1
THERMAL PHYSICS
EXERCISE - 1 : BASIC OBJECTIVE QUESTIONS
Heat, Temperature and Calorimetry
Ans.
(b)
Temperature and Heat
1.
At absolute zero
(a) all substances exist in solid form
(b) molecular motion ceases
Sol.
3
KbT  0
2
t A  30
t 0
 B
180  30 100  0
t A  30 t B

150
100
At critical temperature, the surface tension of a liquid is :
(a) zero
(b) infinity
4.
(c) the same as that at any other temperature
(d) cannot be determined
Ans.
(a)
Sol.
Surface tension decreases with the increases of temperature
as temperature increases the molecular force decreases
between the molecules. Surface tension vanishes at critical
temperature.
3.
 t A H   t A L  t B  H   t B  L

So molecular motion ceases.
2.
t B   t B L
 t A H  180 C ;  t B H  100C
(b)
As T = 0K i.e. absolute zero
K.E. of molecules 

  t A L  180  150  30C ;  t B L  0C
(c) water freezes
(d) None of the above
Ans.
Sol.
t A   t A L
The reading of Centigrade thermometer coincides with that
of Fahrenheit thermometer in a liquid. The temperature of
the liquid is :
(a) –40°C
(b) 0°C
(c) 100°C
Ans.
Sol.
C F  32

5
9
C=F=x

x x  32

5
9
9x = 5x – 32 × 5
4x = –32 × 5
The graph between two temperature scales A and B is
shown in figure.
x
5.
Between upper fixed point and lower fixed point, there are
150 equal divisions on scale A and 100 on scale B. The
relationship for conversion between the two scales is given
by :
(a)
t A  180 t B

100
150
(b)
t A  30
t
 B
150
100
(c)
t B  180 t A

150
100
(d)
t B  40 t A

100
180
(d) 300°C
(a)
32  5
  40C
4
Which of the following parameters does not characterize
the thermodynamic state of matter?
(a) temperature
(b) pressure
(c) work
(d) volume
Ans.
(c)
Sol.
The work does not characterize the thermodynamic state
of matter.
The radius of a ring is R and its coefficient of linear expansion
is . If the temperature of ring increases by , then its
circumference will increase by :
(a)  R
(b) 2 R
6.
(c)  R
Ans. (b)

2
(d)  R

4
2
THERMAL PHYSICS
Sol. L  L0 1  T 
2R1  2R 1  T  = 2R +2R  
Hence cicumfrence increase by
2R
A steel wire of cross–sectional area 0.5 mm2 is held between
two fixed supports. If the wire is just taut at 20°C, determine
the tension when the temperature falls to 0°C. Coefficient
of linear expansion of steel is 1.2 × 10–5/°C and its Young’s
modulus is 2.0 × 1011 N/m2.
(a) 24 N
(b) 36 N
(c) 12 N
(d) 6 N
Ans. (a)
7.
Sol.
10.
Equal masses of three liquids A, B and C have temperatures
10°C, 25°C and 40°C respectively. If A and B are mixed,
the mixture has a temperature of 15°C. If B and C are mixed,
the mixture has a temperature of 30°C. If A and C are mixed,
the mixture will have a temperature of:
(a) 16°C
(b) 20°C
(c) 25°C
(d) 29°C
Ans. (a)
Sol. When A and B are mixed
Tf = 15°C
ms B  25  15  ms A 15  10    s A  2s B
When B and C are mixed
Tf = 30°C
msB (30 – 25) = msC (40 – 30)
5sB = 10sC
sB = 2sC
...(ii)
From (i) & (ii) sA = 4sC
When A and C are mixed
Tf = ?
F
Y A
L
L
F  Y
L
A
L
ms A  Tf  10   msC  40  Tf 
F  Y  T  A
s A  Tf  10   sC  40  Tf 
F   2  1011   1.2  10 5  20  0.5  106
F = 24 N
8. A solid ball of metal has a spherical cavity inside it. If the
ball is heated the volume of cavity will :
(a) increase
(b) decrease
(c) remain unchanged
(d) data insufficient
Ans. (a)
Sol. D  D0 T
Since T is positive therefore
D   ve
Hence volume will increase.
Thermal Expansion
9. An iron ball of mass 0.2 kg is heated to 10°C and put into a
block of ice at 0°C. 2.5 g of ice melts. If the latent heat of
fusion of ice is 80 cal/g, then the specific heat of iron in
cal/g°C is :
(a) 1
(b) 0.1
(c) 0.8
(d) 0.08
Ans. (b)
Sol. m1s1T  m 2 L 2
103  0.2  s iron  10  2.5  80
siron = 0.1 gm°C
4sC  Tf  10   sC  40  Tf 
Tf = 16°C
11.
The temperatures of equal masses of three different liquids
A, B and C are 12°C, 19°C and 28°C respectively. The
temperature when A and B are mixed is 16°C, and when B
and C are mixed, it is 23°C. What will be the temperature
when A and C are mixed ?
(a) 15.6 °C
(b) 23.2 °C
(c) 20.3 °C
(d) 25.8 °C
Ans. (c)
Sol. When A and B are mixed
msA (16 – 12) = msB (19 – 16)
4sA = 3sB
...(i)
When B and C are mixed
msB (23 – 19) = msC (28 – 23)
4sB = 5sC
...(ii)
From equation (i) and (ii)
5
4s A  3  s C
4
When A and C are mixed
ms A  Tf  12   msC  28  Tf 
3
THERMAL PHYSICS
3 5
s C  Tf  12   s C  28  Tf 
4 4
Then for same temperature
15Tf  180  28  16Tf
lB  l0  B T
lA  l0  A T
Tf = 20.3°C
12. 50 g of ice at 0°C is mixed with 50 g of water at 60°C, final
temperature of mixture will be :
(a) 0°C
(b) 40°C
(c) 10°C
(d) 15°C
Ans. (a)
Sol. When 50 gm water at 60°C is converted into 0°C water then
heat released
Qreleased = 50 × sw × (60 – 0)
Qreleased = 3000 cal
Suppose this heat can melt x gm of ice 3000 cal = xL
x
3000
gm  37.5 gm
80
Thus this heat is not sufficient to melt 50 gm of ice. Hence
final temperature will be ‘O°C’.
13.
Two metal rods A and B are having their initial length in
the ratio 2 : 3 and the co-efficients of linear expansion in
the ratio 3 : 4. When they are heated through the same
temperature difference, the ratio of their linear expansion
is
(a) 3 : 4
(b) 1 : 2
(c) 2 : 3
(d) 4 : 3
Ans. (b)
Sol.
l A  l B
lA 2  A 3
 ;

