Transfer of Heat : Conduction , Convection & Radiation

Conduction:

Heat energy is transferred (usually through solids) from one part of the material medium to other without transferring the material particles. -θ°C

(i) Steady State : In this state heat absorption stops and temperature gradient throughout the rod becomes constant i.e. dT/dx = constant

(ii) Before steady State : Temperture of rod at any point changes

Note:
If specific heat of any substance is zero, it can be considered always in steady state.

LAW FOR THERMAL CONDUCTION IN STEADY STATE:

In steady state ΔQ amount of heat passing through a bar of length L and cross-section A in Δt time when its ends are at temperatures T1 and T2 is given by:

$ \displaystyle \frac{\Delta Q}{\Delta t} = KA \frac{T_1 – T_2}{L} $

So rate of flow of heat will be ,

$ \large \frac{\Delta Q}{\Delta t} =  \frac{T_1 – T_2}{L/K A} $

$ \large \frac{\Delta Q}{\Delta t} =  \frac{T_1 – T_2}{R_{Th}} $

$\large Heat \; current = \frac{Temperature \; difference}{Thermal \; Resistance}$

Where $\large R_{Th} = \frac{L}{K A} $ is called Thermal Resistance .

The quantity $ (\frac{dT}{dx} ) $ is called temperature gradient (minus sign indicates that with increase in , temperature decreases) and the constant depends on the nature of metal and is called coefficient of thermal conductivity or simply thermal conductivity and is a measure of the ability of a substance to conduct heat through it.

(i) Two rods joined

$\large \frac{dQ}{dt} = \frac{T_1 -T_2}{R_1 + R_2} $

(ii) Three rods joined to a common point

$\large \frac{T – 100}{R_1} + \frac{T – 20}{R_2} + \frac{T – 5}{R_3} = 0 $

$\large \frac{T – 100}{l_1/K_1 A} + \frac{T – 20}{l_2/K_2A} + \frac{T – 5}{l_3/K_3A} = 0 $

SERIES AND PARALLEL CONNECTION OF RODS IN STEADY STATE:

Series Connection :

Let length l and cross section area A of both rods are same & T1 > T2

$\large R_1 = \frac{l}{K_1 A} \; , R_2 =\frac{l}{K_2 A}$

Thermal Current , $\large \frac{\Delta Q}{\Delta t} = \frac{T_1 – T}{R_1} = \frac{T – T_2}{R_2} = \frac{T_1 – T_2}{R_1+R_2}$

$\large (T – T_2)R_1 = (T_1 – T)R_2 $

$\large T (R_1 + R_2) = T_1 R_2 + T_2 R_1 $

$\large T = \frac{ T_1 R_2 + T_2 R_1}{R_1 + R_2} $

$\large \frac{T_1 -T_2}{R} = \frac{T_1 -T_2}{R_1 + R_2} $

$\large R = R_1 + R_2 $

Two rods together is equivalent to a single rod of thermal resistance (R1 + R2 )

Parallel Connection :

$\large i_1 = \frac{\Delta Q_1}{\Delta t} = \frac{T_1 -T_2}{R_1} $

$\large i_2 = \frac{\Delta Q_2}{\Delta t} = \frac{T_1 -T_2}{R_2} $

$\large i = i_1 + i_2 $

$\large \frac{T_1 -T_2}{R} = (T_1 -T_2) (\frac{1}{R_1} + \frac{1}{R_2} ) $

The system of the two rods is equivalent to a single rod of thermal resistance R is given by

$\large \frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} $

CONDUCTION BEFORE STEADY STATE:

Differential form :

$\large \frac{dQ}{dt} = K A \frac{dT}{dx} $ ;

Where $\large \frac{dT}{dx} = temperature \;gradient $

CONDUCTION IN A SECTION OF MEDIUM BEFORE STEADY STATE:

$\large dQ = dQ_1 – dQ_2 $

(In steady state dQ = 0 , i.e. dQ1 = dQ2 )

$\large m s dT = K A \frac{dQ_1}{dx}dt – K A \frac{dQ_2}{dx}dt$

$\large \frac{dT}{dt} = \frac{K A}{m s} (\frac{dT_1}{dx} – \frac{dT_2}{dx} ) $

dT = increase in temperature of the section in time dt.

Growth Of Ice:

Considering a layer of ice of thickness . The air temperature is -θ °C and water temperature below the ice is 0º C.

Considering unit cross-section area of ice, if a layer of thickness dx grows in time dt .

Then heat given by this layer = mass × latent heat

= 1 . dx . ρ . L

ρ = density of ice , L = latent heat of fusion of ice.

If this quantity of heat is conducted upwards through the ice layer of area A in time dt .

$\large A. dx .\rho .L = \frac{K A (0-(-\theta))}{x} dt $

Time taken , $\large t = \frac{\rho L}{K \theta} \int_{x_1}^{x_2} x dx$

$\large = \frac{\rho L}{2 K \theta} (x_2^2 – x_1^2 ) $

Rate of increase of thickness of the ice layer $\large (\frac{dx}{dt} = \frac{K \theta}{\rho L x} ) $

Convection:

Heat energy is transferred (usually through liquids and gases) by mass movement of molecules from one point to another. (Due to gravity & buoyant force).

