{~cm}, x_2=1.0 \mathrm{~cm}, x_3=25 \mathrm{~cm}\)

⇒ \(k_1=0.125 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}, k_2=1.5 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1},\)

⇒ \(k_3=1 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

k = \(\frac{2.5+1+25}{\frac{2.5}{0.125}+\frac{1}{1.5}+\frac{25}{1}}=0.624 \mathrm{~W} \cdot \mathrm{m}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1}\)

Heat conducted outside through the walls per second

= \(\frac{k A\left(\theta_2-\theta_1\right)}{d}=\frac{0.624 \times 137 \times\{20-(-10)\}}{28.5 \times 10^{-2}}\)

= 8998,7 W ≈ 8999 J s^-1

To keep the temperature of the room fixed the healer must generate 8999 J or 8.999 kJ heat per second.

Hence, power of the heater = 9 kW.

Example 9. The temperature gradient at the earth’s surface is 32 C/km and the average thermal conductivity of earth is 0.005 CGS unit. Find the loss of heat per day from the earth surface taking its radius to be  6000 km.
Solution:

We know, Q = \(k \cdot \Lambda \frac{d \theta}{d x} t \quad\left[\frac{d \theta}{d x}=\frac{\theta_2-\theta_1}{d}\right]\)

Here, k = 0.008 CGS unit, A = 4 x (6000 x 105)² cm²

∴ \(\frac{d \theta}{d x} = 32^{\circ} \mathrm{C} / \mathrm{km}=\frac{32}{10^5}{ }^{\circ} \mathrm{C} / \mathrm{cm},\)

t = \(1 \text { day }=24 \times 60 \times 60 \mathrm{~s}\)

∴ Q = \(\frac{0.008 \times 4 \pi \times 36 \times 10^{16} \times 32 \times 24 \times 60 \times 60}{10^5}\)

= \(1.0006 \times 10^{18} \mathrm{cal}\)

Example 10. Three rods made of material X and three rods of material Y are connected as shown. Each rod has equal length and equal cross-section. If the temperatures of end A and junction E are 60°C and 10°C respectively, find the temperatures of junctions B, C, and D. Coefficients of thermal conductivity of the materials X and Y are 0.92 CGS units and 0.46 CGS units respectively.
Solution:

Let the temperatures of the junctions B, C, and D be θ₁, θ_2, and θ₃ respectively.

Now, heat conducted from A to B

= heat conducted from B to C + heat conducted from B to D

or, 0.46(60 – θ₁) = 0.92(θ₁ -θ₂) + 0.46(θ₁ – θ₃)

or, 60 -θ₁ =2(θ₁ – θ₂) + (θ₁ – θ₃)

or, 4θ₁ -2θ₁-θ₃ = 60 …(1)

Again, heat conducted from B to D

= heat conducted from D to C + heat conducted from D to E

or, \(0.46\left(\theta_1-\theta_3\right)=0.92\left(\theta_3-\theta_2\right)+0.46\left(\theta_3-10\right)\)

or, \(\theta_1-\theta_3=2\left(\theta_3-\theta_1\right)+\left(\theta_3-10\right)\)

or, \(-\theta_1-2 \theta_2+4 \theta_3=10\)….(2)

Similarly, heat conducted from B to C+ heat conducted from D to C= heat conducted from C to E

or, \(0.92\left(\theta_1-\theta_2\right)+0.92\left(\theta_3-\theta_2\right)=0.92\left(\theta_2-10\right)\)

or, \(\theta_1-\theta_2+\theta_3-\theta_2=\theta_2-10\)

or, \(\theta_1-3 \theta_2+\theta_3=-10\)…(3)

Solving (1), (2), and (3), we get, \(\theta_1=30^{\circ} \mathrm{C} ; \theta_2=\theta_3=20^{\circ} \mathrm{C}\)

Example 11. Determine the equivalent thermal conductivity of the system shown. when

1. Heat flows from left to right and
2. Heat flows downwards.

Given, k₁ = 2k₂ and the length and cross section are same for all the slabs.

