NCERT Solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT
SOLUTIONS
CLASS - 11th

aglase .co

Class : 11th
Subject : Physics
Chapter : 11
Chapter Name : Thermal Properties Of matter

Q11.1 The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these
temperatures on the Celsius and Fahrenheit scales.

Kelvin and Celsius scales are related as:

Tc = TK − 273.15 … (i)

Celsius and Fahrenheit scales are related as:

9
TF = TC + 32
5

Celsius and Fahrenheit scales are related as:

9
TF = TC + 32
5

For neon:

Tk =24.57K

∘
∴ Tc = 24.57 − 273.15 = −248.58 C

9
TF = TC + 32
5

9
= (−248.58) + 32
5

∘
= 415.44 F

For carbon dioxide:

Tk = 216.55K
∘
∴ Tc = 216.55 − 273.15 = −56.60 C

9
TF = (TC ) + 32
5

9
= (−56.60) + 32
5

∘
= −69.88 C

Page : 299 , Block Name : Exercise

Q11.2 Two absolute scales A and B have triple points of water de ned to be 200 A and 350 B. What is the
relation between TA and TB ?

Triple point of water on absolute scaleA, T1 = 200A

Triple point of water on absolute scale B, T2 = 350B

Triple point of water on Kelvin scale, Tκ = 273.15K

The temperature 273.15K on Kelvin scale is equivalent to 200A on absolute scale A .

T1 = Tk

200A = 273.15K

273.15
∴ A =
200

The temperature 273.15K on Kelvin scale is equivalent to 350B on absolute scale B.

T2 = Tκ

350B = 273.15

273.15
∴ B =
350

TA is triple point of water on scale A .

TB is triple point of water on scale B.
273.15 273.15
∴ × TA = × TB
200 350

200
TA = TB
350

Therefore, the ratio TA : TB is given as 4 : 7.

Page : 299 , Block Name : Exercise

Q11.3 The electrical resistance in ohms of a certain thermometer varies with temperature according to the
approximate law : R = Ro [1 + α (T – To )] The resistance is 101.6 Ω at the triple-point of water 273.16 K, and
165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4
Ω?

It is given that:

R = R0 [1 + a (T − T0 )] … (i)

Where,

R0 and T0 are the initial resistance and temperature respectively

R and T are the final resistance and temperature respectively

a is a constant

At the triple point of water, T0 = 273.15K

Resistance of lead, R0 = 101.6Ω

At normal melting point of lead, T = 600.5k

Resistance of lead, R = 165.5Ω

Substituting these values in equation (i), we get:

R = R0 [1 + α (T − T0 )]

165.5 = 101.6[1 + α(600.5 − 273.15)]

1.629 = 1 + α(327.35)

0.629 −3 −1
∴ α = = 1.92 × 10 K
327.35

For resistance, R1 = 123.4Ω

R1 = R0 [1 + α (T − T0 )]

Where, T is the temperature when the resis tan ce of lead is 123.4Ω

−3
123.4 = 101.6 [1 + 1.92 × 10 (T − 273.15)]

−3
1.214 = 1 + 1.92 × 10 (T − 273.15)

0.214
= T − 273.15
−3
1.92×10

∴ T = 384, 61K

Page : 299 , Block Name : Exercise

Q11.4 Answer the following :
(a) The triple-point of water is a standard xed point in modern thermometry.
Why ? What is wrong in taking the melting point of ice and the boiling point of water as standard xed points

(as was originally done in the Celsius scale) ?
(b) There were two xed points in the original Celsius scale as mentioned above which were assigned the
number 0 °C and 100 °C respectively. On the absolute scale, one of the xed points is the triple-point of
water, which on the Kelvin absolute scale is assigned the number 273.16 K. What is the other xed point on
this (Kelvin) scale ?
(c) The absolute temperature (Kelvin scale) T is related to the temperature tc on the Celsius scale by tc = T –
273.15 Why do we have 273.15 in this relation, and not 273.16 ?
(d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal
to that of the Fahrenheit scale ?