Radiation , Stefan’s Law & Newton’s Law of cooling

Radiation: 

Heat energy is transferred by electromagnetic waves even in absence of medium.

Absorptive Power (a):

Absorptive power of a body is defined as the fraction of the incident radiation that is absorbed by the body.

Absorptive power , $\large a = \frac{Energy \; absorbed}{Energy \; incident} $

Emissive Power (e)

The emissive power denotes the energy radiated per unit area per unit time per unit solid angle along the normal to the area.

Emissivity (ε)

Emissivity of a surface is the ratio of the emissive power of the surface to the emissive power of black body at the same temperature.

Emissivity, $\large \epsilon = \frac{Emissive \; Power \; of \; the \; surface}{Emissive \; Power \; of \; black \; body \; at \; same \; temperature} $

Black body:
A perfectly black body is one which absorbs completely all the radiation, of whatever wave-length, incident on it. (a = 1)

KIRCHHOFF’S LAW:
It states that the ratio of the emissive power to the absorptive power for radiation of a given wave length is the same for all bodies at the same temperature, and is equal to the emissive power of a perfectly black body at that temperature.

$\large \frac{e}{a} = E $

STEFAN’S LAW OF RADIATION :

The total radiant energy emitted E per unit time by a black body of surface area A is proportional to the fourth power of its absolute temperature and surface area A of the black body .

$\large E \propto A T^4 $

$\large E = \sigma A T^4 $

( Where , σ = Stefan’s constant = 5.67 × 10-8 kg sec-2 k-4)

For a body which is not a black body

$\large E = \epsilon \sigma A T^4 $

Where ε is emissivity of the body

Using Krichoff’s law

$\large \frac{E_{body}}{E_{black\;body}} = a $

$\large \frac{\sigma A T^4}{\epsilon \sigma A T^4} = a $

ε = a

Emissivity and absorptive power have the same value. So, good absorbers are good radiators and bad absorbers are bad radiators.

Net Loss Of Thermal Energy:
If a body of surface area A is kept at absolute temperature T in a surrounding of temperature T0 (T0 < T) . Then energy emitted by the body per unit time

$\large E = \epsilon \sigma A T^4 $

And energy absorbed per unit time by the body

$\large E_0 = \epsilon \sigma A T_0^4 $

Net, loss of thermal energy per unit time

$\large \Delta E = E – E_0 = \epsilon \sigma A (T^4 – T_0^4 )$

This is known as Stefan’s Law .

NEWTON’S LAW OF COOLING:

For a small temperature difference between a body and its surrounding, the rate of cooling of the body is directly proportional to the temperature difference.

If a body of surface area A is kept at absolute temperature T in a surrounding of temperature T0 (T0 < T) . Then net loss of thermal energy per unit time.

$\large \frac{dQ}{dt} = \epsilon \sigma A (T^4 -T_0^4 ) $

If the temperature difference is small ,

$\large T = T_0 + \Delta T$

$\large \frac{dQ}{dt} = \epsilon \sigma A ((T_0 + \Delta T)^4 -T_0^4 ) $

$\large = \epsilon \sigma A (T_0^4(1 + \frac{\Delta T}{T_0})^4 -T_0^4 ) $

$\large = \epsilon \sigma A T_0^4 [(1 + 4\frac{\Delta T}{T_0} + …) – 1 )] $

$\large = 4 \epsilon \sigma A T_0^3 \Delta T $ ….(i)

Now, rate of loss of heat at temperature T

$\large \frac{dQ}{dt} = – m s \frac{dT}{dt}$ ….(ii)

From equation (i) and (ii), we get

$\large m s \frac{dT}{dt} = – 4 \epsilon \sigma A T_0^3 ( T – T_0 ) $

$\large \frac{dT}{dt} = – \frac{4 \epsilon \sigma A T_0^3}{m s} ( T – T_0 ) $

$\large \frac{dT}{dt} = -k (T – T_0) $

$\large \frac{dT}{dt} \propto (T – T_0) $

Wein’s black body radiation :

At every temperature (> 0 K) a body radiates energy in the form of electromagnetic wave of all wavelengths.

According to Wein’s displacement law if the wavelength corresponding to maximum energy is λm

$\large \lambda_m T = b $

Then where b is a constant (Wein’s constant)

T = temperature of body

Intensity at a specific temp. T , $\large I_m \propto T^5 $

This is wein’s fifth power law.

Temperature of the Sun & Solar constant:

• The temperature of the Sun can be determined by assuming it to be a black body.

• Total amount of radiant energy emitted by the sun per second $\large =( 4 \pi R^2 ) \sigma T^4 $ ; Where , R = Radius of the sun.
• let ‘r’ be the mean distance between the sun and the earth and S0 be the solar constant. Then the energy received per second by the sphere of radius (r) is $\large 4 \pi r^2 S_0 $ .

Solar constant:
Solar constant is defined as “the rate at which the radiant energy of the sun received by perfectly black surface, normal to unit area in the absence of atmosphere, when kept at distance equal to the mean distance of earth from the Sun”.

$\large 4 \pi R^2 \sigma T^4 = 4 \pi r^2 S_0 $

$\large T^4 = (\frac{r}{R})^2 (\frac{S_0}{\sigma})$

$\large T = (\frac{r}{R})^2 (\frac{S_0}{\sigma})^{1/4}$

Hence, the surface temperature of the Sun can be determined.

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