lB 3  B 4
As lB  lA ; so will bend with B on outer side.
15.
Ans.
Sol.
1   2
Y1
lA lA  A T  lA   A  2 3

  
    1: 2
lB lB  B T  lB   B  3 4
14.
Ans.
A bi-metallic strip is made of two strips A and B having
co-efficients of linear expansion A and B. If A <B,
then on heating the strip will
(a) bend with A on outer side
(b) bend with B on outer side
(c) not bend at all
(d) None of these
(b)
l1
l
 Y2 2
l1
l2
Y1  1T  Y2   2 T
Y1  2 3

  3: 2
Y2 1 2
16.
Ans.
Sol.
Two rods of different materials having coefficients of
thermal expansion 1 and 2 and Young’s moduli Y1 and
Y2 respectively are fixed between two rigid massive walls.
The rods are heated such that these undergo the same
increase in temperature. There is no bending of the rods.
If 1 : 2 = 2 : 3, the thermal stresses developed in the two
rods are equal, provided Y1 : Y2 is
(a) 2 : 3
(b) 1 : 1
(c) 3 : 2
(d) 4 : 9
(c)
Given stress are equal
Which of the following qualities are best suited for a
cooking utensil
(a) high specific heat and low thermal conductivity
(b) high specific heat and high thermal conductivity
(c) low specific heat and low thermal conductivity
(d) low specific heat and high thermal conductivity
(d)
4
THERMAL PHYSICS
Sol.
Heat flow:
Q
T
 KA
t
x
Heat flow:
Q
K
t
Sol.
Ans.
Sol.
4 3
R
3
V
R
3
V
R
V  3T  V
Heat absorbed : Q  msT
17.
V
For cooking heat flow must be higher than heat absorbed
so K must be high and s must be low for cooking.
What should be the lengths of steel and copper rods at
0°C so that the length of the steel rod is 5 cm longer than
the copper rod at any temperature?
 (Steel) = 1.1 × 10–5 °C–1 and
 (Copper) = 1.7 × 10–5 °C–1
(a) 14.17 cm, 9.17 cm
(b) 9.17 cm, 14.17 cm
(c) 28.34 cm, 18.34 cm
(d) 14.17 cm, 18.34 cm
(a)
Let length of copper and steel rods be lcu and lsteel
lsteel – lcu = 5
...(i)
At any temperature T,
lcu  lcu  cu T  lcu  cu  T  0 
lsteel  lsteel steel T  lsteel steel  T  0 
4
V  3T  R 3
3
V  4R 3T
Numerical Value Type Questions
19.
Steel wire of length L at 40°C is suspended from the ceiling
and then a mass m is hung from its free end. The wire is
cooled down from 40°C to 30°C to regain its original length
L. The coefficient of linear thermal expansion of steel is
10–5/°C. Young’s modulus of steel is 1011N/m2 and radius
of the wire is 1 mm. Assume that length (L) >> diameter
9d) of the wire. Then the value of ‘m’ (in kg to nearest,
integer).
Ans. 3.00
Sol.
lcu  lcu cu T
F
L
mg
Y

 Y   
A
L
A
m
AY   
g
lsteel  lsteel steel T
For constant difference in length, change in length must
be same at all temperature.
lsteel  lcu

20.
18.
Ans.
g
 10 –3   1011  10–5  10
10
   3kg
Two rods of different materials having coefficients of
thermal expansion 1 ,  2 and Young’s moduli Y1, Y2
lsteel  1.1 10  lcu  1.7  10
11 lsteel  17 lcu
r 2 Y   
2
lsteel steel T  lcu  cu T
5

