It was a very cold day, and the lab was only heated by a blower not making much difference. One of the professors climbed onto a table and stood there with his laptop doing his work. After ten minutes, someone asked if he wanted a chair. The professor declined, explaining that he was applying scientific methods. He pointed out that hot air rises, so the warmest spot in the room would be near the ceiling. The next moment everyone in the lab was standing up on their tables and doing their work. Indeed this was an excellent way to apply the concept of heat transfer and use it to solve a problem!
In this document, we will study heat transfer in detail.
The movement of heat across the border of the system due to a difference in temperature between the system and its surroundings.
Interestingly, the temperature difference is said to be a ‘potential’ that causes the transfer of heat from one point to another. Heat can travel from one place to another in several ways. The three different modes of heat transfer include:
Meanwhile, if a temperature difference exists between the two systems, heat will find a way to transfer from the higher to the lower system.
Modes of Heat Transfer
Conduction is that mode of transmission of heat by which heat travels, through an unequally heated body, from the hot end to the cold end, from particle to particle, the particles themselves remaining at their mean positions.
Conduction of heat in a metal rod
Following are the examples of conduction:
The coefficient of thermal conductivity shows that a metal body conducts heat better when it comes to conduction. The rate of conduction can be calculated by the following equation:
Where,
Q1. A 10 cm thick block of ice with a temperature of 0 °C lies on the upper surface of a 2400 cm^{2} slab of stone. The slab is steamexposed on the lower surface at a temperature of 100 °C. Find the heat conductivity of stone if 4000 g of ice is melted in one hour given that the latent heat of fusion of ice is 80 cal/gm.
Solution:
Area of slab, A = 2400 cm^{2}
Thickness of ice, d = 10 cm
Temperature difference, Th – Tc = 100 °C – 0 °C = 100 °C
Time of heat transfer, t = 1 hr = 3600 s
Amount of heat transfer, Q = m L = 4000 × 80 = 320000 cal
Heat transfer rate, q = Q ⁄ t = 320000 cal ⁄ 3600 s = 89 cal ⁄ s
The formula for heat transfer rate is given as:
q = K A (Th – Tc) ⁄ d
Rearrange the above formula in terms of K.
K = q d ⁄ A (Th – Tc)
= (89 × 10) ⁄ (2400 × 100) cal ⁄ cm s °C
= 3.7 × 10^{3} cal/cm s °C
Hence, the thermal conductivity of stone is 3.7 × 10^{3} cal/cm s °C.
It is defined as that mode of transmission of heat by which heat travels from one part of a body to another by the actual motion of the heated particles of the body.
Convection in boiling water
Examples of convection include:
As the temperature of the liquid increases, the liquid’s volume also has to increase by the same factor and this effect is known as displacement. The equation to calculate the rate of convection is as follows:
Where,
Q2. A fluid flows over a plane surface 1 m by 1 m. The surface temperature is 50^{o}C, the fluid temperature is 20^{o}C and the convective heat transfer coefficient is 2000 W/m^{2}C. What is the convective heat transfer between the hotter surface and the colder air?
Solution:
Q = (2000 W/(m^{2 }C)) ((1 m) (1 m)) ((50 ^{o}C)  (20 ^{o}C))
= 60000 (W)
= 60 (kW)
The third means of energy transfer is radiation which does not require a medium. The best known
example of this process is the radiation from sun. All objects radiate energy continuously in the form
of electromagnetic waves.
Following are examples of radiation:
All bodies radiate energy in the form of electromagnetic waves by virtue of their temperature. This energy is called the radiant energy.
As temperature rises, the wavelengths in the spectra of the radiation emitted decrease, and shorter wavelength radiations are emitted. Thermal radiation can be calculated by the StefanBoltzmann law:
Where,
Q3. What is the rate of heat transfer by radiation, with an unclothed person standing in a dark room whose ambient temperature is 22.0ºC? The person has a normal skin temperature of 33.0ºC and a surface area of 1.50 m^{2}. The emissivity of skin is 0.97 in the infrared, where the radiation takes place.