(a) The triple point of water has a unique value of 273.16K . At particular values of

volume and pressure, the triple point of water is always 273.16K . The melting point of

ice and bolling point of water do not have particular values because these points depend

on pressure and temperature.

(b) The absolute zero or O K is the other xed point on the Kelvin absolute scale.
(c) The temperature 273.16K is the triple point of water. It is not the melting point of

∘
ice. The temperature 0 C on Celsius scale is the melting point of ice. Its corresponding

value on Kelvin scale is 273.15K .

Hence, absolute temperature (Kelvin scale) T , is related to temperature tc′ on Celsius

tc = T − 273.15

(d) Let TF be the temperature on Fahrenheit scale and Tκ be the temperature on

absolute scale. Both the temperatures can be related as:

TF −32 TK −273.15
=
180 100

Let TF be the temperature on Fahrenheit scale and Tκ1 be the temperature on absolute
1

scale. Both the temperatures can be related as:

Tr1 −32 Tk1 −273.15
=
180 100

It is given that:

Tk1 − Tk = 1K

Subtracting equation (i) from equation (ii), we get:

TFI −TF TKI −TK 1
= =
180 100 100
1×180 9
TF1 − TF = =
100 5

Triple point of water = 273.16K

¿Triple point of water on absolute scale = 273.16 × 9

5
= 491.69

Page : 299 , Block Name : Exercise

Q11.5 Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following
observations are made :

(a) Triple point of water, T = 273.16K .

5
At this temperature, pressure in thermometer A, PA = 1.250 × 10 Pa

Let T1 be the normal melting point of sulphur.

5
At this temperature, pressure in thermometer Ar P1 = 1.797 × 10 Pa

According to Charles' law, we have the relation:
PA P1
=
T T1

5
P1 T 1.797×10 ×273.16
∴ T1 = = 5
PA 1.250×10

= 392.69K

Therefore, the absolute temperature of the normal melting point of sulphur as read by

thermometer A is 392.69K .

Therefore, the absolute temperature of the normal melting point of sulphur as read by

thermometer A is 392.69K .

5
At triple point 273.16K , the pressure in thermometer B, PB = 0.200 × 10 Pa

5
At temperature T1 , the pressure in thermometer B, P2 = 0.287 × 10 Pa

According to Charles' law, we can write the relation:
PB P1
=
T T1

5 3
0.200×10 0.287×10
=
273.16 T1

5
0.287×10
∴ T1 = 5
× 273.16 = 391.98K
0.200×10

Therefore, the absolute temperature of the normal melting point of sulphur as read by

thermometer B is 391.98K .

(b) The oxygen and hydrogen gas present in thermometers A and B respectively are not

perfect ideal gases. Hence, there is a slight difference between the readings of

thermometers A and B .

To reduce the discrepancy between the two readings, the experiment should be carried

under low pressure conditions. At low pressure, these gases behave as perfect ideal

gases.

Page : 300 , Block Name : Exercise

Q11.6 A steel tape 1m long is correctly calibrated for a temperature of 27.0 °C. The length of a steel rod
measured by this tape is found to be 63.0 cm on a hot day when the temperature is 45.0 °C. What is the actual
length of the steel rod on that day ? What is the length of the same steel rod on a day when the temperature
is 27.0 °C ? Coef cient of linear expansion of steel = 1.20 × 10 K
–5 –1

∘
Length of the steel tape at temperature T = 27 C, I = 1m = 100cm
∘
At temperature T1 = 45 C, the length of the steel rod, l1 = 63cm

−5 −1
coefficient of linear expansion of steel, a = 1.20 × 10 K
n ∘
Let l2 be the actual length of the steel rod and I be the length of the steel tape at 45 C .
′
l = l + αl (T1 − T )
′ −3
∴ l = 100 + 1.20 × 10 × 100(45 − 27)

= 100.0216cm
∘
Hence, the actual length of the steel rod measured by the steel tape at 45 C can be

calculated as:
100.0216
l2 = × 63
100

∘ ∘
Therefore, the actual length of the rod at 45.0 C is 63.0136cm. Its length at 27.0 C is

63.0cm.