Solution:

1. Let, the equivalent thermal conductivity is k

Area of cross-section of each plate = A

Temperature difference between the two ends of each Plate =θ

Then heat conducted, \(Q_1=2 \frac{k_1 A \theta t}{l}+2 \frac{k_2 A \theta t}{l}\)

= \(4 \frac{k A \theta t}{l}\)

or, \(\frac{2 A \theta t}{l}\left(k_1+k_2\right)=\frac{4 k A \theta t}{l}\)

or, \(k_1+k_2=2 k\)

or, \(k=\frac{k_1+k_2}{2}=\frac{2 k_1+k_2}{2}=\frac{3 k_2}{2}=1.5 k_2\)

2. In this case if \(k^{\prime}\) be the equivalent thermal conductivity then,

⇒ \(\frac{l}{k^{\prime}} =\frac{\frac{l}{4}}{k_1}+\frac{\frac{l}{4}}{k_2}+\frac{\frac{l}{4}}{k_1}+\frac{\frac{l}{4}}{k_2}\)

or, \(\frac{1}{k^{\prime}}=\frac{1}{2 k_1}+\frac{1}{2 k_2} \frac{1}{4 k_2}+\frac{1}{2 k_2}=\frac{3}{4 k_2}\)

or, \(k^{\prime}=\frac{4 k_2}{3}=1.33 k_2\)

Conduction of Heat through a Slab of Varying Thickness: The thickness of a slab can vary during thermal conduction.

• In cold countries when temperature falls below 0°C, the water on the surface of the lakes and ponds slowly freeze to ice.
• Heat from rest of the mass of water is conducted to the atmosphere through the top frozen layer and the thickness of the ice layer on the surface gradually increases.
• So this is a fine example of conduction of heat through a slab of varying thickness. When the temperature of the atmosphere falls below 0°C, the uppermost surface of water loses its latent heat to the atmosphere and forms a thin sheet of ice.
• In this way, as the thickness of the layer of ice increases, heat will have to be conducted through a thicker layer too.

Let the coefficient of conductivity of ice = k, the density of ice at 0°C = ρ, the temperature of air above the ice surface = -θ °C and the uppermost surface area of the water body = A

After a time t s from the beginning of the formation of ice, let the thickness of ice formed on the water body be x.

If in a small interval of time dt, the increase in thickness of ice is dx, then the volume of ice formed in dt is Adx.

∴ The mass of that ice =Aρdx.

Therefore, the heat lost by water or the heat conducted to air in time dt through ice of thickness x is,

dQ = Aρdx- L [where L = latent heat of ice]

∴ Rate of flow of heat, \(\frac{d Q}{d t}=A \rho L \frac{d x}{d t}=\frac{k A[0-(-\theta)]}{x}=\frac{k A \theta}{x}\)

or, \(xdx  =\frac{k}{\rho L} \theta d t\)…(1)

2. Let us assume that at t = 0, x = x₁ and at t = t₁, x = x₂. This means that in t₁ s time if the thickness of ice slab has increased from x₁ to x₂, integrating the equation (1) we get,

⇒ \(\int_{x_1}^{x_2} x d x=\int_0^{t_1} \frac{k \theta}{\rho L} d t\)

or, \(\frac{1}{2}\left(x_2^2-x_1^2\right)=\frac{k \theta}{\rho L} t_1\)

∴ \(t_1=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

1. If x₁ = 0 when t = 0 and x2-x when t = t₁

∴ \(t_1=\frac{\rho L}{2 k \theta} x^2\)….(3)

This equation gives the time (t₁) required for the deposi¬tion of a layer of ice of thickness x.

Note that in both equations (1) and (2), t₁ is independent of the surface area A of the lake. So in the same weather conditions, water bodies of all sizes from ice in this same rate.

Transmission Of Heat Conductivity Of Ice Numerical Examples

Example 1. A 3 cm thick layer of ice is formed over a water reservoir. Temperature over the reservoir is -20°C. In what time, the thickness of the ice layer increase by 1 mm? The conductivity of ice = 0.005 CGS unit; latent heat of fusion of ice = 80 cal · g^-1; density of ice at 0°C = 0.91 g · cm^-3.
Solution:

We shall use the equation, \(t=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

where ρ = 0.91 g · cm^-3, L = 80 cal · g^-1,

k = 0.005 CGS unit, θ = 0 – (-20) = 20°C , x₁ = 3cm, x₂ = 3.1 cm

∴ t = \(\frac{0.91 \times 80}{2 \times 0.005 \times 20}\left[(3.1)^2-(3)^2\right]\)

= \(\frac{0.91 \times 80}{2 \times 0.005 \times 20} \times 0.61\)

= 222.04 s = 3 min 42 s

Example 2. The surface of a lake is covered with a layer of ice of a thickness 10 cm. For increase in thickness of the ice layer by 1 mm, time taken is found to be 49 min9s. Conductivity of ice = 0.005 CGS unit, latent heat of fusion of ice = 80 cal · g^-1, and density of ice = 0. 917 g • cm^-3. Find the outside temperature.
Solution:

Let the temperature of air outside = -θ

It is known that, t = \(\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right) \quad \text { or, } \theta=\frac{\rho L}{2 k t}\left(x_2^2-x_1^2\right)\)

In this case, t = 49 min 9 s = 2949 s, x₁ = 10.1 cm, x₂ = 10 cm, ρ = 0.917 g · cm^-3, L = 80 cal · g^-1, k = 0.005 CGS unit

∴ \(\theta =\frac{0.917 \times 80}{2 \times 0.005 \times 2949}\left[(10.1)^2-(10)^2\right]\)

= \(\frac{0.917 \times 80 \times 2.01}{2 \times 0.005 \times 2949}=5\)

∴ The outside temperature is -5°C

Example 3. Water in a tank at 0°G is in contact with a surrounding temperature of -20°C. Prove that the rate of increase of thickness x (in cm) ofice on the surface, is related to time t (in s) as x² = 0.00273t. The density of ice =0.917 g · cm^-3, latent heat of fusion of ice =80 cal • g^-1, and conductivity of ice =0.005 CGS unit.
Solution:

Let, A be the surface area of the tank, x be the initial thickness of ice floating on the tank surface and dx be the increase in thickness in time dt.

Hence, mass of ice formed in the time dt = Adx x ρ [ρ = density of ice]

Heat released by water = Adx x ρ xL [L = latent heat of fusion of ice]

This is the amount of heat that is conducted from water in the tank through the ice of thickness x to the surroundings.

∴ Rate of release of heat, \(\frac{d Q}{d t}=A \rho L \frac{d x}{d t}=\frac{k A[0-(-\theta)]}{x}\)

[-θ°C being the temperature of the surroundings]

or, \(x d x=\frac{k \theta}{\rho L} d t\)

Integrating, \(\int_0^x x d x=\frac{k \theta}{\rho L} \int_0^t d t\)

or, \(x^2=\frac{2 k \theta}{\rho L} t=\frac{2 \times 0.005 \times 20 \times t}{0.917 \times 80}\)

or, \(]y=0.00273 t\) (Proved).

Example 4. A 14.9 cm thick layer of ice floats on a deep lake. The temperature of the upper surface of the ice layers is the same as that of the surrounding air. If this temperature remains constant at -1°C then determine the time required for the layer to increase by 2mm in thickness. Given, latent heat of melting of ice =80 cal · g^-1, density of ice = 0.9 g cm^-3 and coefficient of thermal conductivity of ice = 0.006 CGS unit.
Solution:

We know, \(t=\frac{\rho L}{2 k \theta}\left(x_2^2-x_1^2\right)\)

Here, \(\rho=0.9 \mathrm{~g} \cdot \mathrm{cm}^{-3}, L=80 \mathrm{cal} \cdot \mathrm{g}^{-1}\)

k = \(0.006 \mathrm{CGS} \text { unit }, \theta=1^{\circ} \mathrm{C}\)

∴ \(x_1 =14.9 \mathrm{~cm}, x_2=14.9+0.2=15.1 \mathrm{~cm}\)

t = \(\frac{0.9 \times 80}{2 \times 0.006 \times 1}\left[(15.1)^2-(14.9)^2\right]\)

= \(\frac{0.9 \times 80}{0.012} \times 30 \times 0.2=36000 \mathrm{~s}=10 \mathrm{~h}\)

 

Energy Distribution In Black Body Radiation Wien’s Displacement Law

Radiation from a hot body consists of electromagnetic waves of different wavelengths (λ). However the energy is not evenly distributed in all wavelengths of radiation.

The intensity of wavelengths in black body radiation, obtained by experimental observations, reveals a pattern as shown by the graph.

The abscissa of the graph denotes the wavelengths A of radiated heat and the ordinate denotes radiation per unit area per unit time (Eλ) for different wavelengths.

Interpretation of the graph:

1. At a temperature T₁, the source

1. Energy distribution is unequal for different wavelengths and
2. The intensity is maximum at a certain wavelength \(\lambda_{m_1}\) as represented by point P on graph A.

2. Radiations from the body at different temperatures T₁, T₂, T_3, and the corresponding intensity distribution can be compared from graphs A, B, C. Clearly

1. As temperature increases the total intensity, represented by the area under the curve, also increases,
2. The wavelength of maximum intensity shifts towards the lower wavelength side, with the increase in temperature of the source.