Solution:
The StefanBoltzmann law is given by:
Given:
ε = 0.97 (emissivity of skin),
A = 1.50 m^{2} (surface area),
T = 22.0ºC + 273.15 = 295.15 K (ambient temperature),
Tc = 33.0ºC + 273.15 = 306.15 K (skin temperature).
Substituting these values into the equation, we have:
Q = 0.97 * (5.67 x 10^{8} W/(m^{2}K^{4})) * 1.50 m^{2} * ((295.15 K)^{4}  (306.15 K)^{4})
Calculating this expression will give us the rate of heat transfer by radiation (Q) in watts (W).
Q ≈ 0.97 * (5.67 x 10^{8}) * 1.50 * ((295.15)^{4}  (306.15)^{4}) W
Please note that due to the complexity of the calculation, the final numerical value of Q may be quite small.
It is defined as the rate of change of temperature of the different crosssection with distance.
Mathematically, the temperature gradient (∇T) can be defined as:
∇T = dT/dx
where ∇T represents the temperature gradient, dT represents the temperature change, and dx represents the change in distance along the coordinate axis (usually the xaxis).
The coefficient of thermal conductivity (K) of the material of a rod is defined as the heat current (amount of heat flowing per second) flowing per unit area between two crosssections of the rod each of area 1 m^{2} and separated 1 m apart.
Mathematically, the thermal conductivity (k) is defined as:
K = Ql/(A(θ_{1}θ_{2})t)
Dimension of K: [K] = [M^{1}L^{1}T^{3}K^{1}]
Unit: C.G.S cal cm^{1}s^{1} ºC^{1}
S.I.: Wm^{1}K^{1}
Rate of flow of heat = dQ/dt
Heat current, through a conducting rod, is defined as the amount of heat conducted across any crosssection of the rod in one second.
H= dQ/dt
H depends upon the following factors:
The total heat Q crossing from one crosssection to the other in time t:
Q = KA(θ_{1}θ_{2})t/l, where l is the length.
Or K = Ql/A(θ_{1}θ_{2})t
Q4. Calculate the thermal conductivity of a copper rod if a heat flux of 1000 W is conducted through a 2meter length of the rod. The temperature difference between the ends of the rod is 50 degrees Celsius, and the crosssectional area is 0.05 square meters.
Solution:
Heat flux (Q) = 1000 W
Length of the rod (l) = 2 m
Temperature difference (θ1  θ2) = 50°C
Crosssectional area (A) = 0.05 m²
We can use the formula for thermal conductivity:
K = Ql / (A * (θ1  θ2))
Plugging in the values:
K = (1000 W * 2 m) / (0.05 m² * 50°C)
Simplifying:
K = 40 W/(m·°C)
Therefore, the thermal conductivity of the copper rod is 40 W/(m·°C).
It is defined as heat current per unit temperature difference.
σ_{h}= KA/l
σ_{h}= H/dθ
Unit S.I WK^{1}
The thermal resistance, of a conductor, is defined as the temperature difference between its two crosssections when a unit heat current flows through it.
Reciprocal thermal conductance is known as the thermal resistance of the substance.
R_{h}= 1/σ_{H} = l/KA = dθ/H
Units of R_{h}: S.I – W^{1}K
The thermal resistance of a conductor of length d: R_{TH} = d/KA
H= (θ_{1} θ_{2})/(l/KA) = (θ_{1} θ_{2})/R_{h}
K= m(θ_{4} θ_{3})d/A(θ_{1} θ)t
The ratio of thermal and electrical conductivities is the same for the metals at a particular temperature and is proportional to the absolute temperature of the metal. If T is the absolute temperature, then
K/σ ∝ T or K/ σT = constant
K_{1}/K_{2} = l_{1}^{2}/ l_{2}^{2}
(a)Thermal resistance in series: The thermal resistance of the composite slab is equal to the sum of their thermal resistances.
