THERMAL RADIATION

Rate of Heat loss (RH) and Rate of Cooling (RC)

* If temp of surrounding is T0 then
dQ
Rate of Emission = σAT 4
dt e
dQ
Rate of absorption :- dt a
= σAT04
dQ dQ
Net Rate :- −
dt e dt a

dQ
Rate of heat loss :- RH = = σA T 4 − T04
dt

where T0  Temp of surrounding

Rate of heat loss (RH) depends on surface area
more the area more the heat loss
 THERMAL RADIATION
dQ
Rate of heat loss :- RH = = σA T 4 − T04
dt

As we know dQ = msdT
dQ dT
Q = ms
dt dt

dT
ms = sA (T4 – T04 )
dt
dT σA
 Rate of cooling RC = = ms T 4 − T04
dt

Rate of cooling (RC ) :

depends on area , mass and specific heat of substance
THERMAL RADIATION
 THERMAL RADIATION
Q. Sun and a star at 2000 K and 1000 K respectively and radius of star is half the
radius of sun. Find the ratio of heat loss.
1 16 64 1
(1) (2) (3) (4)
16 1 1 64

SUN STAR

2000 K 1000 K
R R
2
 THERMAL RADIATION
Q. Sun and a star at 2000 K and 1000 K respectively and radius of star is half the
radius of sun. Find the ratio of heat loss.
1 16 64 1
(1) (2) (3) (4) Sol. Rate of heat loss :-
16 1 1 64 dQ
RH = dt = σA T 4 − T04

dQ
= σAT 4
SUN STAR dt

QS 4πR2 2000 4
2000 K 1000 K  =
QSt πR2 1000
R R
2
QS𝑈𝑁 64
⇒
QStar 1
 THERMAL RADIATION
Q. A sphere, a cube and a thin circular plate, all made of the same material and having the
same mass are initially heated to a temperature of 1000°C. Which one of these will cool first
?
(1) Plate (2) Sphere
(3) Cube (4) None of these

Sphere Plate
3D Cube
2D
3D
 THERMAL RADIATION

Sol.

Sphere Plate
3D Cube
2D
3D

all made of the same material(s) and having the same mass(m)

σeA(T 4 − T04
RC =
ms
RC  (Area)
So Plate will cool first
 THERMAL RADIATION
Q. The rate of cooling at 600 K, if surrounding temperature is 300 K is H. The rate of cooling at
900K is:-
16
(1) 3
𝐻 (2) 2H
3 σA T 4 − T04
(2) (3) 3H (4) 𝐻 Sol. R C =
3
ms
𝑅1 (𝑇1 4 −𝑇𝑜 4 T1 = 600 K
= T2 = 900 K
𝑅2 (𝑇2 4 −𝑇𝑜 4
T0 = 300 K
𝑅1 (6004 −3004
= 𝑅1 = H
𝑅2 (9004 −3004
𝑅2 = ?
𝑅1 (24 −14
𝑅2
= (34 −14
𝐻 15
=
𝑅2 80
16𝐻
𝑅2 = 3
 THERMAL RADIATION
Q. If 2 Sphere of radius 𝑅 𝑎𝑛𝑑 2𝑅 but have same density same specific heat and both
have same temperature. T and surrounding temperature is Ts. Then comment on :-

(i) Rate of Heat loss

(ii) Rate of cooling
 THERMAL RADIATION
dQ
Sol. Rate of heat loss :- RH = = σA T 4 − T04
dt
dT σA
Rate of cooling RC = = T 4 − T04
dt ms
dQ
(1) = RH rea
dT
RH1 𝑅2 1
(RH  r2) so R = 4𝑅2 = 4
H2
dT A
(ii) RC = 
dt m
4𝜋𝑅2
RC  4 = R …….radius only
3
𝜋𝑅3

RC1 𝑅 1
RC2 = 2𝑅 = 2
THERMAL RADIATION
 THERMAL RADIATION
NEWTON'S LAW OF COOLING
dq
Rate of cooling is directly proportional to excess of
dt
temperature (q – q0) of the body over that of surrounding.