respectively are fixed separately between two rigid walls.
The rods are heated such that they undergo the same
increase in temperature. There is no bending of the rods.
5
...(ii)
From equation (i) and (ii)
lsteel = 14.17 cm
lcu = 9.17 cm
The radius of a metal sphere at room temperature T is R,
and the coefficient of linear expansion of the metal is .
The sphere is heated a little by a temperature T so that its
new temperature is (T + T). The increase in the volume
of the sphere is approximately :
(a) 2R T
(b) R2 T
(c) 4R3 T/3
(d) 4R3 T
(d)
If 1 :  2 = 2 : 3; then thermal stress developed in both
Y1
rods would be equal, if Y is
2
Ans.
1.50
Sol.
Thermal stress is given by Y   
Here  is same and 1 :  2  2 : 3
So, Y11  Y2  2 for the stress to remain the same then we
Y1  2 3
have Y    2
2
1
5
THERMAL PHYSICS
21.
As a result temperature rise of 32°C, a bar with a crack at
its center buckles upwards. If the fixed distance L0 is 4 m,
and the coefficient of linear expansion of bar is 25 × 10–6
°C–1. Find the rise x (in cm, to the nearest integer) of the
center.
Sol.
As the track won’t be allowed to expand linearly, the rise
in temperature would lead to developing thermal stress in
track.
 Stress 
y
 Y or   YT
Energy stored per unit volume 
1 2
2 Y
 Energy stored per unit length 

A2
2Y
A
 Y 2 T 2
2
10 –2 1011  10 –10  100
 5J m
2
Calorimetry
24.
Relation between molar and principal specific heat of gases
(a) Cp = Mcp
(b) Cp = M + cp
(c) cp = MCp
(d) Cp = M – cp
Ans. (a)
Sol. dQ = mcpdT
...(i)
where cP = principal specific heat of substance
m = mass of substance
dQ = nCpdT
...(ii)
Where Cp = molar specific heat at constant pressure
n = number of moles
from equation (i) and (ii)
mcpdT = nCpdT

Ans.
8.00
Sol.
2L  L0 1  T   4 1.0008  L  2.0016
By Pythagoras theorem,
 x2 
2.0016  22  x 2  2  1  
8 


22.
x2
 0.0008  x  0.8m
8
Two vessels connected at the bottom by a thin pipe with a
sliding plug contain liquid at 20°C and 80°C respectively.
The coefficient of cubic expansion of liquid is 10–3 K–1.
 h 20 
The ratio of heights columns in the vessel  h  is nearest
 80 
Ans.
Sol.
to which integer?
1.00

23.
Ans.
25.
Because the sliding plug stay in the connecting pipe, the
pressure in both the vessels at he level of hte pipe must be
the same.
h 20 d 20  h 80 d 80 
d 0 1 – 80 
d 0 1 – 20 
h 20 d80

h 80 d 20
 0.94
The area of cross - section of a railway track is 0.01 m2.
The temperature variation is 10°C. Coefficient of linear
expansion of material of track is 10–5/°C. The energy stored
per meter in the track is _____ J/m.
(Young’s modulus of material of track is 1011 Nm–2)
5.00
m
Cp
M
Cp = Mcp
Liquids at temperature 60°C and 20°C, respectively, have
mass ratio 3 : 4 and their specific heats in the ratio 4 : 5. If
the two liquid mixed, the resultant temperature
(a) 70°C
(b) 0°C
(c) 35°C
(d) 40°C
(c)
Let us consider final temperature is T.
Heat given = Heat taken
mc p 
Ans.
Sol.
m1s1T1  m 2 s 2 T2
m1 s1
  60  T    T  20 
m2 s2
3 4
  60  T    T  20 
4 5
180 – 3T = 5T – 100
8T = 280°C
T = 35°C
6
THERMAL PHYSICS
26.
Ans.
Sol.
27.
Heat given to a body which raises its temperature by 1°C
is
(a) water equivalent
(b) thermal capacity
(c) specific heat
(d) temperature gradient
(b)
Thermal capacity
If m is the mass,  is temp. and ‘s’ is specific heat, then
thermal capacity K is given by
(a) K = ms 
(b) K = m 
(c) K 
Ans.
(d)
Sol.
K
ms

...(1)
 Q  ms  T 
...(2)
Sol.
Q  mL
 L 
Q
Since lenght of horizontal line (time)
m
for A is greater than B and heat is supplied at the same rate
therefore QA < QB
Q  mL  L 
Q  msdT  S 
Q
for same time
mdT
dTA  dTB  SA  SB
Q
T
 Q  K  T 
(d)
 LA  LB
30.
Ans.
29.
Sol.
(d) K = ms
From (1) & (2)
K = ms
Dimensions of latent heat are
(a) [M1 L2 T–2]
(b) [M0 L2 T–2]
(c) [M1 L1 T–1]
(d) [M1 L 1 T–2]
(b)
28.
Ans.
Ans.
Sol.
31.
 Q  ML2 T 2   M0 L2 T 2 