(l_{1} +l_{2})/KA = (l_{1}/K_{1}A) + (l_{2}/K_{2}A)
R_{c}_{omb} = R_{h}+R_{h}^{'}
If l_{1}=l_{2}=l, then, K = 2K_{1}K_{2}/K_{1}+K_{2}
Temperature of the interface:
θ_{0}= [θ_{1}R_{h}^{'} + θ_{2}R_{h}]/ [R_{h}+ R_{h}^{'}] or θ_{0} = [θ_{1}K_{1}l_{2}+ θ_{2} K_{2}l_{1}]/ [K_{1}l_{2}+ K_{2}l_{1}]
(b) Thermal resistance in parallel: Reciprocal of the combination thermal resistance is equal to the sum of the reciprocals of individual thermal resistances.
1/R_{comb} = 1/R_{h} + 1/R_{h´}
The reflecting power (r) of a substance is defined as the ratio between the amount of heat reflected by the substance to the total amount of heat incident upon it.
The transmitting power (t) of a substance is defined as the ratio between the amount of heat transmitted by the body to the total amount of heat incident upon it.
Radiant emittance of a body at a temperature T is defined as the total amount of energy (for all wavelengths) radiated per unit time, per unit area by the body.
Unit  S.I: Jm^{2}s^{1}
C.G.S: erg cm^{2}s^{1}
The total energy density (U) at any point is defined as the radiant energy per unit volume, around that point, for wavelengths taken together.
Now, let us define a few important terms related to Heat Transfer
A bolometer is a device for measuring the power of incident electromagnetic radiation via the heating of a material with a temperaturedependent electrical resistance.
If R_{t} and R_{0} are the resistances of the conductor at 0ºC and tºC, then, R_{t} = R_{0}(1+αt), Here α is the temperature coefficient of change of resistance with temperature.
Absorptive power (a) of the substance is defined as the ratio between amounts of heat absorbed by it to the total amount of heat incident upon it.
Q5. A substance absorbs 400 J of heat when 800 J of heat is incident upon it. Calculate the absorptive power (a) of the substance.
Solution:
Given:
Heat absorbed (Q_absorbed) = 400 J
Total heat incident (Q_incident) = 800 J
The absorptive power (a) can be calculated using the formula:
a = Q_absorbed / Q_incident
Substituting the given values:
a = 400 J / 800 J
Calculating:
a = 0.5
Therefore, the absorptive power (a) of the substance is 0.5 or 50%. This means that the substance absorbs 50% of the total amount of heat incident upon it.
Absorptive power refers to all wavelengths (total radiant energy). However, any surface will have different values of absorptive powers for different wavelengths.
a =
aλ ≤ 1
For a perfectly black body a_{λ }= 1
Q6. 200 units of energy are incident on a surface. Out of this, 50 units are of wavelength λ_{1}, 70 units are of wavelength λ_{2}, and the remaining 80 units correspond to other wavelengths. The surface absorbs a total of 150 units of energy. Within this absorbed energy, 30 units are of wavelength λ_{1} and 40 units are of wavelength λ_{2}. Find the absorptive power (a), spectral absorptive power for λ_{1} (a_{λ}_{1}), and spectral absorptive power for λ_{2} (a_{λ2}).
Solution:
Given:
Total incident energy = 200 units
Energy absorbed by λ1 = 30 units
Energy absorbed by λ2 = 40 units
To find a, the total absorptive power:
a = (Energy absorbed) / (Total incident energy) = 150 units / 200 units = 0.75
To find a_{λ1}, the spectral absorptive power for _{λ1}:
a_{λ1} = (Energy absorbed by λ1) / (Energy incident at λ1) = 30 units / 50 units = 0.6
To find a_{λ2}, the spectral absorptive power for λ2:
aλ2 = (Energy absorbed by λ2) / (Energy incident at λ2) = 40 units / 70 units = 0.57
From this example, we can observe that the total absorptive power (a) of the surface is 0.75. The spectral absorptive power for λ1 (a_{λ1}) is 0.6, indicating that the surface is a good absorber of wavelength λ1. On the other hand, the spectral absorptive power for λ2 (a_{λ2}) is 0.57, which is less than the total absorptive power (0.75). This suggests that the surface is a relatively poorer absorber of wavelength λ2 compared to λ1.
