Thermal properties of Matter

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.

Solution:

Relation between the Celsius and Fahrenheit scale is:

TF = 9/5 TC+32 -----------(1)

Relation between Kelvin and Fahrenheit scale is:

TC = TK -273.15 ----(2)

For carbon dioxide:

TK = 216.55 K

TC = 216.55 – 273.15 = -56.600C

TF = �

� (TC) +32

= �

� (-56.60) +32

= -69.880F

For neon:

TK = 24.57 K

TC = 24.57 – 273.15 = -248.580C

TF = �

� (TC) +32

TF = �

� (-248.58) +32 = - 415.44

0F

2) Two absolute scales A and B have triple points of water defined to be 200 A and

350 B. what is the relation between TA and TB?

Solution:

Triple point of water on the Kelvin scale, TK = 273.15 K

According to the problem,

The temperature 273.15 K on the Kelvin scale is equivalent to 20 A on the absolute scale

A.

200 A = 273.15 K

One degree is scale A, A =273.15/200 K

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

B.

350 B = 273.15 K

One degree in scale B, B = 273.15/350 K

If a body temperature is measured as TA in the absolute scale A and TB in the absolute

scale B, then, the temperature the body

���.��

��� x TA =

���.��

��� x TB

� = 200/350 = 4/7

So the ratio between TA and TB is given as 4:7.

3) The electrical resistance in ohms of a certain thermometer varies with

temperature according to the approximate law: R = R0 [1 + α (T –T0)]

The resistance is 101.6 ohms at the triple point of water 273.15 K, and 165.5 ohms

at the normal melting point of lead (600.5K). What is the temperature when the

resistance is 123.4 ohms?

Solution:

Variation in resistance with temperature is given as:

R = R0 [1+α (T-T0)] ------------(1)

Where,

R0 and T0 are the initial resistance and temperature respectively and R and T are the final

resistance and temperature respectively.

‘α’ is temperature coefficient of resistance

At the triple point of water, T0 = 273.15 K, resistance of thermometer,

R0 = 101.6 ohms

On substituting theses values in equation (1), we get

R = R0 [ 1+α (T-T0)]

165.5 = 101.6 [1+α (600.5 – 273.15)

1.629 = 1+ α (327.35)

α =�.���

���.�� 192 x 10

-3 K

-1

Let us consider the resistance to be R1 = 123.4 ohms at temperature T1

R = R0 [1+α (T-T0)]

123.4 = 101.6 [1+1.92 x 10-3

(T1 – 273.15)

1.214 = 1+1.92 x 10-3

(T1 – 273.15)

�.��

�.������ = T1 -273.15

T1 = 384.61 K

Thus, the temperature when the resistance is 123.4 ohms is 384.6 K.

4) Answer the following:

(a) The triple point of water is a standard fixed point in modern thermometry. Why?

What is wrong in taking the melting point of ice and the boiling point of water as

standard fixed points (as we originally done in the Celsius scale)?

(b) There were two fixed point in the original Celsius scale as mentioned above

which were assigned the number 00C and 100

0C respectively. On the absolute scale,

one of the fixed points is the triple point of water, which on the Kelvin absolute

scale is assigned the number 273.16 K. what is the other fixed point on this (Kelvin)

scale?

Solution:

(a) The triple point of water is a standard fixed point in modern thermometry because it is

a unique value which is 273.16 K and at particular values of volume and pressure. The

melting point of ice and boiling point of water do not have particular values because these

points depend on pressure.

(b) the other fixed point on the Kelvin absolute scale is absolute zero or 0 K.

5) Answer the following:

(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?

Solution:

(c) The temperature 273.15 K is the melting point of ice where as the temperature 273.16

K is the triple point of water. The temperature 00C on Celsius scale is the melting point of

ice, and its corresponding value on Kelvin scale is 273.15 K. the temperature of triple

point of water on Celsius scale is 0.010C.