M
 m
Equal masses of two liquids A and B contained in vessels
of negligible heat capacity are supplied heat at the same
rate. The (temperature vs time) graphs for the two liquids
are shown in figure. If S represents specific heat and L
represents latent heat of liquid, then
If mass-energy equivalence is taken into account, when water
is cooled to form ice, the mass of water should
(a) increase
(b) remain unchanged
(c) decrease
(d) first increase then decrease
(c)
According to mass energy equivalence, when water is
cooled to form ice, it lose some energy, hence its mass
will decrease accordingly.
Three copper blocks of masses M 1, M 2 and M 3 kg,
respectively are brought into thermal contact till they reach
equilibrium. Before contact, they were at temperatures T1,
T2, T3 (T1 > T2 > T3). Assuming there is no heat loss to the
surroundings, the equilibrium temperature T is :
(s is specific heat of copper)
(a) T 
T1  T2  T3
3
(b) T 
M1T1  M 2 T2  M 3 T3
M1  M 2  M 3
(c) T 
M1T1  M 2 T2  M 3 T3
3  M1  M 2  M 3 
(d) T 
M1T1s  M 2 T2 s  M 3 T3s
M1  M 2  M 3
Ans.
(b)
Sol.
 Q  system = 0
Q1  Q 2  Q3  0
M1S  T  T1   M 2S  T  T2   M3S  T  T3   0
M1T  M1T1  M 2 T  M 2 T2  M3 T  M 3 T3  0
(a) SA > SB ; LA < LB
(c) SA < SB ; LA < LB
(b) SA > SB ; LA > LB
(d) SA < SB ; LA > LB
T
M1T1  M 2 T2  M 3T3
M1  M 2  M 3
7
THERMAL PHYSICS
Numerical Value Type Questions
32.
In two experiments with a continuous flow - calorimeter
to determine the specific heat capacity of a liquid, an input
power of 16 W produced a rise of 10 K in the liquid.
When the power was doubled, the same temperature rise
was achieved by making the rate of flow of liquid three
times faster. Find the power lost (in W) to the surroundings
in each case, assuming that heat lost to surroundings
depends on temp of liquid and atmospheric temp. only.
Ans. 8.00
Sol.
35.
cal
A 50 gm lead bullet (specific heat 0.020 gC ) is initially
Ans.
Sol.
at 30°. It is fired vertically upwards with a speed of 84 m/
s. On returning to the starting level, it strikes a slab of ice
kept at 0°C. (A × 100) mg of ice is melted due to this. Find
the value of ‘A’. (Take:- Lice = 80 cal/gm and 1 cal = 4.2 J)
9.00
Heat lost = Heat gain
Let power lost to surrounding be Q.
1
mv 2  ms  mice Lice
2
 dm 
16 – Q  
 S 10 
 dt 
1
 50  10 –3  84  84  4.2  50  0.02  30 
2
 4.2  mice  80 
 dn 

and 32 – Q  3  s  S 10  
dt



On simplification
m ice  0.9 gm  900 mg
32 – Q

 3  Q  8W
16 – Q
33.
Ans.
Sol.
34.
 A9
A copper ball of mass 100 gm is at a temperature T. It is
dropped in a copper calorimeter of mass 100 gm filled
with 170 gm of water at room temperature. Subsequently,
the temperature of the system is found to be 75° C. (given
room temperature = 30°C, specific heat of copper = 0.1
cal/gm°C. Temperature T (in °C) is:
885.00
100 × 0.1 × (t – 75) = 100 × 0.1 × 45 + 170 × 1 × 45
100t – 750 = 450+ 7650
10t = 8850
t = 885°C
Two liquids A and B are at 32° C and 24°C respectively.
When mixed in equal masses the temperature of the mixture
36.
Ans.
Sol.
 SA 
if found to be 28°C. The ratio of specific heats  S  is
 B
Ans.
1.00
Sol.
Temperature of mixture mix 
L
 270
S
37.
A SA  BSB
SA  SB
Ans.
Sol.
32  SA  24  SB
 28 
SA  Ss
 28SA  28SB  32SA  24SB 
SA 1

SB 1
A liquid at 30°C is poured very slowly into an open
Calorimeter that is at temperature of 110°C. The boiling
temperature of the liquid is 80°C. It is found that the first
5 gm of the liquid completely evaporated. After pouring
another 80 gm of the liquid the equilibrium temperature is
found to be 50 °C. The ratio of the Latent heat of the liquid
to its specific heat will be ____ °C. (Neglect the heat
exchange with surrounding)
270.00
Calorimeter is open and after evaporation liquid escapes
5 × S × (80 – 30) + 5L = W × (110 – 80)
80 × S × (50 – 30) = W × (80 – 50)
5 × S × 50 + 5L = 80 × S × 20
5L= 1350 S
How much heat is required to convert 8.0 g of ice at –
15°C to steam at 100°C? (Given, Cice = 0.53 cal/g–°C, Lf
= 80 cal/g and Lv = 539 cal/g, and cwater = 1 cal/g–°C)
Give the answer in kcal correct to one decimal place.
5.8
Q1  mcice  Tf – Ti 
8
THERMAL PHYSICS
  8.0  0.53   0 –  –15    63.6 cal
39.
Q 2  mL f   8 80   640cal
Q3  mcwater  Tf – t i 
  8.0  0.1 100 – 0  800cal
Q 4  mL v   8.0  539   4312 cal
 Net heat required, Q  Q1  Q 2  Q3  Q 4
Ans.
Sol.
 5815.6cal
38.
Ans.
Sol.
Ti   227  273  500 K
The temperature of equal masses of three different liquids
A, B and C are 12°C, 19°C and 28°C respectively. the
temperature when A and B are mixed is 16° C and when B
and C are mixed it is 23°C. What should be the temperature
when A and C are mixed (in °C, correct to two decimal
places)?
20.26
Let m be the mass of each liquid and SA, SB, SC specific
heats of liquids A, B and C respectively. When A and B
are mixed. The final temperature is 16°C.
 Heat gained by A = Heat lost by B
And final temperature of container,
Tf   27  273  300 K
Now, heat gained by the ice cuibe = heat lost by the
container
  0.1  8  10 4    0.1 103   27 
 –m 
i.e., ms B  23 –19   msC  28 – 23
4
i.e.,SC  SB
5
300
Ans.
After substituting the values of A and B and the proper
limits, we get
m=0.495 kg
m grams of steam at 100°C is mixed with 200g of ice at its
melting point in a thermally insulated container. If it
produces liquid water at 40°C [heat of vaporization of
water is 540 cal/g and heat of fusion of ice is 80 cal/g], the
value of m (in grams) is _______.
40.00
Sol.
Here, heat absorbed by ice = mice Lf  mice Cw  40 – 0 
40.
...(2)
4 4
16
From Eqs. (i) and (ii), SC   SA  SA
5 3
15
 A  BT  dT
BT 2 