Q7. 100 units of energy are incident on a surface. In this 20 units are of wavelength λ_{1}, 30 units are of wavelength λ_{2} and rest 50 units are other wavelengths. A total of 60 units of energy is absorbed by the surface. Of these 60 units, 5 units are of λ_{1} and 25 units are of λ_{2}. Find a, and .
Solution:
Total absorptive power a = 60/100 = 0.6
and spectral absorptive power for λ_{1} = = 5/20 = 0.25
Spectral absorptive power for λ_{2} = = 25/30 = 0.83
From this example, it is clear that the total absorption power of the surface is only 0.6 whereas is 0.83 (> 0.6) i.e. the surface is a good absorber of wavelength λ_{2}.
Emissive Power (e) refers to the amount of radiant energy emitted per second per unit area of a surface in the form of electromagnetic radiation.
For a black body e = σ T^{4}
(Do not confuse it with the emissivity e which is different from it, although both have the same
symbol e).
Q8. A surface with a temperature of 500 K has an emissive power of 1000 W/m². Calculate the emissive power (e) for a black body at the same temperature.
Solution:
Temperature of the surface (T) = 500 K
The emissive power of the surface (e) = 1000 W/m²
The emissive power for a black body (e) can be calculated using the formula:
e = σ * T^4
where σ is Stefan's constant.
Substituting the given values:
e = 5.735 x 10^8 W/m²K^4 * (500 K)^4
Calculating:
e ≈ 5.735 x 10^8 W/m²K^4 * 62500000 K^4
e ≈ 3.584375 W/m²
Therefore, the emissive power (e) for a black body at the same temperature is approximately 3.584375 W/m². This indicates the amount of radiant energy emitted per second per unit area by the black body at the given temperature.
Emissivity (e) is a property of a surface that indicates its relative emittance compared to a perfectly black body. The emissivity value can be represented as ε = e/E, where ε ranges between 0 and 1.
ε = e/E, 0 ≤ ε ≤1
Spectral emissive power is defined as the rate at which radiation of wavelength l is emitted in all directions form a surface per unit wavelength dl about l. In Simpler terms, It is emissive power of the surface for a particular wavelength λ.
e_{λ} is the maximum for a perfectly black body.
Q9. The spectral emissive power E for a body at temperature T is planted against the wavelength and the area under the curve is found to be A. At a different temperature, T_{2 }the area is found to be 9A. Then λ_{1}/λ_{2 }=
Solution:
Radiant emittance or the energy radiated per second per unit area by a perfect black body varies directly as the fourth power of its absolute temperature.
E = σT^{4}
Here σ is Stefan’s constant and its value is 5.735×10^{8 }Wm^{2}K^{4}
Q10. A black body has an absolute temperature of 300 K. Calculate the radiant emittance of the body.
Solution:
Given:
Absolute temperature (T) = 300 K
Stefan's constant (σ) = 5.735 × 10^{8} Wm^{2}K^{4}
Using Stefan's Law:
E = σT^{4}
Substituting the values:
E = (5.735 × 10^{8} Wm^{2}K^{4})(300 K)^{4}
Calculating:
E ≈ 12.93 W/m^{2}
Therefore, the radiant emittance of the black body is approximately 12.93 W/m^{2}.
Q11. The radiant emittance of a black body is 50 W/m^2. Find the absolute temperature of the body.
Solution:
Given:
Radiant emittance (E) = 50 W/m^{2}
Stefan's constant (σ) = 5.735 × 10^{8} Wm^{2}K^{4}
Using Stefan's Law:
E = σT^{4}
Rearranging the formula:
T^{4} = E / σ
Substituting the values:
T^{4} = (50 W/m^{2}) / (5.735 × 10^{8 }Wm^{2}K^{4})
Calculating:
T^{4} ≈ 8.719 × 10^{15 }K^{4}
Taking the fourth root of both sides:
T ≈ 238.6 K
Therefore, the absolute temperature of the black body is approximately 238.6 K.
A body that absorbs all the radiation incident upon it and has an emissivity equal to 1 is called a perfectly black body.