Page : 300 , Block Name : Exercise

Q11.7 A large steel wheel is to be tted on to a shaft of the same material. At 27 °C, the outer diameter of the
shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry
ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coef cient of linear
expansion of the steel to be constant over the required temperature range : αsteel = 1.20 × 10 K –5 –1

∘
The given temperature, T = 27 C can be written in Kelvin as:

27 + 273 = 300K

Outer diameter of the steel shaft at T , d1 = 8.70cm

Dlameter of the central hole in the wheel at T , d2 = 8.69cm

−5 −1 −1
coefficient of linear expansion of steel, a steel = 1.20 × 10 K K

After the shaft is cooled using 'dry ice, its temperature becomes T1 .

The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 − 8.70

= −0.01cm

Temperature T1, can be calculated from the relation:

Δd = d1 a steel (T1 − T )

−5
0.01 = 8.70 × 1.20 × 10 (T1 − 300)

(T1 − 300) = 95.78

∴ T1 = 204.21K

= 204.21 − 273.16
∘
= −68.95 C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is —690C.

Page : 300 , Block Name : Exercise

Q11.8 A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 °C. What is the change in
the diameter of the hole when the sheet is heated to 227 °C? Coef cient of linear expansion of copper = 1.70
× 10 K
–5 –1

∘
Initial temperature, T1 = 27.0 C

Diameter of the hole at T11 , d1 = 4.24cm
∘
Final temperature, T2 = 227 C

Diameter of the hole at T2 = d2

−5 −1
Co-efficient of linear expansion of copper, acu = 1.70 × 10 K

For co-efficient of superficial expansion β , and change in temperature ΔT , we have the

relation:
Change in area (ΔA)
= βΔT
Original area (A)
2 2
d d
2 1
(π −π )
4 4
ΔA
=
2 A
d
1
(π )
4

2 2
d −d
ΔA 2 1
∴ =
2
A d
1

But β = 2α
2 2
d −d
2 1
∴ = 2αΔT
2
d
1

2
d
2
− 1 = 2α (T2 − T1 )
2
d
1
2
d
2 −5
= 2 × 1.7 × 10 (227 − 27) + 1
(4.24)2

2
d = 17.98 × 1.0068 = 18.1
2

∴ d2 = 4.2544cm

Change in diameter = d2 − d1 = 4.2544 − 4.24 = 0.0144cm

−2
Hence, the diameter increases by 1.44 × 10 cm.

Page : 300 , Block Name : Exercise

Q11.9 A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire
is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-
ef cient of linear expansion of brass = 2.0 × 10 K ; Young’s modulus of brass = 0.91 × 10 Pa –5 –1 11

∘
Initial temperature, T1 = 27 C

Length of the brass wire at T1, /I = 1.8m

∘
Final temperature, T2 = −39 C

−3
Diameter of the wire, d = 2.0mm = 2 × 10 m

Tension developed in the wire = F

−5 −1
coefficient of linear expansion of brass, a = 2.0 × 10 K

11
Young's modulus of brass, Y = 0.91 × 10 Pa

Young's modulus is given by the relation:

F

Stress A A
Y = = =
Strain Strain ΔL

L

F ×L
ΔL =
A×Y

Where,

F = Tension developed in the wire

A = Area of cross-section of the wire.