–m
AT

or 10700

2  500

...(1)
When B and C are mixed.
Heat gained by B = Heat lost by C
300
500
i.e.ms A 16 –12   ms B 19 –16 
4
i.e.s B  SA
3
An ice cube of mass 0.1 kg at 0°C is placed in an isolated
container which is at 227°C. The specific heat s of the
container varies with temperature T according to the
empirical relation s = A + BT, where A = 100 cal/kg-K and
B = 2 × 10–2 cal/kg-K2. If the final temperature of the
container is 27°C, determine the mass of the container (in
kg, correct to one decimal place). (Latent heat of fusion
for water = 8 × 104 cal/kg, specific heat of water = 103 cal/
kg-K).
0.5
Let m be the mass of the container.
Initial temperature of container,
When A and C are mixed, let the final temperature be 
Heat released by steam  msteam L v  msteam Cw 100 – 40 
Heat gained by A = Heat lost by C
Heat absorbed = heat released
 ms a   –12   msC  28 –  
mice Lf  mice Cw  40 – 0   msteam L v  msteam C w 100 – 40 
i.e.  – 12 
16
 28 –  
15
By solving, we get

628
 20.26C
31
 200  80 cal g  200 1cal g  C   40 – 0 
 m  540cal g  540  1cal g C  100 – 40 
 200 80   40 1  m 540   60 1
m  40 g
9
THERMAL PHYSICS
41.
1 kg ice at –10°C is mixed with 0.2 kg of steam at 200°C.
If final temperature (°C) of mixture at equilibrium is
K1
T

K 2 100  T 
 255 
Teq  
 , then fill the value of x.
 x 
Ans.
Sol.
Latent heat of fusion of ice = 80 cal/gram, latent heat of
vaporization of water = 540 cal/gram, specific heat of water
= 1 cal/gm K
specific heat capacity of ice  specific heat of steam =
0.5 cal/gram-K.
6.00
44.
1
1  10  1 80  1 Teq
2
1
 0.2   100  0.2  540  0.2  1100 – Teq 
2
Teq 
Ans.
Sol.
265
6
Heat Transfer
Heat Transfer, Conduction and Convection
42.
Dimension of co-efficient of thermal conductivity are
(a) [L0M1T–3K–1]
(b) [L1M1T–3K–1]
1
1 –3
(c) [L M T K]
(d) [L1M–1T–2K–1]
Ans. (b)
Sol.
2
T

3 100  T 
45.
Q
dT
 KA
t
dx
K1 K 2
(a) K  K
 ML2 T 2   L
K    2
 L   K  T 
1
Ans.
Sol.
Two metal rods A and B of equal lengths and equal cross
sectional areas are joined end-to-end. The co-efficients of
thermal conductivity of A and B are in the ratio 2 : 3. When
the free end of A is maintained at 100°C and the free end
of B is maintained at 0°C, the temperature of the junction
is
(a) 30°C
(b) 40°C
(c) 50°C
(d) 60°C
(b)
Let temperature of the junction be T.
 K1 : K2 = 2 : 3
Q K1A 100  T  K 2 A  T  0 


t
L
L
k1
k2
l
l
2
K1  K 2
(c)
2
 K    ML T 3   K –1 
43.
200 – 2T = 3T
5T = 200
T = 40°C
In steady state
(a) temperature does not change with time
(b) all parts of the body are at same temperature
(c) there is no flow of heat
(d) all of the above
(a)
- In steady state, temperature of the region is constant w.r.t.
to time.
- The temperature may be different for different parts of
same body.
- The heat flow will not be zero but equal in both direction.
Two metallic plates of equal thicknesses and thermal
conductivities K1 and K2 are put together face to face and
a common plate is constructed, figure. The equivalent
thermal conductivity will be:
Ans.
2K1 K 2
(b) K  K
1
(d)
2
K  K 
2
1
2K1 K 2
(b)
Sol.
l
 R1
K1A
Req = R1 + R2
2l
l
l


K eq A K1 A K 2 A
2
l
l


K eq K1 K 2
K eq 
2K1K 2
K1  K 2
l
K2A
2 3/ 2
2
 R2
10
THERMAL PHYSICS
46.
Two metallic plates of equal lengths and thermal
conductivities k1 and k2 are put together such that their
ends coincide. If their cross-sectional areas are the same,
then the equivalent thermal conductivity of the combination
will be :
k1 k 2
(a) k  k
1
k1
A
k2
2
k1  k 2
2
(c)
Ans.
A
 2 – 100 
 312.5C
 Temperature gradient on 100°C side

2k1k 2
(b) k  k
1
(d)
2
25l
25  1

2KA 2  400  10 –4
206.25 –100
12
 212.5 C m
48.
Three rods made of same material and having same cross
- section have been joind as shown in the figure. Each rod
is of same length. The left end is kept at 0°C and both the
right ends are kept at 90°C. The temperature of the junction
of the three rods (in degree Celsius) will be
Ans.
60.00
Sol.
Let the junction temperature be  in equilibrium
k1 k 2
(c)
Sol.
1
1
1


R e q R1 R 2
2k eq A
l

k1A k 2 A

l
l
 dQ 
 dQ 
 dQ 

 
 

 dt  rod1  dt  rod 2  dt rod 3
2 K eq  k1  k 2
Ans.
k1  k 2
2
The ends of a copper rod of length 1m and area of cross
2
section 1 cm are maintained at 0°C and 100°C. At the
centre, power is supplied at a constant rate of 25 J/s. The
temperature gradient on higher temperature side of the rod
in steady state (in °C/m) will be
(K = 400 J/m-K-s)
212.50
Sol.
Let  be the temperature at the middle in steady state.
K eq 
47.