It states that at any temperature, the ratio of emissive power e_{λ} of a body to its absorptive power a_{λ}, for a particular wavelength, is always constant and is equal to the emissive power of a perfect black body for that wavelength.
e_{λ}/a_{λ} = Constant = E_{λ}
This implies the ratio between e_{λ} and a_{λ }for anybody is a constant quantity (= E_{λ}).
Q12. At a particular temperature and wavelength, the spectral emissive power and monochromatic absorptive power of a body are 10 and 8 units respectively. What will be the emissive power of a black body at the same temperature?
Solution: According to Kirchoff's law.
e_{λ}/a_{λ} = Constant = E_{λ}
here, we are given that e_{λ}= 10 and a_{λ }= 8
So, E_{λ}= 10/8 = 1.25
Q13. At a temperature of 500 K, a body has a spectral emissive power of 20 W/m²·nm and a monochromatic absorptive power of 15 W/m²·nm at a specific wavelength. According to Kirchoff's Law of Heat Radiation, what is the value of the emissive power of a black body at the same temperature and wavelength?
Solution:
To find the emissive power of a black body at the same temperature and wavelength, we can use Kirchoff's Law of Heat Radiation, which states that the ratio of spectral emissive power (eλ) to monochromatic absorptive power (aλ) at a given temperature and wavelength is constant.
Given:
Temperature (T) = 500 K
Spectral emissive power (eλ) = 20 W/m²·nm
Monochromatic absorptive power (aλ) = 15 W/m²·nm
Using Kirchoff's Law, we have:
eλ/aλ = Constant = Eλ (emissive power of a black body)
Substituting the given values:
Eλ = eλ/aλ = 20 W/m²·nm / 15 W/m²·nm = 1.33
Therefore, the emissive power of a black body at the same temperature and wavelength is 1.33 W/m²·nm.
If a body having temperature T is kept in an environment of temperature To (< T), the body will lose energy at a rate
and absorb energy at a rate
Thus, the net rate of loss of energy
If m is the mass of the body and c is its specific heat, the rate of loss of heat at temperature T must be
Thus, the rate of cooling (dT/dt) depends on A, m, c, T, To, and e ( i.e. nature of the radiating surface). Hence greater the emissivity faster the cooling.
The rate of cooling of a body is directly proportional to the difference in temperature of the body over its surroundings.
If a body at temperature θ1 is placed in surroundings at lower temperature θ2, the rate of cooling is given by:
dQ/ dt ∝ (θ_{1} −θ_{2}) where dQ is the quantity of heat lost in time dt.
Newton’s law of cooling gives
dQ /dt = −k(θ_{1}−θ_{2}) where k is constant.
If a body of mass m and specific heat s loses a temperature dθ in time dt, then
dQ /dt =ms(dθ/dt) =−k(θ_{1}−θ_{2})
Q14. A liquid cools from 70℃ to 60℃ in 5 minutes. Calculate the time taken by the liquid to cool from 60℃ to 50℃, if the temperature of the surroundings is constant at 30^{0}C. (JEE Mains, 2022)
Solution:
Q15. A body at a temperature of 40ºC is kept in a surrounding of constant temperature of 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC. (JEE Mains, 2022)
Solution:
From Newton’s law of cooling, q_{f }= q_{i} e^{kt}
Now, for the interval in which temperature falls from 40 to 35^{o}C.
(35 – 20) = (40 – 20) e^{k.10}
e^{10k} = 3/4
k = [ln 4/3]/10 . . . . (a)
Now, for the next interval,
(30 – 20) = (35 – 20)e^{kt}
e^{kt} = 2/3
kt = ln 3/2 . . . . (b)
From equations (a) and (b),
t = 10 × [ln(3/2)/ln(4/3)]= 14.096 min.
When things are not too hot (below around 600℃), the heat radiation they give off can't be seen by our eyes because it's mostly in longer wavelengths than what we can see. The graph illustrates how the energy from this heat radiation changes with temperature and wavelength. As the temperature rises, two things happen. First, the highest point in the energy distribution moves to shorter wavelengths. This shift follows a rule called Wien's displacement law.
It states that the wavelength of radiation that is emitted with maximum intensity varies inversely to the absolute temperature of the body.