The observations referred to above led to the formulation of Wien’s displacement law which can be stated as: With the increase in the temperature of a body, the wavelength corresponding to the maximum intensity shifts towards the lower wavelength side.

The mathematical representation of Wien’s displacement law is, λ_mT = b (constant), where λ_m is the wavelength having maximum intensity at a temperature T of the source.

The value of Wien’s constant b, is about 0.0029 m K in SI unit. Thus, if λ_m is known, the temperature of the source T can be calculated. This method is widely used in estimating the temperature of stars. For example, if the temperature of a black body (star) is 1000 K,

∴ \(\lambda_m=\frac{0.0029}{1000}=2.9 \times 10^{-6} \mathrm{~m}\)

= \(29000 \times 10^{-10} \mathrm{~m}=29000\)Å

This wave lies in the infrared region of the electromagnetic spectrum.

Transmission Of Heat Energy Distribution In Black Body Radiation Numerical Examples

Example 1. The wavelength of the radiation of maximum Intensity from the solar surface Is 4.9 x 10^-7 m. From Wien’s displacement law, find the surface temperature of the sun. [ b = 0.0029 m • K]
Solution:

As per Wien’s displacement law, \(\lambda_m T=b=0.0029 \mathrm{~m} \cdot \mathrm{K}\)

Hence, \(T=\frac{0.0029}{4.9 \times 10^{-7}}=5918 \mathrm{~K}\).

Example 2. Assume that a star, which has a surface temperature of 5 x 10⁴ K, is a black body. Calculate the length of maximum intensity In Its radiation [b = 0.0029 m · K]
Solution:

From Wien’s displacement law, \(\lambda_m T=b\); given, \(T=5 \times 10^4 \mathrm{~K}\) and \(b=0.0029 \mathrm{~m} \cdot \mathrm{K}\).

∴ \(\lambda_m=\frac{b}{T}=\frac{0.0029}{5 \times 10^4}=\frac{29}{5} \times 10^{-8}=5.8 \times 10^{-8} \mathrm{~m}\).

 

Transmission Of Heat Long Answer Type Questions

The capability of a material to conduct heat is called its thermal conductivity.

• The quantity of heat conducted per second perpendicularly across the opposite faces of a unit cube, when the difference of temperature between its opposite faces is unity, is called the coefficient of thermal conductivity or simply thermal conductivity (k) of that material.
• In case of an ideal conductor k → ∞ and in case of an ideal insulator k = 0. So in reality, 0 < k < ∞.

Dimension of k: \(\mathrm{MLT}^{-3} \Theta^{-1}\)

• In the pre-steady state, the rate of increase in temperature of any layer of a conducting rod depends on the coefficient of thermal conductivity, the density and the specific heat of the material of the rod.
• In steady state, no layer of the rod has any increase in temperature and hence, the transmission of heat along the rod depends only on the coefficient of thermal conductivity.
• The ratio of the thermal conductivity of a substance to the thermal capacity per unit volume of that substance is called the thermal diffusivity of the substance.

In the radiation process, heat energy spreads all around in the form of electromagnetic waves. This thermal wave is called thermal radiation or radiant heat.

Prevost’s theory of heat exchange: The rise and fall of temperature of a body depends on the heat exchange of the body with its surroundings.

• The body which is a good absorber is also a good radiator of heat. Conversely, a bad absorber is also a bad radiator of heat.
• The body which absorbs all the incident radiations without reflecting or transmitting any part of it is called a perfectly black body. Since a perfectly black body is an ideal absorber of heat, it is also an ideal radiator of heat.

Kirchhoff’s law: At a particular temperature, the ratio of the emissive power and the absorptive power of a substance is always a constant and is equal to the emissive power of a perfectly black body at that temperature.

Stefan’s law: The total radiation of all wavelengths emitted per unit time per unit area of a perfect blackbody is directly proportional to the fourth power of its absolute temperature.

Newton’s law of cooling: For a small difference in temperature, heat lost per unit time by a body is directly proportional to the difference in temperature of the body with its surroundings.

Men’s law: With the increase in temperature of a body, the wavelength of the radiation corresponding to the maximum intensity shifts towards the lower wavelengths.

