NEET Physics Class 11 Chapter 8 Heat Transfer

NEET Physics Class 11 Chapter 8 Heat Transfer Introduction

Heat is energy in transit that flows due to temperature difference; from a body at a higher temperature to a body at a lower temperature. This transfer of heat from one body to the other takes place through three processes.

1. Conduction
2. Convection
3. Radiation

NEET Physics Class 11 Chapter 8 Conduction

The process of transmission of heat energy in which heat is transferred from one particle of the medium to the other, but each particle of the medium stays at its position is called conduction, for example, if you hold an iron rod with one of its end on a fire for some time, the handle will get heated.

• The heat is transferred from the fire to the handle by conduction along the length of the iron rod. The vibrational amplitude of atoms and electrons of the iron rod at the hot end takes on relatively higher values due to the higher temperature of their environment.
• These increased vibrational amplitudes are transferred along the rod, from atom to atom during collision between adjacent atoms. In this way, a region of rising temperature extends itself along the rod to your hand.

Consider a slab of face area A, Lateral thickness L, whose faces have temperatures THand TC(TH> TC).

Now consider two cross sections in the slab at positions A and B separated by a lateral distance of dx. Let the temperature of face A be T and that of face B be T + ΔT. Then experiments show that Q, the amount of heat crossing the area A of the slab at position x in time t is given by

⇒ \(\frac{Q}{t}=-K A \frac{d T}{d x}\)

Here K is a constant depending on the material of the slab and is named the thermal conductivity of the material, and the quantity \(\left(\frac{d T}{d x}\right)\) is called temperature gradient. The (–) sign in the equation heat flows from high to low temperature (ΔT is a –ve quantity)

NEET Physics Class 11 Chapter 8 Steady State

If the temperature of a cross-section at any position x in the above slab remains constant with time (remember, it does vary with position x), the slab is said to be in a steady state.

• Remember steady-state is distinct from thermal equilibrium for which temperature at any position (x) in the slab must be the same.
• For a conductor in a steady state, there is no absorption or emission of heat at any cross-section. (as the temperature at each point remains constant with time).
• The left and right faces are maintained at constant temperatures TH and TC respectively, and all other faces must be covered with adiabatic walls so that no heat escapes through them and the same amount of heat flows through each cross-section in a given Interval of time.

Hence Q₁= Q = Q₂. Consequently, the temperature gradient is constant throughout the slab.

Hence, \(\frac{d T}{d x}=\frac{\Delta T}{L}=\frac{T_f-T_i}{L}=\frac{T_C-T_H}{L}\)

and \(\frac{Q}{t}=-\mathrm{KA} \frac{\Delta T}{L} \Rightarrow \frac{Q}{t}\)

⇒ \(\mathrm{KAQ}\left(\frac{T_H-T_C}{L}\right)\)

Here Q is the amount of heat flowing through a cross-section of slab at any position in a time interval of t.

Question 1. One face of an aluminum cube of edge 2 meter is maintained at 100ºC and the other end is maintained at 0ºC. All other surfaces are covered by adiabatic walls. Find the amount of heat flowing through the cube in 5 seconds. (The thermal conductivity of aluminum is 209 W/m–ºC)
Answer:

Heat will flow from the end at 100ºC to the end at 0ºC.

Area of a cross-section perpendicular to the direction of heat flow,

A = 4m²

then \(\frac{Q}{t}=\mathrm{KA} \frac{\left(T_H-T_C\right)}{L}\)

Q = \(\frac{\left(209 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}\right)\left(4 \mathrm{~m}^2\right)\left(100^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}\right)(5 \mathrm{sec})}{2 \mathrm{~m}}\)

= 209 KJ

NEET Physics Class 11 Chapter 8 Thermal Resistance To Conduction

If you are interested in insulating your house from cold weather or for that matter keeping the meal hot in your tiffin box, you are more interested in poor heat conductors, rather than good conductors. For this reason, the concept of thermal resistance R has been introduced.