Therefore, the absolute temperature (Kelvin scale) T, is related to temperature tc, on the

Celsius scale as:

tc = T – 273.15

(d) Let the temperatures on the Fahrenheit scale and absolute scale be TF and TK

respectively. Both the temperatures can be related as:

����

��� =

�����.��

��� ---(1)

Let TF1 be the temperature on the Fahrenheit scale and TK1 be the temperature on the

absolute scale. Both temperatures can be related as:

����

��� =

����

��� ----------(2)

Subtracting equation (1) from equation (2), we get, the unit interval size of the Fahrenheit

scale:

����

��� =

����

���

If ��� - �� = 1 then,

����

��� = 1/100

��� − �� = 1 x 180/100 = 9/5

Since the triple point of water on the Kelvin scale= 273.16 K, its value on the absolute

scale, which has its unit interval size that of a Fahrenheit scale, is given as:

273.16 x 9/5 = 491.69

6) Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The

following observations are made:

(a) What is the absolute temperature of normal melting point of sulphur as read by

thermometers A and B?

(b) What do you think is the reason behind the slight difference in answers of

thermometers A and B? (The thermometers are not faulty). What further procedure

is needed in the experiment to reduce the discrepancy between the two readings?

Solution:

(a) Given data is:

For thermometer A

Triple point of water, T = 273.16 K

At this temperature, pressure in thermometer A, PA = 1.250 x 105 Pa

On an absolute scale, let T1 be the normal melting point of sulphur

At this temperature, pressure in thermometer A, P1 = 1.797 x 105 Pa

According to the Charles’s law, we have the relation:

PA/T = P1/T1

T1 =P1T/PA = �.���������.��

�.������

- 392.69 K

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

thermometer A is 392.69 K.

For thermometer B

At triple point 273.16 K, the pressure in thermometer B, PB = 0.200 x 105 Pa

At temperature T1, the pressure in thermometer B, P2 = 0.287 x 105 Pa

According to Charles’ law we can write the relation:

PB/T =P1/T1

=�.������

���.�� =

�.������

�

T1 = �.������

�.������ x 273.16 = 391.98 K

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

thermometer B is 391.98 K.

(b) There is a slight difference between the readings of thermometers A and B

Because, they are not perfect ideal gases.

To reduce this error the experiment should be conducted at different lower pressure

conditions. Then we have to plot a graph between the temperature and pressure. By

extrapolating the graph to the limiting value of P= 0,

The temperature value we get is the accurate measure because real gas behaves like an

ideal gas under at low pressure and high temperature.

7) A steel tape 1 m long is correctly calibrated for a temperature of 270C. 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 45.00C. 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 27.00C?

Coefficient of linear expansion of steel = 1.20 x 10-5

K-1

.

Solution:

Given data is length of the steel tape at temperature t = 270C, l = 1 m= 100 cm

At temperature t1 = 450C, let the length of the steel rod be, L1 = 63 cm, and the coefficient

of the linear expansion of steel, α = 1.20 x 10-5

K-1

Let l1 be the actual length of the steel rod and l be the length of the steel tape at 450C.

So length of the steel tape is given as

l’ = l + α(t1 – t)

l’ = 100 + 1.20 x 10-5

x 100(45 – 27)

=100.0216 cm

Hence the actual length of the steel rod measured by the steel tape at 450C can be

calculated as:

l1’ = 100.0216/100 x 63

= 63.0136 cm

So, the actual length of the rod at 450C is 63.0136 cm. its length at 27.0

0C is 63.0 cm.

8) A large steel wheel is to be lifted on to a shaft of the same material. At 270C, 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 coefficient of linear expansion of the

steel to be constant over the required temperature range:

αsteel = 1.20 x 10-5

K-1

Solution:

Given data is:

Outer diameter of the steel shaft at t = 270C is d1 = 8.70 cm

Diameter of the central hole in the wheel at t=270C, d2 = 8.68 cm

Coefficient of linear expansion of steel, αsteel = 1.20 x 10-5

K-1

After the shaft is cooled using ‘dry ice’ let us assume that its temperature becomes t1. The

wheel will slip on the shaft, if the change in diameter, ∆d = 8.69 – 8.70

,∆d = - 0.01 cm

Temperature t1, can be calculated from the relation:

∆d = d1 αsteel(t1 – t)

- 0.01 = 8.70 x 1.20 x 10-5

(t1 – 27)

(t1-27) = -95.8

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

9) A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.00C.

What is the change in the diameter of the hole when the sheet is heated to 2270C?