Transmission Of Heat Useful Relations For Solving Numerical Problems

If the temperature of the two faces of a rectangular plate of cross-sectional area A and of thickness d be θ₂ and θ₁ (θ₂ > θ₁) then the amount of heat conducted in time t from the hot to the cold face of the plate will be

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

Thermal resistance of conductor = \(\frac{1}{k} \cdot \frac{d}{A} .\)

Thermal resistivity of a substance = \(\frac{1}{k}\)

Thermal diffusivity of a substance, \(h=\frac{k}{\rho s}\)

[k = coefficient of thermal conductivity, ρ = density, s = specific heat]

If the equivalent thermal conductivity of a combined slab is k, then, \(\frac{x_1+x_2+\cdots+x_n}{k}=\frac{x_1}{k_1}+\frac{x_2}{k_2}+\cdots+\frac{x_n}{k_n}\)

where, x_n = thickness of n-th slab, k_n= thermal conductivity of the n-th slab. The temperature of the interface,

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

Emissive power (e) of a surface = amount of radiation emitted at a particular temperature from unit area of the surface in unit time.

Relative emittance of a surface (e) = \(\begin{gathered}
=\frac{\text { the amount of heat radiated by that surface }}{\text { the amount of heat radiated by the same surface area }} \\
\text { of a black body at the same temperature } \\
\text { in an equal interval time }
\end{gathered}\)

Absorptive power (a) of a surface = \(\frac{\text { the amount of heat absorbed }}{\text { the amount of heat incident on the surface }}\)

If the emissive power and absorptive power of a surface at a particular temperature be e and a respectively and the emissive power of a perfectly black body is E, then according to Kirchhoff’s law, \(\frac{e}{a}\) = E.

According to the Stefan’s law, E = σT⁴ (where σ = Stefan’s constant]

If the absolute temperature of an ideal black body and its surroundings be T and T₀, then by Prevosfs theory of heat exchange, the net rate of radiation of heat from the black body, \(E=\sigma\left(T^4-T_0^4\right)\)

Let mass of a body be m, its specific heat s, and the temperature of the surroundings θ₀. If at time t the temperature of the body drops from θ₁ to θ₂, then

t = \(\frac{1}{C} \ln \left(\frac{\theta_1-\theta_0}{\theta_2-\theta_0}\right)\)

According to the Wien’s law, λ_mT= constant.

[where, λ_m = wavelength of the radiation corresponding to the maximum intensity emitted from a black body kept at an absolute temperature T.]

Solar temperature T = \(\left[\left(\frac{R}{r}\right)^2 \times \frac{s}{\sigma}\right]^{1 / 4}\)

(where r = radius of the sun, R = mean distance of the earth from the sun, s = solar constant and σ = Stefan’s constant)

 

Transmission Of Heat Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 Is true, statement 2 Is true; statement 2 Is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 Is false, and statement 2 Is true.

Question 1.

Statement 1: The equivalent thermal conductivity of two plates of same thickness in contact is less than the smaller value of thermal conductivity.

Statement 2: For two plates of equal thickness contact, the equivalent thermal conductivity is given by \(\frac{2}{k}=\frac{1}{k_1}+\frac{1}{k_2} \text {. }\)

Answer: 4. Statement 1 Is false, statement 2 Is true.

Question 2.

Statement 1: If the thermal conductivity of a rod is 5 units, then its thermal resistivity is 0.2 units.

Statement 2: Thermal conductivity = \(\frac{1}{\text { thermal resistivity }}\)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: When the temperature difference across the two sides of a wall is increased, its thermal conductivity increases.

Statement 2: Thermal conductivity depends on the nature of material of the wall.

Answer: 4. Statement 1 Is false, statement 2 Is true.

Question 4.

Statement 1: If the temperature of a star is doubled then the rate of loss of heat from it becomes 16 times.

Statement 2: The specific heat varies with temperature.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 5.

Statement 1: Radiant heat is an electromagnetic wave.

Statement 2: Heat from the sun reaches the earth by convection.

Answer: 3. Statement 1 is true, statement 2 is false.

Transmission Of Heat Match Column 1 With Column 2

Question 1. On average, the temperature of the earth’s crust increases 1°C for every 30 m of depth. The average thermal conductivity of the earth’s crust is 0.75 J • m^-1 • K^-1 • s^-1. Solar constant is 1.35 kW • m^-2.

Answer: 1. A, 2. 3. 4. E

Question 2. Three rods of equal length of the same material are joined form an equilateral triangle ABC as shown. Area of a cross-section of rod AB is S, of rod BC is 2S and that of AC is S.