dq
− dt  (q – q0)

all temperatures in °C
(when (q – q0) ≯ 35°C
q = temperature of body
qo = temperature of surrounding,
q – q0 = excess of temperature (q > q0)
 THERMAL RADIATION

If the temperature of body is decreased by dq in time dt then rate of

dθ
fall of temperature ∝ − θ − θ0
dt

dθ
= −K θ − θ0
dt
Where negative sign indicates that the rate of cooling is decreasing
with time.
 THERMAL RADIATION
NEWTON'S LAW OF COOLING qor T
dθ
= −K θ − θ0 exponantional
dt decay graph
dθ
= −Kdt
θ − θ0 q0
integrating both side
t t
dθ
= − Kdt
θ − θ0
0
loge(q – q0) = –Kt + C
θ − θ0 = e−Kt+C
θ = e−Kt+C + θ0
 THERMAL RADIATION

For Numerical Problems, Newton's Law of cooling

If the temperature of body decreases from q1 to q2 and temperature
of surroundings is q0 then
θ1 +θ2
average excess of temperature = − θ0
2

θ1 − θ2 θ1 + θ2
 = +K − θ0
t 2
 THERMAL RADIATION
Q. A body takes 10 min. to cool from 90°C to 80°C. If the room
temperature is 25°C. How much time will it take to cool from 80°C to
70°C :-
(1) 12 min. (2) 15 min.
(3) 10 min. (4) 8 min.
 THERMAL RADIATION
θ1 − θ2 θ1 + θ2
Sol.  = +K − θ0
t 2

θ1 = 90°C
θ2 = 80°C 10 170
=𝐾 − 25 ....(1)
θ0 = 25°C 10 2
Time =10 min
θ1 = 80°C
θ2 = 70°C 10
=𝐾
150
− 25 ....(2)
θ0 = 25°C 𝑡 2

Time =? min
𝑡 60
= 50
10
t = 12 min
 THERMAL RADIATION
Q. Assuming Newton's law of cooling to be valid. The temperature of body changes from
60°C to 40°C in 7 minutes. Temperature of surroundings being 10°C, Find its
temperature after next 7 minutes.

(1) 48°C
(2) 38°C
(3) 28°C
(4) 18°C
 THERMAL RADIATION
Sol. θ1 − θ2 θ1 + θ2
 = +K − θ0
t 2
θ1 = 60°C
θ2 = 40°C 60−40 60+40
=𝐾 − 10 ....(1)
θ0 = 10°C 7 2
Time =7 min 1
K=
14

θ1 = 40°C
40−x 40+x
θ2 = ? =𝐾 − 10 ....(2)
7 2
θ0 = 10°C
Time =7 min
 40 – x = (40 + x – 20)  160 – 4x = 20 + x
 5x = 140  x = 28°C
 THERMAL RADIATION
Q. The initial temperature of a body is 80°C. If its temperature falls to 64°C in 5 minutes
and in 10 minutes to 52°. Find the temperature of surrounding :-
(1) 24°C (2) 60°C
(3) 36°C (4) 49°C
 THERMAL RADIATION
Sol. θ1 − θ2 θ1 + θ2
 = +K − θ0
t 2
θ1 = 80°C 80−64 80+64
θ2 = 64°C =𝐾 − x ....(1)
5 2
θ0 = ?
Time =5 min 16
= 𝐾 72 − x ....(1 )
5

θ1 = 80°C 80−52 80+52

=𝐾 − x ....(2)
10 2
θ2 = 52
θ0 = ? 28
= 𝐾 66 − x ....(2)
Time =10 min 10

16×2 72−x
= 66−x ⇒ (66 − x (8 = (72 − x )(7)
28
X= 24. the temperature of surrounding to 52°C
 THERMAL RADIATION
Important points of Newton's Law of Cooling
• Temperature difference should not exceed 35°C : (q – q0) ≯ 35°C
• Newton's law of cooling is used to calculate the specific Heat of
liquid.
• This law is an extended form of Stefan's law.
• all temperatures in °C
q = temperature of body
qo = temperature of surrounding,
q – q0 = excess of temperature (q > q0)
 THERMAL RADIATION
Spectral Energy distribution curve of Black Body radiations
Practically given by : Lumers and Pringshem
Mathematically given by : Plank
At any temperature the spectral
energy distribution curve for
surface of an ideal black body is
always continuous
 THERMAL RADIATION
Results from these Graphs