Coefficient of linear expansion of copper = 1.70 x 10-5

K-1

.

Solution:

Given data is:

Initial temperature, T1 = 270C or 300 K

Diameter of the hole at T1, d1 = 4.24 cm

Final temperature, T2 = 2270C or 500 K

Let the diameter of the hole at T2 be d2

Coefficient of linear expansion of copper, αCu = 1.70 x 10-5

K-1

Change in diameter of the hole is, ∆d = dα∆T

Where α is the coefficient of linear expansion and ∆T is the change in temperature.

∆d = 4.24 x 1.7 x 10-5

x (500 – 300)

∆d = 0.0144 cm

Thus, the diameter increases by 1.44 x 10-2

cm.

10) A brass wire 1.8 m long at 270C is held taut with little tension between two rigid

supports. If the wire is cooled to a temperature of -390C, what is the tension

developed in the wire, if its diameter is 2.0 mm? Coefficient of linear expansion of

brass =2.0 x 10-5

K-1

, Young’s modulus of brass = 0.91 x 1011

Pa

Solution:

Given data is:

Initial temperature, T1 = 270C or 300 K

Length of the brass wire at T1, l = 1.8 m

Final temperature, T2 = -390C or 234 K

Diameter of the wire, d = 2.0 mm = 2 x 10-3

m

Radius of the wire, r= 1 x 10-3

m

Coefficient of linear expansion of brass, α = 2.0 x 10-5

K-1

Tension developed in the wire = F

Young’s modulus of brass, Y = 0.91 x 1011

Pa

When cooled, the brass wire acquires tensile strain due to external forces provided by

rigid support.

The relation between Young’s modulus, stress ans strain is given by the relation:

��

� =

�

� --------------(a)

Where,

A = area of cross section of the wire,

F = tension developed in the wire

!" = change in the length, if the wire is free to contract

We know that, !" = l α∆T ----------(b)

Equating equations (a) and (b), we get:

Tension in the string, F = YA α(∆T)

F = 0.91 x 1011

x π x (1 x 10-3

) x 2 x 10-5

x (300 – 234)

F = 377.17 N

So, the tension developed in the wire is 377.17 N

11) 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

2500C, if the original lengths are at 40.0

0C? is there a ‘thermal stress’ developed at

the junction? The ends of the rod are frees to expand (Coefficient of linear

expansion of brass = 2.0 x 10-5

K-1

, steel = 1.2 x 10-5

K-1

)

Solution:

Given data is:

Initial temperature, T1 = 400C or 313 K

Final temperature, T2 = 2500C or 523 K

Change in temperature, ∆T = T2 – T1 = 210 K

Length of the steel rod at T1, ls = 50 cm

Diameter of the steel rod at T1, ds = 3.0 mm

Length of the steel rod at T1, ls = 50 cm

Diameter of the steel rod at T1, ds = 3.0 mm

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

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

Coefficient of linear expansion of brass, αb = 2.0 x 10-5

K-1

Coefficient of linear expansion of steel, αs = 1.2 x 10-5

K-1

The expansion in the brass rod, is given by:

∆lb = lb αb ∆T

∆lb = 50 x 2.0 x 10-5

x 210

∆lb = 0.21 cm

The expansion in the steel rod, is given by:

∆ls = ls αs ∆T

∆ls = 50 x 1.2 x 10-5

x 210

∆ls = 0.126 cm

Total change in the length of the combined rod is given by:

∆l = ∆ls + ∆lb

∆l = 0.21 + 0.126

∆l = 0.336 cm

Thermal stress will not be developed at the junction because the rods are free to expand.

12) The coefficient of volume expansion of glycerin is 49 x 10-5

K-1.

What is the

fractional change in its density for a 300C rise in temperature?

Solution:

The temperature should be converted into Kelvin because the coefficient of expansion is

in Kelvin.

Given data is:

Rise in temperature, ∆T = 300C or 30 K

Coefficient of volume expansion of glycerin, αv = 49 x 10-5

K-1

Fractional change in volume is given by:

�#

# = αv ∆T

V2 – V1 = V1αv ∆T

$

%& -

$

%� =

$

%� αv ∆T

Where,

m = Mass of glycerin

ρ1 = Initial density at T1

ρ2 = Final density at T2

%��%&

%& = αv∆T

Where,

Fractional change in density, %��%&

%& = αv∆T

%��%&

%& = 49 x 10

-5 x 30

Therefore, fractional change in density is 1.47 x 10-2

13) 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. Specific heat of aluminium = 0.91 J g-1

K-1

.