Answer: 1. C, 2. D, 3. 1, 4. B

Question 3. A copper rod (initially at room temperature 20°C) of the non-uniform cross section is placed between a steam chamber at 100°C and an ice water chamber at 0°C. A and B are cross sections as shown. Then match the statements in Column 1 with results in Column 2 using comparing only cross sections A and B.

Answer: 1. A, C, 2. D, 3. B

Transmission Of Heat Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A body cools in a surrounding of constant temperature 30°C. Its heat capacity is 2 J • °C^-1. The initial temperature of die body is 40°0. Assume Newton’s law of cooling is valid. The body cools to 38°C in 10 minutes.

1. In further 10 min it will cool from 38 °C to

1. 36°C
2. 36.4 °C
3. 37°C
4. 37.5°C

Answer: 2. 36.4 °C

2. The temperature of the body in T. denoted by θ. The variation of θ versus time t is best denoted as

Answer: 1

3. When the body temperature has reached 38°C, it is heated again so that it reaches 40°C in 10 min. The heat required from a heater by the body is

1. 3.6J
2. 0.364J
3. 8J
4. 4J

Answer: 3. 4

Question 2. Two insulated metal bars each of length 5 cm and rectangular cross-sections with sides 2 cut and 3 cm are wedged between two walls one held at 100°C and the other at 0°C. The bars are lead (Pb) and silver (Ag) k_Pb = 350 W m^-1 K^-1, k_Ag = 425 W m^-1 K^-1

1. Thermal current through lead bar is

1. 210W
2. 420W
3. 510W
4. 930W

Answer: 2. 420W

2. Total thermal current through the two-bar system is

1. 210 W
2. 420W
3. 510 W
4. 930W

Answer: 4. 930W

3. Equivalent thermal resistance of the two-bar system is

1. 0.1 W
2. 0.23 W
3. 0.19 W
4. 0.42W

Answer: 1. 0.1 W

Question 3. Assume that the thermal conductivity of copper is twice that of aluminum and four times that of brass. Three metal rods made of copper, aluminum, and brass are each 15 cm long and 2 cm in diameter. These rods are placed end to end, with aluminum between the other two. The free ends of the copper and brass rods are maintained at 100°C and 0°C, respectively. The system is allowed to reach the steady state condition. Assume there is no loss of heat anywhere,

1. Under steady-state condition, the equilibrium temperature of the copper-aluminum junction will be

1. 86°C
2. 18.8°C
3. 57°C
4. 73°C

Answer: 1. 86°C

2. When steady state condition is reached everywhere,

1. No heat is transmitted across the copper-aluminium or aluminium-brass junction
2. More heat is transmitted across the copper-aluminium junction than across the aluminium-brass junction
3. More heat is transmitted across the aluminium-brass junction than the copper-aluminium junction
4. Equal amount of heat is transmitted at the copper-aluminum and aluminium-brass junction

Answer: 4. Equal amount of heat is transmitted at the copper-aluminium and aluminium-brass junction

3. Under steady-state conditions, the equilibrium temperature of the aluminum-brass junction will be

1. 57°C
2. 35°C
3. 18.8°C
4. 28.5°C

Answer: 1. 57°C

Transmission Of Heat Integer Type Questions And Answers

In this type, the answer to each of the questions is a single-digit integer ranging from 0 to 9.

Question 1. A metal rod AB of length 10x has its one end A in. ice at 0°C and the other end B in water at 100°C if a point P on the rod is maintained at 400°C, then it is found that equal amounts of water and ice evaporate and melt per unit time. The latent heat of evaporation of water is 540 cal · g^-1 and the latent heat of melting of ice is 80 cal · g^-1. If the point P is at a distance of λx from the ice end A, find the value of λ. (Neglect any heat loss to the surroundings.)
Answer: 9

Question 2. Two spherical bodies A (radius 6 cm] and B (radius IS cm) are at temperature T₁ and T₂, respectively. The maximum intensity in the emission spectrum of A is at 500 runs and of B is at 1500 nm. Considering them to be blackbodies, what will be the ratio of the rare of total energy radiated by A to that of B?
Answer: 9

Question 3. A liquid takes 5 minutes to cool from 80°C to 50°C- How much time (in min) will it take to cool from 60°C to 30 °C? The temperature of the surroundings is 20°C
Answer: 9

Question 4. The ends of the two rods of different materials with their thermal conductivities, radii of cross-section, and lengths in the ratio 1:2 are maintained at the same temperature difference. If the rate of flow of heat through the larger rod is 4 cal · s^-1, what is the rate of flow of heat (in cal • s^-1) through the shorter rod?
Answer: 1