For a slab of cross-section A, Lateral thickness L, and thermal conductivity K,

Resistance \(R=\frac{L}{K A}\)

In terms of R, the amount of heat flowing through a slab in steady-state (in time t)

⇒ \(\frac{Q}{t}=\frac{\left(T_H-T_L\right)}{R}\)

If we name as thermal current ir

then, \(i_T=\frac{T_H-T_L}{R}\)

This is mathematically equivalent to OHM’s law, with temperature donning the role of electric potential. Hence results derived from OHM’s law are also valid for thermal conduction.

Moreover, for a slab in the steady state, we have seen earlier that the thermal current it remains the same at each cross-section. This is analogous to Kirchoff’s current law in electricity, which can now be very conveniently applied to thermal conduction.

Question 2. Three identical rods of length 1m each, having cross-section area of 1cm2 each and made of Aluminium, copper, and steel respectively are maintained at temperatures of 12ºC, 4ºC, and 50ºC respectively at their separate ends. Find the temperature of their common junction.

[ KCu = 400 W/m-K , K_Al = 200 W/m-K , K_steel = 50 W/m-K ]

Answer:

⇒ \(\mathrm{R}_{\mathrm{Al}}=\frac{L}{K A}=\frac{1}{10^{-4} \times 209}=\frac{10^4}{209}\)

Similarly = \(R_{\text {steel }}=\frac{10^4}{46} \text { and } R_{\text {copper }}=\frac{10^4}{385}\)

Let the temperature of common junction = T then from Kirchoff’s Junction law.

i_Al + i_steel + iCu = 0

⇒ \(\frac{T-12}{R_{A I}}+\frac{T-51}{R_{\text {steel }}}+\frac{T-u}{R_{C u}}=0\)

⇒ (T – 12) 200 + (T – 50) 50 + (T – 4) 400 = 0

⇒ 4(T – 12) + (T – 50) + 8 (T – 4) = 0

⇒ 13T = 48 + 50 + 32 = 130

⇒ T = 10ºC

 

NEET Physics Class 11 Chapter 8 Growth Of Ice On Ponds

When atmospheric temperature falls below 0°C the water in the lake will start freezing. Let at any time t, the thickness of ice in the lake be y and atmospheric temperature is –θ°C. The temperature of water in contact with the lower surface of ice will be 0ºC.

the area of the lake = A

heat escaping through ice in time dt is

Now due to the escaping of this heat if the thickness of water in contact with the lower surface of ice freezes,

⇒ \(d Q_1=K A \frac{[0-(-\theta)]}{y} d t\)

dQ₂= mL = ρ(dy A)L [as m = ρV = ρA dy]

But as dQ₁= dQ₂, the rate of growth of ice will be

⇒ \(\frac{d y}{d t}=\frac{K \theta}{\rho L} \times \frac{1}{y}\)

and so the time taken by ice to grow a thickness y, \(t=\frac{\rho L}{K \theta} \int_0^y y \quad d y=\frac{1}{2} \frac{\rho L}{K \theta} \quad y^2\)

Time taken to double and triple the thickness will be in the ratio t₁: t₂: t₃:: 1² : 2²: 3², i.e., t₁: t₂: t₃:: 1 : 4: 9 and the time intervals to change thickness from 0 to y, from y to 2y and so on will be in the ratio Δt₁: Δt2: Δt₃: : (1² – 0² ) : (2² – 1² ) : (3² – 2² ), i.e., Δt₁: Δt₂: Δt₃:: 1 : 3: 5.

Can you now see how the following facts can be explained by thermal conduction?

1. In winter, iron chairs appear to be colder than the wooden chairs.
2. Ice is covered in gunny bags to prevent melting.
3. Woolen clothes are warmer.
4. We feel warmer in a fur coat.
5. Two thin blankets are warmer than a single blanket of double the thickness.
6. Birds often swell their feathers in winter.
7. A new quilt is warmer than an old one.
8. Kettles are provided with wooden handles.
9. Eskimos make double-walled ice houses.
10. Thermos flask is made double-walled.

NEET Physics Class 11 Chapter 8 Convection

When heat is transferred from one point to the other through the actual movement of heated particles, the process of heat transfer is called convection.