49.
 dQ   dQ 
Then, P   dt    dt 

1 

KA  0 –  
2KA
 2 – 100 
I
Ans.
KA  90 –  


KA  90 –  

   60
Two rectangular blocks, having identical dimensions, can
be arranged either in configuration I or in configuration II
as shown in the figure. One of the blocks has thermal
conductivity K and other 2K. The temperature difference
between the ends along the x-axis is the same in both the
configurations. It takes 9 s to transport certain amount of
heat from the hot end in the configuration I. The time (in
seconds) to transport the same amount of heat in the
configuration II is:
KA
KA
or 25  1 2   –100   1 2   – 0 



2.00
11
THERMAL PHYSICS
Sol.
51.
Q T

i (R = thermal resistance)
t
R

Q
t

T R
...(1)
Since Q and T are the same 
t
constant. Hence
R
The temperature of the two outer surface (end surfaces
shown) of a composite slab, consisting of two materials
having coefficients of thermal conductivity K and 2K,
thickness ‘x’ and ‘4x’ are T2 and T1(T2 > T1). The rate of
heat transfer through the slab in steady state is
 A  T2 – T1  K  1

 , where ‘f’ is equal to
x

f
t1
T
 2
R1 R 2
(all quantities measured in S.I unit)
Configuration 1: R 1 
1
1
3


KA 2KA 3KA
1
2KA KA 3KA
Configuration 2: R      
2

3KA
Using these in (i)
 R2 
50.
t
9
 2  t  2s
3

2KA 3KA
Figure shows a metal rod of uniform cross section area A,
with variable thermal conductivity given by
  
k  x   k 0 sec 
x  . If the end A is maintained at
 6L 
Ans.
Sol.
3.00
Let temperature of common interface be T°C. Rate of heat
flow
H
Q KAT

t

 Q  2KA  T – Ti 
 H1    
4x
 t 1
temperature T0 , the rod carries a thermal current I0 (from
KA  T2 – T 
Q
and H 2    
x
 t 2
I0 L

B to A) in steady state and k AT  3 ; find the
0
1
In steady state, the rate of heat flow should be same in
whole system i.e.
temperature of the end B of the rod. Let’s say this
temperature is kT0, find integer value k.
H1  H 2

or
2KA  T – Ti 
4x

KA  T2 – T 
x
T – Ti
 T2 – T
2
or T – T1  2T2 – 2T
Ans.
Sol.
or T 
2.00
I0 
k0A
dT


 dx
cos 
x
 6L 
6I L
  
 T  T0  0 sin 
x
k 0 A
 6L 
 T  2T0 at x  L
2T2  T1
3
...(1)
Hence heat flow from composite slab is
H

KA  T2 – T 
x
KA
 T2 – T1 
3x

2T  T1 
KA 
T2 – 2

x 
3 
...(2)
12
THERMAL PHYSICS
52.
Two conducting cylinders of equal length but different radii
are connected in series between two heat baths kept at
temperature T1 = 300 K and T2 = 100 K, as shown in the
figure. The radius of the bigger cylinder twice that of the
smaller one and the thermal conductivities of the materials
of the smaller and the larger cylinders are K1 and K2
respectively. If the temperature at the junction of the two
cylinders in the steady state is 200 K, then K1/K2 =
_______.
Sol.
Ans.
Thermal radiations, are electromagnetic waves Thus
radiation travels with speed of light.
Heat is transferred most rapidly by the process of
(a) Conduction
(b) Convection
(c) Radiation
(d) Combustion
(c)
Sol.
Heat radiation  T 4 i.e. Fastest.
56.
Two circular discs A and B with equal radii are blackened.
They are heated to same temperature and then cooled under
identical conditions. What inference do you draw from
their cooling curves as shown in figure?
55.
B
A
R
4.00
Since rate of heat flow is same, we can say
( – 0)
K1 A 2

4
K 2 A1
Radiation
53.
A sphere, a cube and a thin circular plate, all of same
material and same mass, are initially heated to same high
temperature. Choose the correct statement.
(a) The plate will cool fastest and cube the slowest.
(b) The sphere will cool fastest and cube the slowest.
(c) The plate will cool fastest and sphere the slowest.
(d) The cube will cool fastest and plate the slowest.
Ans. (c)
Sol. For same mass and density, surface area of plate will be
largest and surface area of sphere will be smallest. The
more surface area the more is rate of cooling.
54.
Velocity of heat radiation v as related to the velocity of
light c is
(a) v > c
(b) v = c
(c) v < c
(d) no definite relation
Ans. (b)
Ans.
Sol.
57.
(a)
(b)
Temperature