∞
Area = 0 Eλ dλ
Area = E = σT 4
A1 T1 4
hence A2
= T2

Eλm ∝ T 5

Area between curve and l axis gives

𝑸
the emissive power of body E = 𝑨𝒕 = σT 4
 THERMAL RADIATION
Wein's Displacement Law
The wavelength ( λm ) corresponding to maximum emission of
1
radiation decrease with increasing temperature λm ∝ T . This is
known as Wein's displacement law.
E
λm T = b
where b is Wein's constant
b= 2.89 x 10–3 m-K.
Dimensions of b : = M0 L1 T0 q1 λ 2.0 2.5 l in µm
0 1.0 1.5
m l

On increasing the temperature of a black body, wavelength for

maximum emission Shifts towards smaller wavelength
THERMAL RADIATION
 THERMAL RADIATION
Results from these Graphs

1
λm ∝
T
λm T = b
λm1 T1 = λm2 T2

Wein's Displacement Law

 THERMAL RADIATION

Wein's Displacement Law

1
λm ∝
T

V I B G O R
𝝀𝒎 : 𝐼𝑁𝐶𝑅𝐸𝐴𝑆𝐸 𝑻∶ 𝐷𝐸𝐶𝑅𝐸𝐴𝑆𝐸
 THERMAL RADIATION
Wein's Displacement Law
1
λm ∝
T

E
λb < λg < λr
Tb > Tg > Tr

λ
λb λg λr
 THERMAL RADIATION
Wein's Displacement Law hc
𝐸𝑛𝑒𝑟𝑔𝑦 = = h𝜈
λ
λm wavelength νm frequency
νm ∝ 𝑇

1
λm ∝
T

T T
 THERMAL RADIATION
Q. In the figure, the distribution of spectral energy of the radiation emitted by a black
body at a given temperature is shown. The possible temperature of the black body is :-
(b = 3 × 10–3 m-K)

(1) 1500 K
(2) 2000 K
(3) 2500 K E
(4) 3000 K

0 1.0 1.5 2.0 2.5 l in µm

 THERMAL RADIATION
Sol.
As we know
lm T = b
E
(1.5 × 10–6)T =b

1.5 × 10–6T = 3 × 10–3

3×10−3
T= 1.5×10−6

T = 2 × 103 0 1.0 1.5 2.0 2.5 l in µm

T = 2000 K
 THERMAL RADIATION
Q. A black body is at a temperature of 5760 K. The energy of radiation emitted by the
body
at wavelength 250 nm is U1
at wavelength 500 nm is U2 and
at wavelength 1000 nm is U3.
Wien's constant b = 2.88 × 10–3 m-K.
Which of the following is correct?

(1) U1 = 0 (2) U3 = 0
(3) U1 > U2 (4) U2 > U1
 THERMAL RADIATION
Sol. temperature of 5760 K
at wavelength 250 nm is U1
at wavelength 500 nm is U2 and
at wavelength 1000 nm is U3.
Wien's constant b = 2.88 × 10–3 m-K.
We know that
lmT = b E
𝑏 2.88×106
lm = 𝑇 = 𝑛𝑚
5760
lm = 500 nm
So U2 > U1
U2 > U3
λ 2.0 2.5 l in µm
0 1.0 1.5
m l
 THERMAL RADIATION
Solar constant 'S'

EM-Waves
SUN

Earth

The solar radiant energy received per unit area per unit time by a black surface held at
right angles to the Sun's rays and placed at the mean distance of the Earth (in the
absence of atmosphere) is called solar constant.