ΔL = Change in the length, given by the relation:

ΔL = aL (T2 − T1 ) … (ii)

Equating equations ( (i) and (ii), we get:
FL
αL (T2 − T1 ) = 2
d
π( ) ×Y
2

2
d
F = α (T2 − T1 ) πY ( )
2

2
−3
−3 11
2 × 10
F = 2 × 10 × (−39 − 27) × 3.14 × 0.91 × 10 × ( )
2

2
= −3.8 × 10 N

(The negative sign indicates that the tension is directed inward.)

2
Hence, the tension developed in the wire is 3.8 × 10 N .

Page : 300 , Block Name : Exercise

Q11.10 A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and
diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °C? Is
there a ‘thermal stress’ developed at the junction ? The ends of the rod are free to expand (Co-ef cient of
linear expansion of brass = 2.0 × 10 K , steel = 1.2 × 10 K ).
–5 –1 –5 –1

∘
Initial temperature, T1 = 40 C
∘
Final temperature, T2 = 250 C
∘
Change in temperature, ΔT = T2 − T1 = 210 C

Length of the brass rod at T1 , l1 = 50cm

Diameter of the brass rod at T1, d1 = 3.0mm

Length of the steel rod at T2 , I2 = 50cm

Diameter of the steel rod at T2 , d2 = 3.0mm

−5 −1
Coefficient of linear expansion of brass, a1 = 2.0 × 10 K

−5 −1
Coefficient of linear expansion of steel, a2 = 1.2 × 10 K

For the expansion in the brass rod, we have:
Change in length (Δ/ )
1
= α1 ΔT
Original length (l1 )

−5
∴ ΔI1 = 50 × (2.1 × 10 ) × 210

= 0.2205cm

For the expansion in the steel rod, we have:
Change in length (ΔI2 )
= α2 ΔT
Original length (l2 )

−5
∴ ΔI2 = 50 × (1.2 × 10 ) × 210

= 0.126cm

Total change in the lengths of brass and steel,

Δl = ΔI1 + Δ/2

= 0.2205 + 0.126

= 0.346cm

Total change in the length of the combined rod = 0.346cm

since the rod expands freely from both ends, no thermal stress is developed at the

junction.

Page : 300 , Block Name : Exercise

Q11.11 The coef cient of volume expansion of glycerine is 49 × 10 . What is the fractional change in its
–5 –1
K

density for a 30 °C rise in temperature ?

−5 −1
Coefficient of volume expansion of glycerin, aV = 49 × 10 K
∘
Rise in temperature, ΔT = 30 C

ΔV
Fractional change in its volume =
V

This change is related with the change in temperature as:
ΔV
= αV ΔT
V

VT − VT = VT αV ΔT
2 1 1

m m m
− = αv ΔT
ρTz ρτ ρT
1 1

Where,

m = Mass of glycerine

ρT = Initial density at T
1 1

ρr = Final density at T
2 2

ρT −ρT
1 2
= αv ΔT
ρT
i

Where,
ρT −ρT
1 2
= Fractional change in density
ρT
2

−5 −2
∴ Fractional change in the density of glycerin = 49 × 10 × 30 = 1.47 × 10

Page : 301 , Block Name : Exercise

Q11.12 2 A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8.0 kg. How
much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the
machine itself or lost to the surroundings. Speci c heat of aluminium = 0.91 J g K –1 –1

3
Power of the drilling machine, P = 10kW = 10 × 10 W

3
Mass of the aluminum block, m = 8.0kg = 8 × 10 g

Time for which the machine is used, t = 2.5min = 2.5 × 60 = 150s

−1 −1
Specific heat of aluminium, c = 0.91Jg K

Rise in the temperature of the block after drilling = δT

Total energy of the drilling machine = Pt

3
= 10 × 10 × 150

6
= 1.5 × 10 J

It is given that only 50% of the power is useful.