L1
L2

K1A1 K 2 A 2
Time
(c)
(d)
Time
Ans.
Sol.
(c)
In convection
dQ
  T2  T1 
dt
dT 
Time
1
t
Temperature
R1  R 2 
(a) A and B have same specific heats
(b) specific heat of A is less
(c) specific heat of B is less
(d) nothing can be said
(c)
As the slope of A is less than slope of B. so rate of cooling
will be slow in A so it has larger specific heat
The temperature of coffee in a cup with time is most likely
given by the curve in figure.
Temperature
300 – 200 200 –100

R1
R2
Temperature
Ans.
Sol.
Time
13
THERMAL PHYSICS
A block of steel heated to 100°C is left in a room to cool.
Which of the curves shown in the figure represents the
decrease of temperature with time?
Temperature
58.
B
Sol.
dT
 K  T  T0 
dt
D
C
K
A
62.
Ans.
(a) A
(c) C
(a)
Sol.
dT 
59.
A body cools from 50°C to 46°C in 5 minutes and to 40°C
in the next 10 minutes. The surrounding temperature is :
(a) 30°C
(b) 28°C
(c) 36°C
(d) 32°C
(b)
Sol.
(b) B
(d) D
1
t
dT / dt 0.2C / min

 0.01 / min
T  T0
20C
Ans.
Newton’s law of cooling is applicable for
(a) Any excess of temperature over the surrounding
(b) Small excess of temperature over the surrounding
(c) Large excess of temperature over the surrounding
(d) Very large excess of temperature over the surrounding
(b)
Newton’s law of cooling of only for small excess of
temperature otherwise value of K will change law
Newton’s law of cooling leads us to the following
expression,
(a) ( – 0) = Kt + C
(b) log ( – 0) = Kt + C
(c) log  = Kt + C
(d)  = K0 + C
(b)
Sol.
From
Time
Ans.
dT
 K  T  T0 
dt
Ans.
Sol.
63.
msdT1
 K  TAV1  TS 
t1
dQ
 K    0 
dt
dQ
       K d t
0
ms  4
 K  48  TS 
5
64.
msdT2
 K  TAV 2  TS 
t2
ms  6
 K  43  TS 
10
60.
Ans.
Sol.
61.
Ans.
log    0   Kt  C
...(i)
Ans.
Two spheres of the same material have radii 1 m and 4 m
and temperatures 4000 K and 2000 K respectively. The
ratio of the energy radiated per second by the first sphere
to that by the second is
(a) 1 : 1
(b) 16 : 1
(c) 4 : 1
(d) 1 : 9
(a)
Sol.
E  AT 4
...(ii)
TS = 28°C
Newton’s law of cooling is used in laboratory for
determining:
(a) Specific heat of gases (b) Specific heat of liquids
(c) Latent heat of gases
(d) Latent heat of liquids
(b)
Newton’s law of cooling is used to find specific heat of
liquids.
If the rate of change of temperature is 0.2°C/ min and
excess temperature of a body over surrounding is 20°C,
the constant of proportionality is
(a) 0.1
(b) 0.01
(c) 1
(d) 0.001
(b)
E1 4R12 T14 1  4000 


 1:1
E 2 4R 22 T24  4  2  2000 4
2
65.
Ans.
Sol.
66.
4
Which of the following is more close to a black body ?
(b) Black board paint
(b) Green leaves
(c) Black holes
(d) Red roses
(c)
Black hole is black body.
Infrared radiations are detected by
(a) spectrometer
(b) pyrometer
(c) nanometer
(d) photometer
14
THERMAL PHYSICS
Ans.
Sol.
67.
(b)
Infrared radiations are detected by pyrometer.
A liquid in a beaker has temperature (t) at time t and 0 is
temperature of surroundings, then according to Newton’s
law of cooling, the correct graph between loge ( – 0) and
t is :
Sol.
70.
Time of cooling increase with decrease in temperature.
Assuming the sun to be a spherical body of radius R at a
temperature of T K, evaluate the total radiant power,
incident on earth, at a distance r from the sun, where r0 is
radius of earth
(a)
4r02 R 2 T 4
r2
(b)
r02 R 2 T 4
r2
(c)
r02 R 2 T 4
4r 2
(d)
R 2 T 4
r2
Ans.
(b)
Sol.
Power emitted by sun  AT 4  4R 2 T 4
Energy per area per unit time at earth

4R 2 T 4 R 2 T 4

4r 2
r2
Energy power incident on earth.

71.
Ans.
Sol.
R 2 T 4
 r02 
r2
A piece of metal is heated to temperature  and then
allowed to cool in a room which is at temperature 0. The
graph between the temperature T of the metal and time t
will be closed to :
T
(d)
According to Newton’s law of cooling
(a) T
O
In    0   kt  C
68.
So, graph in option - (d) is correct.
The maximum wavelength of radiations emitted at 900 K
is 4 m. What will be the maximum wavelength of
radiations emitted at 1200K:
(a) 3 m
(b) 0.3 m
Ans.
(c) l m
(a)
Sol.
1T1   2 T2
(d) none of these
4m  900   2  1200
 2  3 m
69.
Ans.
A bucket full of hot water is kept in a room and it cools
from 75°C to 70°C in T1 minutes, from 70°C to 65°C in T2
minutes and from 65°C to 60°C in T3 minutes then
(a) T1 = T2 = T3
(b) T1 < T2 > T3
(c) T1 > T2 > T3
(d) T1 < T2 < T3
(d)
(b)
t
O
T
T
(c)
(d)
O
Ans.
Sol.
t
t
O
t
(c)
According to Newton’s law of coding temperature decrease
exponentially with time and temperature becomes equal
to room temp. at infinity time.
15
THERMAL PHYSICS
72.
Ans.
If the temperature of the sun were to increase from T to 2T
and its radius from R to 2R, then the ratio of the radiant
energy received on earth to what it was previously, will be
(a) 4
(b) 16
(c) 32
(d) 64
(d)
  4R  T
2
Sol.
Energy received by earth 
 4r 
2
4
75.
Ans.
Sol.
T
 K  T – T0 
t
 r  t
2
0
For cooling from 60°c to 40°C