• In liquids and gases, some heat may be transported through conduction. But most of the transfer of heat in them occurs through the process of convection.
• Convection occurs through the aid of the earth’s gravity. Normally the portion of fluid at greater temperature is less dense, while that at lower temperature is denser. Hence hot fluids rise while colder fluids sink, accounting for convection. In the absence of gravity, convection would not be possible.
• Also, the anomalous behavior of water (its density increases with temperature in the range of 0-4ºC) gives rise to interesting consequences vis-a-vis the process of convection. One of these interesting consequences is the presence of aquatic life in temperate and polar waters. The other is the rain cycle.

Can you now see how the following facts can be explained by thermal convection?

1. Oceans freeze top-down and not bottom-up. (this fact is singularly responsible for the presence of aquatic life in temperate and polar waters.)
2. The temperature at the bottom of deep oceans is invariably 4ºC, whether it is winter or summer.
3. You cannot illuminate the interior of a lift in free fall or an artificial satellite of earth with a candle.
4. You can Illuminate your room with a candle.

NEET Physics Class 11 Chapter 8 Radiation

The process of the transfer of heat from one place to another place without heating the intervening medium is called radiation. The term radiation used here is another word for electromagnetic waves. These waves are formed due to the superposition of electric and magnetic fields perpendicular to each other and carry energy.

Properties of Radiation:

1. All objects emit radiation simply because their temperature is above absolute zero, and all objects absorb some of the radiation that falls on them from other objects.
2. Maxwell based on his electromagnetic theory proved that all radiations are electromagnetic waves and their sources are vibrations of charged particles in atoms and molecules.
3. More radiations are emitted at higher temperatures of a body and less at lower temperatures.
4. The wavelength corresponding to the maximum emission of radiation shifts from a longer wavelength to a shorter wavelength as the temperature increases. Due to this, the color of a body appears to be changing. Radiations from a body at NTP have predominantly infrared waves.
5. Thermal radiation travels with the speed of light and moves in a straight line.
6. Radiations are electromagnetic waves and can also travel through a vacuum.
7. Similar to light, thermal radiations can be reflected, refracted, diffracted, and polarized.
8. Radiation from a point source obeys the inverse square law (intensity α ).

NEET Physics Class 11 Chapter 8 Prevost Theory Of Heat Exchange

According to this theory, all bodies radiate thermal radiation at all temperatures. The amount of thermal radiation radiated per unit of time depends on the nature of the emitting surface, its area, and its temperature. The rate is faster at higher temperatures.

• Besides, a body also absorbs part of the thermal radiation emitted by the surrounding bodies when this radiation falls on it. If a body radiates more than what it absorbs, its temperature falls.
• If a body radiates less than what it absorbs, its temperature rises. And if the temperature of a body is equal to the temperature of its surroundings it radiates at the same rate as it absorbs.

NEET Physics Class 11 Chapter 8 Perfectly Black Body And Black Body Radiation (Fery’s Black Body)

A perfectly black body absorbs all the heat radiations of whatever wavelength, is incident on it. It neither reflects nor transmits any of the incident radiation and therefore appears black whatever the color of the incident radiation.

In actual practice, no natural object possesses strictly the properties of a perfectly black body.

• But the lamp-black and platinum black are a good approximation of the black body. They absorb about 99 % of the incident radiation. The most simple and commonly used black body was designed by Fery.
• It consists of an enclosure with a small opening which is painted black from inside. The opening acts as a perfect black body.
• Any radiation that falls on the opening goes inside and has very little chance of escaping the enclosure before getting absorbed through multiple reflections. The cone opposite to the opening ensures that no radiation is reflected directly.

NEET Physics Class 11 Chapter 8 Absorption, Reflection, And Emission Of Radiations

Q = Q_r+ Q_t+ Q_a

1 = \(\frac{Q_r}{Q}+\frac{Q_t}{Q}+\frac{Q_a}{Q}\)

where r = reflecting power , a = absorptive power

and t = transmission power.

1. r = 0, t = 0, a = 1, perfect black body
2. r = 1, t = 0, a = 0, perfect reflector
3. r = 0, t = 1, a = 0, perfect transmitter

Absorptive Power :

In particular, the absorptive power of a body can be defined as the fraction of incident radiation that is absorbed by the body.

a = \(\frac{\text { Energy absorbed }}{\text { Energy incident }}\)

As all the radiation incident on a black body is absorbed, a = 1 for a black body.

Emissive Power:

Energy radiated per unit time per unit area along the normal to the area is known as emissive power.