50 6 5
Useful energy, ΔQ = × 1.5 × 10 = 7.5 × 10 J
100

But ΔQ = mcΔT

ΔQ
⋅ΔT =
mc
5
7.5 × 10
=
3
8 × 10 × 0.91
∘
= 103 C

Therefore,in 2.5 minutes of drilling, the rise in the temperature of the block is 103 C ∘

Page : 301 , Block Name : Exercise

Q11.13 A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500 °C and then placed on a
large ice block. What is the maximum amount of ice that can melt? (Speci c heat of copper = 0.39 J g K ;
–1 –1

heat of fusion of water = 335 J g ). –1

Mass of the copper block, m = 2.5kg = 2500g
∘
Rise in the temperature of the copper block, Δθ = 500 C

−1 −1
Specific heat of copper, C = 0.39Jg C

−1
Heat of fusion of water, L = 335Jg

The maximum heat the copper block can lose, Q = mCΔθ

Let m1 g be the amount of ice that melts when the copper block is placed on the ice

block.

The heat gained by the melted ice, Q = m1 L
Q 487500
∴ m1 = = = 1455.22g
L 335

Hence, the maximum amount of ice that can melt is 1.45kg .

Page : 301 , Block Name : Exercise

Q11.14 In an experiment on the speci c heat of a metal, a 0.20 kg block of the metal at 150 °C is dropped in a
copper calorimeter (of water equivalent 0.025 kg) containing 150 cm of water at 27 °C. The nal
3

temperature is 40 °C. Compute the speci c heat of the metal. If heat losses to the surroundings are not
negligible, is your answer greater or smaller than the actual value for speci c heat of the metal ?

Mass of the metal, m = 0.20kg = 200g
∘
Initial temperature of the metal, T1 = 150 C
∘
Final temperature of the metal, T2 = 40 C

′
Calorimeter has water equivalent of mass, m = 0.025kg = 25g

3
volume of water, V = 150cm
∘
Mass (M) of water at temperature T = 27 C :

150 × 1 = 150g

Fall in the temperature of the metal:

∘
ΔT = T1 − T2 = 150 − 40 = 110 C
∘
Specific heat of water, Cv = 4.186J/g/ K

Specific heat of the metal = C

Heat lost by the metal, θ = mCΔT … (i)

Rise in the temperature of the water and calorimeter system:
′ ∘
ΔT = 40 − 27 = 13 C

Heat gained by the water and calorimeter system:

′′
Δθ = m1 Cv ΔT

′ ′
= (M + m ) Cn ΔT … (ii)

Heat lost by the metal = Heat gained by the water and colorimeter system

′ ′
mCΔT = (M + m ) Cv ΔT

200 × C × 110 = (150 + 25) × 4.186 × 13

175×4.186×13 −1 −1
∴ C = = 0.43Jg K
110×200

If some heat is lost to the surroundings, then the value of C will be smaller than the

actual value.

Page : 301 , Block Name : Exercise

Q11.15 Given below are observations on molar speci c heats at room temperature of some common gases

The measured molar speci c heats of these gases markedly different from those for monatomic gases,
Typically, molar speci c heat of a monatomic gas is 2.92 cal/mol K. Explain this difference. What can you
infer from the somewhat larger (than the rest) value for chlorine?

Answer. The gases listed in the given table are diatomic. Besides the translational degree Of
freedom, they have other degrees of freedom (modes of motion).
Heat must be supplied to increase the temperature of these gases. This increases the
average energy Of all the modes Of motion. Hence, the molar speci c heat Of diatomic
gases is more than that of monatomic gases.
If only rotational mode of motion is considered, then the molar speci c heat of a diatomic
5
gas = R
2
5 −1 −1
= × 1.98 = 4.95 cal mol K
2

5
With the exception of chlorine, all the observations in the given table agree with ( R)
2

This is because at room temperature, chlorine also has vibrational modes of motion

besides rotational and translational modes of motion.