So, E  R 2 T 4
E f  2R   2T 

 64
So,
EL
R 2T4
2
73.
A body cools in 7 minutes from 60°C. What will be its
temperature (in °C) after the next 7 minutes? The
temperature of surroundings in 10°C.
28.00
Rate of cooling
60 – 40
20
1
 60  40

 K
–10   K 

7
7  40 14
 2

For cooling from 40°C to T
4

In the figure, the distribution of energy density of the
radiation emitted by a black body at a given temperature
is shown. The possible temperature of the black body is
3
n × 10 K. Write n as the nearest integer.
–3
(Take:- b = 2.898 × 10 mK)
40 – T
40 – T 1  40  T – 20 
 40  T

 K
–10  
 

7
7
14 
2
 2


 160 – 4T  20  T
 140  5T  T 
140
 23C
5
 T  28C
76.
A solid cube of side a, density d and specific heat ‘s’ is at
temperature 400 K. It is placed in an ambient temperature
of 200 K.
–3
3
3
Take: a = 0.9 m, d = 4.8 ×10 kg/m , s = 2.0 × 10 J/kg K.
Stefan’s constant   6  10 –8 W K 4 m 2 . Consider the
Ans.
2.00
Ans.
cube to be a black body. If the time for the temperature of
the cube to drop by 5 K is 1000x second, find x in nearest
integer.
5.00
Sol.
 m T  b where b  2.898  10 –3 mK
Sol.
  T 4 – TS4    6a 2  t   d.a 3  s.T
T
74.
b
2.898  10–3

 1932K  1.9  103 K
m
1.5  10–6
So, nearest integer value of n will be 2.
Two starts (say S1 and S2) emit their maximum intensity
–6
–7
radiations at wavelengths 10 m and 0.4 × 10 m
t
77.
 T2 
respectively. The ratio of their surface temperature  T 
 1
Ans.
Sol.
is
25.00
T

10 –6
 m T  constant  2  1 
 25
T1  2
0.4  10 –7
Ans.
Sol.
dos T
6.  T 4 – Ts4 

4.8 103  0.9  2.0  103  5
6  6 10 –8   400 4 – 200 4 
 5000s.
x5
Two planets X and Y revolving around the sun is circular
orbits, have temperature of their surfaces as T1 and T2. If
their distance from the sun are in the ratio of 1 : 4, then
find T1 : T2. Assume the planets to be in the steady state,
and the sun and the planets to be black-bodies. Neglect
the energy exchange between the two planets.
2.00
dQsum
 eA s T 4
dt
dQ rod.

dt
dQsum
dt
4d 2
R 2p 
16
THERMAL PHYSICS
sol.
dQ rod.
In steady state
dt

From Newton’s law of cooling
dT
 k  T – Tsurrounding 
dt
dQ
radiated  4R 2p tP 2
dt
...(1)
Integration equation (1)
Tf – Tsurrounding
14
 As 
Tp  T  
2 2 
 16 d 
Ti – Tsurrounding
 e – kt
...(2)
From given data in question in first 7 minutes
1
2
p
Tp d  constant.
Tf  40C, Ti  60C, Tsurrounding  10C
 T1 : T2  2: 1
78.
A glass of boiling water at 100°C cools down to 90°C in
10 minutes when placed in surrounding temperature of
30°C. It will cool down to 80°C in an additional time
t  10
...(3)
Now the temperature decrease further due to temperature
gradient from given data in question in next 7 minutes
nx – n  x –1
Tf  ?, Ti – 40C, Tsurrounding  10C
n x  1 – nx minutes. Assume Newton’s law of


Ans.
cooling to hold. Find x
6.00
Sol.
T – T0   Ti – T0  e – kt  60  70e –10k  k 
applying equation (1)
Tt – 10
 e– k 7
40 – 10
1 7
n  
10  6 
Tf –10 3

40 – 10 5
T – T0   Ti – T0  e – kt  50  60e – kt
6
6
 kt  n    t  10 5
7
5
n
6
79.
Ans.
A body cools in 7 minutes from 60°C to 40°C. What will
be its temperature in °C after next 7 minutes? The temperature of surrounding is 10°C.
28.00
...(4)
substituting value of e – k 7 , from equation (3)
For further cooling,
n
40 – 10
 e– k 7
60 – 10
Tf  10  18  28C
80.
Ans.
Sol.
The emissive power of a black body at T = 300 K is 460
W/m2. Consider a body B of area A = 100 cm2 , coefficient
of reflectivity r = 0.3 and coefficient of transmission t =
0.5. Its temperature is 300 K. Then the power radiated by
B is ____ W.
0.92
Since, e = a = 0.2
(Since, a = (1–r–t) = 0.2 for the body B)
E = (460) (0.2) = 92 W/m2
P = EA = 92 × 100 × 10–4 = 0.92W