⇒ \(\frac{Q}{\Delta A \Delta t}\)

(Notice that, unlike absorptive power, emissive power is not a dimensionless quantity).

Spectral Emissive Power (Eλ) :

Emissive power per unit wavelength range at wavelength λ is known as spectral emissive power, Eλ. If E is the total emissive power and Eλ is spectral emissive power, they are related as follows,

⇒ \(\mathrm{E}=\int_0^{\infty} E_\lambda \mathrm{d} \lambda\)

and \(\frac{\mathrm{dE}}{\mathrm{d} \lambda}=\mathrm{E}_\lambda\)

Emissivity:

⇒ \(\mathrm{e}=\frac{\text { Emissive power of } \mathrm{a} \text { body at temperature } \mathrm{T}}{\text { Emissive power of } \mathrm{a} \text { black body at same temperature } \mathrm{T}}\)

⇒ \(\frac{E}{E_0}\)

NEET Physics Class 11 Chapter 8 Kirchoff’s Law

The ratio of the emissive power to the absorptive power for the radiation of a given wavelength is the same for all substances at the same temperature and is equal to the emissive power of a perfectly black body for the same wavelength and temperature.

⇒ \(\frac{E(\text { body })}{a(\text { body })}\) = E(black body)

Hence we can conclude that good emitters are also good absorbers.

Applications Of Kirchhoff’s Law

If a body emits strongly the radiation of a particular wavelength, it must also absorb the same radiation strongly.

1. Let a piece of china with some dark painting on it be first heated to nearly 1300 K and then examined in a dark room. It will be observed that the dark paintings appear much brighter than the white portion. This is because the paintings being better absorbers also emit much more light.
2. The silvered surface of a thermos flask does not absorb much heat from outside. This stops ice from melting quickly. Also, the silvered surface does not radiate much heat from the inside. This prevents hot liquids from becoming cold quickly.
3. A red glass appears red at room temperature. This is because it absorbs green light strongly. However, if it is heated in a furnace, it glows with green light. This is because it emits green light strongly at a higher temperature.
4. When white light is passed through sodium vapors and the spectrum of transmitted light is seen, we find two dark lines in the yellow region. These dark lines are due to absorption of radiation by sodium vapors which it emits when heated.

Fraunhofer lines are dark lines in the spectrum of the sun. When white light emitted from the central core of the sun (photosphere) passes through its atmosphere (chromosphere) radiations of those wavelengths will be absorbed by the gases present there which they usually emit (as a good emitter is a good absorber) resulting in dark lines in the spectrum of sun.

At the time of a solar eclipse, direct light rays emitted from the photosphere cannot reach the earth and only rays from the chromosphere can reach the earth’s surface. At that time we observe bright Fraunhofer lines.

NEET Physics Class 11 Chapter 8 Nature Of Thermal Radiations : (Wien’s Displacement Law)

From the energy distribution curve of black body radiation, the following conclusions can be drawn:

1. The higher the temperature of a body, the higher the area under the curve i.e. more amount of energy is emitted by the body at a higher temperature.
2. The energy emitted by the body at different temperatures is not uniform. For both long and short wavelengths, the energy emitted is very small.
3. For a given temperature, there is a particular wavelength (λ_m) for which the energy emitted (Eλ) is maximum.
4. With an increase in the temperature of the black body, the maxima of the curves shift towards shorter wavelengths.

From the study of the energy distribution of black body radiation discussed above, it was established experimentally that the wavelength (λ_m) corresponding to the maximum intensity of emission decreases inversely with an increase in the temperature of the black body. i.e.

λ_m ∝ or λ_m T = b

This is called Wien’s displacement law.

Here b = 0.282 cm-K, is the Wien’s constant.

Question 1. Solar radiation is found to have an intensity maximum near the wavelength range of 470 nm. Assuming the surface of the sun to be perfectly absorbing (a = 1), calculate the temperature of the solar surface.
Solution :

Since a =1, the sun can be assumed to be emitting as a black body from Wien’s law for a black body

λ_m. T = b

⇒ T = \(\frac{0.282(\mathrm{~cm}-\mathrm{K})}{\left(470 \times 10^{-7} \mathrm{~cm}\right)}\)

= ~ 6125 K.