Page : 301 , Block Name : Exercise

Q11.16 A child running a temperature of 101°F is given an antipyrin (i.e. a medicine that lowers fever) which
causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to 98 °F in

20 minutes, what is the average rate of extra evaporation caused, by the drug. Assume the evaporation
mechanism to be the only way by which heat is lost. The mass of the child is 30 kg. The speci c heat of
human body is approximately the same as that of water, and latent heat of evaporation of water at that
temperature is about 580 cal g–1

∘
Initial temperature of the body of the child, T1 = 101 F
∘
Final temperature of the body of the child, T2 = 98 F

5
Change in temperature, ΔT = L(101 − 98) × ]∘
9 C

Time taken to reduce the temperature, t = 20min

3
Mass of the child, m = 30kg = 30 × 10 g

Specific heat of the human body = Specific heat of water = c
∘
= 1000cal/kg/ C

−1
Latent heat of evaporation of water, L = 580calg

The heat lost by the child is given as:

Δθ = meΔT

5
= 30 × 1000 × (101 − 98) ×
9

= 50000cal

Let m1 be the mass of the water evaporated from the child's body in 20mi

Loss of heat through water is given by:

Δθ = m1 L

Δθ
∴ m1 =
L

50000
= = 86.2g
580

∴ Average rate of extra evaporation caused by the drug

86.2
= = 4.3g/min
200

Page : 301 , Block Name : Exercise

Q11.17 A ‘thermocol’ ice box is a cheap and an ef cient method for storing small quantities of cooked food in
summer in particular. A cubical icebox of side 30 cm has a thickness of 5.0 cm. If 4.0 kg of ice is put in the
box, estimate the amount of ice remaining after 6 h. The outside temperature is 45 °C, and coef cient of
thermal conductivity of thermocol is 0.01 J s–1 m K . [Heat of fusion of water = 335 × 10 J kg ]
–1 –1 3 –1

Side of the given cubical ice box, s = 30cm = 0.3m

Thickness of the ice box, l = 5.0cm = 0.05m

Mass of ice kept in the ice box, m = 4kg

Time gap, t = 6h = 6 × 60 × 60s
∘
Outside temperature, T = 45 C

−1 −1 −1
coefficient of thermal conductivity of thermacole, K = 0.01Js m K

3 −1
Heat of fusion of water, L = 335 × 10 Jkg

′
Let m be the total amount of ice that melts in 6h .

The amount of heat lost by the food:

KA(T −0)t
θ =
l

Where,

2 2 3
A = Surface area of the box = 6s = 6 × (0.3) = 0.54m

0.01×0.54×(45)×6×60×60
θ = = 104976J
0.05
′
But θ = m L

′ θ
∴ m =
L

′ θ
∴ m =
L

104976
= = 0.313kg
3
335×10

Mass of ice left = 4 − 0.313 = 3.687kg

Hence, the amount of ice remaining after 6h is 3.687kg .

Page : 301 , Block Name : Exercise

Q11.18 A brass boiler has a base area of 0.15 m and thickness 1.0 cm. It boils water at the rate of 6.0 kg/min
2

when placed on a gas stove. Estimate the temperature of the part of the ame in contact with the boiler.
Thermal conductivity of brass = 109 J m K ; Heat of vaporisation of water = 2256 × 10 J kg .
–1 –1 3 –1

2
Base area of the boiler, A = 0.15m

Thickness of the boiler, / = 1.0cm = 0.01m

Boiling rate of water, R = 6.0kg/min

Mass, m = 6kg

Time, t = 1min = 60s

−1 −1 −1
Thermal conductivity of brass, K = 109Js m K
3 −1
Heat of vaporisation, L = 2256 × 10 Jkg

The amount of heat flowing into water through the brass base of the boiler is given by:

KA(T1 −T2 )t
θ =
l

Where,

T1 = Temperature of the flame in contact with the boiler
∘
T2 = Boiling point of water = 100 C

Heat required for boiling the water:

θ = mL … (ii)

Equating equations (i) and (ii), we get:

KA(T1 −T2 )t
∴ mL =
l

mLl
T1 − T2 =
KAt
3
6 × 2256 × 10 × 0.01
=
109 × 0.15 × 60
∘
= 137.98 C

Therefore, the temperature of the part of the ame in contact with the boiler is
237.98oC..