Solution:

Given data is:

Mass of the aluminium block, m = 8.0 kg

Power of the drilling machine, P = 10 k W = 10 x 103 W

Specific heat of aluminium, s = 0.91 J g-1

K-1

= 910 J kg-1

K-1

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

Energy consumed by the drilling machine = Pt

= 10 x 103 x 150

= 1.5 x 106 J

Only 50% of the power is utilized to heat the aluminium block.

Useful energy, ∆Q = ��

��� x 1.5 x 10

6

∆Q = 7.5 x 105 J

But, ∆Q = ms ∆T

∆T = ∆Q/ms

= 7.5 x 105/8x910

= 103 K

So the rise in temperature of the block is 1030C or 103 K.

14) A copper block of mass 2.5 kg is heated in a furnace to a temperature of 5000C

and then placed on a large ice block. What is the maximum amount of ice that can

melt?

(Specific heat of copper = 0.39 J g-1

K-1

; heat of fusion of water = 335 J g-1

)

Solution:

Given data is:

Temperature of the copper block, ⍬1 = 5000C or 773 K

Specific heat of copper, s = 0.39 J g-1

K-1

Mass of the copper block, m = 2.5 kg = 2500 g

Latent heat of fusion of water, L = 335 J g-1

When the copper block is placed in contact with ice, it losses heat energy and ice absorbs

this heat energy.

Heat lost by the copper block, Q = ms ∆⍬ = 2500 x 0.39 x 500

Q =487500 J

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 ice will be utilized to melt it so, Q’ =m1L

Q’ = 335 m1

According to the principle method of mixtures,

Heat lost by the body = heat gain by the cold body or Q’ = Q

335 m1 = 487500

m1 = 1455.2 g

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

15) In an experiment on the specific heat of a metal, a 0.20 kg block of the metal at

1500C is dropped in a copper calorimeter (of water equivalent 0.025 kg ) containing

150 cm3 of water at 27

0C. The final temperature is 40

0C . Compute the specific heat

of the metal. If heat losses to the surroundings are not negligible, is your answer

greater or smaller than the actual value for specific heat of the metal?

Solution:

Given data is:

Water equivalent of calorimeter = m1 = 0.025 kg = 25 g

Mass of the metal, m = 0.20 kg = 200 g

Initial temperature of the metal, t1 = 1500C

Final temperature of the metal, t2 = 400C

Volume of water at 270C, V = 150 cm

3

Mass of water, M= 150 cm3 x 1 g cm

-3 = 150 g

Fall in the temperature of the metal:

∆t = t1 – t2 = 150 – 40 = 1100C

Specific heat of water, sw = 4.186 J g-1

K-1

Specific heat of the metal = s

Heat lost by the metal, Q = ms∆t ---(1)

Rise in the temperature of the water and calorimeter system:

∆t’ = 40 – 27 = 130C

Heat gained by the water and calorimeter system:

∆Q’ = (M + m1) sw ∆t’ ----------(2)

Since, heat lost by the metal = heat gained by the water and calorimeter system

So, ms ∆t = (M + m1) sw ∆t’

200 x s x 110 = (150+25) x 4.186 x 13

s = 175 x 4.186 x 13/110 x 200

= 0.43 J g-1

K-1

Thus, the specific heat of the metal is 0.43 J g-1

K-1

If heat losses to the surroundings are not negligible then the heat gained by the calorimeter

and water will be less than the heat lost by the metal. Hence, the value of specific heat

obtained from the above calculation will be smaller than its actual value.

16) Given below are observations on molar specific heats at room temperature of

some common gases.

The measured molar specific heats of these gases are markedly different from those

for monatomic gases. Typically, molar specific 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?

Solution:

A monatomic gas can only have degrees of freedom associated with translator motion

only. The gases in the above table are diatomic and have degrees of freedom other than

translational which are commonly known as rotational degrees of freedom. So, more heat

must be supplied to these diatomic molecules in order to increase their temperature.