The solar constant S is taken to be 1340 watt/m2

 THERMAL RADIATION
Temperature of the Sun
Let R be the radius of the Sun and 'd' be the radius of Earth's orbit around the Sun.
Let E be the energy emitted by the Sun per second per unit area.
The total energy emitted by the Sun in one second = E.A = E × 4pR2.
d

A'
 THERMAL RADIATION
The total energy emitted by the Sun in one second = E.A = E × 4pR2
This energy is falling on a sphere of radius equal to the radius of the Earth's orbit around
the Sun i.e., on a sphere of surface area 4pd2
d
So, The energy falling per unit time per unit area of
Earth
4πR2 × E ER2
Solar constant S = = R
4πd2 d2
ER2
Solar constant S= 2 T
d
𝑄
By Stefan's Law 𝐴𝑡
= 𝐸 = sT4 A
R = 7× 108m , d = 1.5 × 1011 m, A'
s = 5.7 × 10–8 W m–2 K–4
 HEAT TRANSFER
Conduction : -
It is a process of heat transfer in which heat is transferred from high temperature to low
temperature without the actual motion of particles in a solid. Let we heat a rod from its one
end as shown in fig.

(1) Variable state : - In this state temperature of each and every part of the rod increases
with time.
(2) Steady state : - After some time a variable state comes when no heat is absorbed by
any part. So the temperature of every part is constant and decreases uniformly from
hotter end to colder end.

Note : In steady state each point have different temperature but it remain constant.
 HEAT TRANSFER
Rate of heat flow :-
In given diagram, in steady state the heat flow through the rod is constant. The rate of heat
dQ
flow dt
depends on :-

(1) Cross section Area (A) of rod

𝒅𝑻
(2) rate of change of temperature with distance − = temperature gradient
𝒅𝒙

dQ KA
 = T1 − T2 Rate of heat flow
dt L Q = 0
dx (T1>T2)
T1 T2
HOT COLD
T (T–dT)
L
 HEAT TRANSFER
Q = 0

dQ
∝A
dQ dT
∝ − dx dx (T1>T2)
dt dt T1 T2
On Combining the relations
HOT COLD
dQ dT
∝A − T (T–dT)
dt dx
L
dQ dT
 = −KA
dt dx

where K  coefficient of conductivity

if we take the whole length of rod then
dQ T2 −T1
 = −KA
dt L
dQ KA
 dt
=
L
T1 − T2 Rate of heat flow
 HEAT TRANSFER
Note : Thermal conductivity (K) :
It depends on nature of material.
Order of thermal conductivity Ag > Cu > Au > Al
SI Unit : J s–1 m–1 K–1 Dimensions : M1 L1 T–3 q–1

• For cooking the food, low specific heat and high conductivity utensils are most suitable.
• In winter, the iron chairs appear to be colder than the wooden chairs.
• Cooking utensils are made of aluminium and brass whereas their handles are made of
wood.
• Two thin blankets are warmer than a single blanket of double the thickness.
 HEAT TRANSFER
Thermal resistance :-
It is the obstruction offered to the flow of heat through the conductor.
When we compare the flow of heat to the flow of electricity.
T1 > T2
i
T
1
T
2
V1 V2

dQ KA V1 −V2
iT = = T1 − T2 i=
dt L R

(iT = Thermal current)

L
On comparing we get that is the value of thermal resistance
KA

L
So R=
KA
 HEAT TRANSFER
Q. Find thermal resistance of a rod having area 10–3 m2,
W
K = 200 , ℓ = 2m
m°C

ℓ 2
Sol. R= ⇒ = 10
KA 200×10−3
 HEAT TRANSFER
Q. Find the rate of heat flow in the following figure. :-
(1) 50 J/s (2) 100 J/s (3) 150 J/s (4) 200 J/s

W
Sol. K = 200 m°C , ℓ = 2m
A = 10–2 m2
Q KA ΔT
iT = =
t ℓ
200×10−2 ×100
= 2
 100 J/sec.