Page : 301 , Block Name : Exercise

Q11.19 Explain why :
(a) a body with large re ectivity is a poor emitter
(b) a brass tumbler feels much colder than a wooden tray on a chilly day
(c) an optical pyrometer (for measuring high temperatures) calibrated for an ideal black body radiation gives
too low a value for the temperature of a red hot iron piece in the open, but gives a correct value for the
temperature when the same piece is in the furnace
(d) the earth without its atmosphere would be inhospitably cold
(e) heating systems based on circulation of steam are more ef cient in warming a building than those based
on circulation of hot water

(a) A body with a large re ectivity is a poor absorber of light radiations. A poo absorber will in turn be a poor
emitter of radiations. Hence, a body with a large re ectivity is a poor emitter.
(b) Brass is a good conductor of heat. When one touches a brass tumbler, heat is conducted from the body to
the brass tumbler easily. Hence, the temperature of the body reduces to a lower value and ond ond ond ond
ond one feels cooler. Wood is a poor conductor of heat. When one touches a wooden tray, very little heat is
conducted from the body to the wooden tray. Hence, there is only a negligible drop in the temperature of the
body and one does not feel cool. Thus, a brass tumbler feels colder than a wooden tray on a chilly day.
(c) An optical pyrometer calibrated for an ideal black body radiation gives too low a value for temperature of
a red hot iron piece kept in the open. Black body radiation equation is given by: E = σ(T − T ) Where, E =
4

0
4

Energy radiation T=Temperature of optical pyrometer τ =Temperature of open space σ= constant.
0

Hence, an increase in the temperature of open space reduces the radiation energy When the same plece of
iron is placed in a furnace, the radiation energy, E = σT . 4

(d) Without its atmosphere, earth would be inhospitably cold. In the absence of atmospheric gases, no extra
heat will be trapped. All the heat would be radiated back from earth's surface.
(e) A heating system based on the circulation of steam is more ef cient in warming a building than that
based on the circulation of hot water. This is because steam contains a surplus heat in the form of latent heat
(540cal/g).

Page : 301 , Block Name : Exercise

Q11.20 A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C.
The temperature of the surroundings is 20 °C.

According to Newton's law of cooling, we have:

dT
− = K (T − T0 )
dt

dT
= −Kdt
K(T −T0 )

Where,

Temperature of the body = T

∘
Temperature of the surroundings = T0 = 20 C

K is a constant
∘ ∘
Temperature of the body falls from 80 C to 50 C in time, t = 5min = 300s

Integrating equation ( i ), we get:

80 dT ∞0
∫ = −∫ Kdt
50 K(T −T0 ) 0

30 300
[log (T − T0 )] = −K[t]
e 50 0

2.3026 80−20
log = −300
K 10 50−20

2.3026
log 2 = −300
K 10

−2.3026
log 2 = K
300 10

∘ ∘ ′
The temperature of the body falls from 60 C to 30 C in time = t

Hence, we get:
2.3026 60−20
log = −i
K 10 30−20

−2.3026
log 4 = K
t 10

Equating equations (ii) and (iii), we get:
−2.3026 −2.3026
log 4 = log 2
t 10 300 10

∴ i = 300 × 2 = 600s = 10min
∘ ∘
Therefore, the time taken to cool the body from 60 C to 30 C is 10 minutes.

Thus,CO condenses into the solid state directly, without going through the liquid state.
2

(b) At 4 atm pressure, CO2 lies below 5.11 atm (triple point C). Hence, it lies in the

region of vaporous and solid phases. Thus, it condenses into the solid state directly,

without passing through the liquid state.