λ_{m}×T = Constant
The value of this constant in S.I. units is 2.898 × 10^{−3} mK
Have you ever wondered how scientists calculate the temperature of the sun and other stars? It is through this law.
Q16. A black body has a maximum intensity wavelength of 500 nm. According to Wien's Displacement Law, if the absolute temperature of the body is 4000 K, what would be the constant value?
Solution:
According to Wien's Displacement Law, the product of the wavelength of radiation emitted with maximum intensity (λm) and the absolute temperature (T) of a black body is a constant value.
Given:
Maximum intensity wavelength (λm) = 500 nm
Absolute temperature (T) = 4000 K
Using the formula:
λm × T = Constant
Substituting the values:
Constant = 500 nm × 4000 K
Solving for the constant value:
Constant = 2,000,000 nm·K
Therefore, the constant value is 2,000,000 nm·K.
Q17. What is the wavelength of the brightest part of the light from our next closest star, Proxima Centauri? Proxima Centauri is a red dwarf star about 4.2 light years away from us with an average surface temperature of 3,042 Kelvin. (JEE Mains, 2021)
Solution:
λ_{max} = T b
We don’t need the distance to solve this. All we need is the surface temperature to plug into our Wien’s law equation.
Wavelength λ_{max} in meters = 0.0029mK / 3042K; which is 0.000000953 meters.
We can convert this to nanometers and we get a peak wavelength of 953 nm.
Q18. The maximum intensity wavelength of radiation emitted by a black body is 700 nm. According to Wien's Displacement Law, if the constant value is 3,000,000 nm·K, what would be the absolute temperature of the body?
Solution:
According to Wien's Displacement Law, the product of the wavelength of radiation emitted with maximum intensity (λm) and the absolute temperature (T) of a black body is a constant value.
Given:
Maximum intensity wavelength (λm) = 700 nm
Constant value = 3,000,000 nm·K
Using the formula:
λm × T = Constant
Substituting the values:
700 nm × T = 3,000,000 nm·K
Solving for the absolute temperature:
T = 3,000,000 nm·K / 700 nm
T ≈ 4285.7 K
Therefore, the absolute temperature of the black body is approximately 4285.7 K.
Q19. Two bodies A and B have thermal emissivity of 0.1 and 0.81 respectively. The outer surface areas of the two bodies are identical. These two bodies emit total radiative power at the same rate. The wavelength λ_{B} corresponding to the maximum spectral radiancy in the radiation from B is 1.0 µm larger than the wavelength λ_{A} corresponding to the maximum spectral radiancy in the radiation from A. If the temperature of body A is 5802 K, find (JEE Mains, 2021)
(a) the temperature of (B)
(b) λ_{B}
Solution:
Wien's displacement law states that the black body radiation curve for different temperatures peaks at a wavelength inversely proportional to the temperature.
E_{λ}dλ = (A/λ^{5}) f(λT) dλ = (A/λ^{5}) e^{a}^{/λT} dλ
The rate of cooling of water increases with a fall in temperature.
The solar constant is how much sunlight the Earth gets in one minute on each square centimeter of a surface directly facing the sun at an average distance from it. If we ignore the absorption of sunlight by the Earth's atmosphere, the solar constant, denoted as 'S,' is calculated to be 1.94 cal/cm^{2}/min.
The temperature of the sun, T, is given as follows :
Where S is the solar constant, σ is Stefan’s constant, R is the mean distance of the earth from the sun and r is the radius of the sun.
NCERT Exemplar: Thermal Properties of Matter Doc  18 pages 
Revision Notes: Kinetic Theory of Gases Doc  6 pages 
Revision Notes: Heat Phenomena Doc  5 pages 
1. What is heat transfer? 
2. What is Stefan’s Law? 
3. What is Newton’s Law of Cooling? 
4. What is Kirchhoff’s law of Heat Radiation? 
5. What is Wien’s Displacement Law? 
NCERT Exemplar: Thermal Properties of Matter Doc  18 pages 
Revision Notes: Kinetic Theory of Gases Doc  6 pages 
Revision Notes: Heat Phenomena Doc  5 pages 

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