(c) When the temperature of a mass of solid C02 (at 10 atrn pressure nd at —650C) is increased, it changes to
the liquid phase and then to the vaporous phase. It forms a line parallel to the temperature axis at 10 atm.
The fusion and boiling points are given by the intersection point where this parallel line cuts the fusion and
vaporisation curves.
∘
(d) If CO2 is heated to 70 C and compressed isothermally, then it will not exhibit any

∘
transition to the liquid state. This is because 70 C is higher than the critical temperature

of CO2 . It will remain in the vapour state, but will depart from its ideal behaviour as

pressure increases.

Page : 302 , Block Name : Exercise

Q11.21 Answer the following questions based on the P-T phase diagram of carbon dioxide:
(a) At what temperature and pressure can the solid, liquid and vapour phases of CO co-exist in equilibrium
2

?
(b) What is the effect of decrease of pressure on the fusion and boiling point of CO ? 2

(c) What are the critical temperature and pressure for CO ? What is their signi cance ?
2

(d) Is CO solid, liquid or gas at (a) –70 °C under 1 atm, (b) –60 °C under 10 atm, (c) 15 °C under 56 atm ?
2

Answer. (a) The P-T phase diagram' for CO is shown in the following gure.
2

C is the triple point of the CO phase diagram. This means that at the temperature and
2

pressure corresponding to this point (i.e., at —56.60C and 5.11 atm), the solid, liquid,
and vaoorous Dhases of CO. co-exist in eauilibrium.
(b) The fusion and boiling points of CO decrease with a decrease in pressure.
2

(c) The critical temperature and critical pressure of CO are 31. IOC and 73 atm
2

respect vely. Even if it IS compressed to a pressure greater than 73 atm, CO will not liquefy above the
2

critical temperature.
(d) It can be concluded from the P − T phase diagram of CO2 that:

∘
(a) CO2 is gaseous at − 70 C , under 1 atm pressure

∘
(b) CO2 is solid at − 60 C , under 10 atm pressure

∘
(c) CO2 is liquid at 15 C, under 56 atm pressure

Page : 302 , Block Name : Exercise

Q11.22 Answer the following questions based on the P – T phase diagram of CO : 2

(a) CO at 1 atm pressure and temperature – 60 °C is compressed isothermally. Does it go through a liquid
2

phase ?
(b) What happens when CO at 4 atm pressure is cooled from room temperature at constant pressure ?
2

(c) Describe qualitatively the changes in a given mass of solid CO at 10 atm pressure and temperature –65
2

°C as it is heated up to room temperature at constant pressure.
(d) CO is heated to a temperature 70 °C and compressed isothermally. What changes in its properties do
2

you expect to observe ?

(a) It condenses to solid directly.

(b) It condenses to solid directly.

(c) The fusion and boiling points are given by the intersection point where this parallelel

line cuts the fusion and vaporisation curves.

(d) It departs from ideal gas behaviour as pressure increases.

Explanation:

(a) The P − T phase diagram for CO2 is shown in the following figure.

∘ ∘
At 1 atm pressure and at − 60 C, CO2 lies to the left of − 56.6 C (triple point C). Hence, it

lies in the region of vaporous and solid phases.

Thus, CO2 condenses into the solid state directly, without going through the liquid state.

(b) At 4 atm pressure, CO2 lies below 5.11 atm (triple point C). Hence, it lies in the

region of vaporous and solid phases. Thus, it condenses into the solid state directly,

without passing through the liquid state.

(c) When the temperature of a mass of solid CO (at 10 atm pressure and at —650C) is
2

increased, it changes to the liquid phase and then to the vaporous phase. It forms a line
parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the
intersection point where this parallel line cuts the fusion and vaporisation curves.
∘
(d) If CO2 is heated to 70 C and compressed isothermally, then it will not exhibit any

∘
transition to the liquid state. This is because 70 C is higher than the critical temperature

of CO2 . It will remain in the vapour state, but will depart from its ideal behaviour as

pressure increases.

Page : 302 , Block Name : Exercise