This gives us an explanation of why, the molar specific heat of diatomic gases is more

than that of monatomic gases. Usually, at room temperatures, a diatomic molecule has

both translational and rotational degrees of freedom.

If only the rotational mode of motion is considered, then the molar specific heat of a

diatomic gas = 5/2 R

= 5/2 x 1.98 = 4.95 cal mol-1

K-1

This value is very near to values given in the table for all the gases except chlorine. The

higher value for chlorine indicates that it has vibration modes at room temperature.

17) 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 CO2

coexist in equilibrium?

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

CO2?

(c) What are the critical temperature and pressure for CO2? What is their

significance?

(d) Is CO2 solid, liquid or gas at (a) -700C under 1 atm, (b) – 60

0C under 10 atm,

(c) 150C under 56 atm?

Solution:

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

Point C on the graph indicates the triple point of the CO2 phase diagram.

This means that at the temperature and pressure corresponding to this point (i.e,. at -

56.60C and 5.11 atm), all the three states solid, liquid, and vapour phases of CO2 coexist in

equilibrium.

(b) Both the fusion and boiling points of CO2 will decrease with a decrease in pressure.

(c) The critical temperature and critical pressure of CO2 are 31.10C and 73 atm

respectively.

Its significance is CO2, which will not liquefy above the critical temperature, even if it is

compressed to a pressure greater than 73 atm.

(d) Conclusion which can be drawn from the P-T phase diagram of CO2 is:

(a) CO2 is gaseous at -700C, under 1 atm pressure

(b) CO2 is solid at -600C, under 10 atm pressures

(c) CO2 is liquid at 150C, under 56 atm pressures

18) Answer the following questions based on the P-T phase diagram of carbon

dioxide:

(a) CO2 at 1 atm pressure and temperature -600C is compressed isothermally. Does

it go through a liquid phase?

(b) What happens when CO2 at 4 atm pressure is cooled from room temperature at

constant pressure?

(c) Describe qualitatively the changes in a given mass of solid CO2 at 10 atm

pressure and temperature -650C as it is heated up to room temperature at constant

pressure.

(d) CO2 is heated to a temperature 700C and compressed isothermally. What

changes in its properties do you expect to observe?

Solution:

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

The sublimation curve extents from -273.160C to -56.6

0C. If CO2 is compressed

isothermally when it is at -600C, it condenses to solid start without undergoing liquid

phase. Thus, CO2at temperature -600C when compressed isothermally condenses into the

solid state directly, without going through the liquid state.

(b) 4 atm pressure is in the region of vapour and solid phases. So, CO2 condenses into a

solid directly, without passing through the liquid state.

(c) As the temperature of a mass of solid CO2 (at 10 atm pressure and at -650C) is

increased. It changes to the liquid phase and then to the vapour phase. It forms a line

parallel to the temperature axis at 10 atm. The fusion and boiling points are given by the

intersection point of this parallel line and the fusion and vaporization curves.

(d) Behaviour of CO2will deviate from its ideal behaviour as it is heated to 700C and

compressed isothermally, because 700C is higher than the critical temperature of CO2 and

it will not be condensed to liquid state.

19) A child running a temperature of 1010F 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 980F in 20 min, 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 specific 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

.

Solution:

Given data is:

Initial temperature of the body of the child, T1 = 1010F

Final temperature of the body of the child, T2 = 980F

In degree Celsius, the change in temperature is, ∆Q = [(101-98) x 5/9]0C

∆T = 1.6670C

Time in which the temperature reduced t = 20 min

Mass of the child, m = 30 kg = 30 x 103 g

Since only evaporation is taken as the way to heat loss, the specific heat capacity of a

child should be equal to that of water

The specific heat of a human body = specific heat of a water = 1 cal g-1

C-1

Latent heat of evaporation of water, L= 580 cal g-1

The heat lost by the child is, ∆Q = ms ∆t

∆Q = 30000 x 1 x 1.667

∆Q = 50000 cal

If m1 is the mass of the water evaporated from the child’s body

Then heat gained by water to convert its state is given by:

∆Q’ = m1L = 538 m1

Heat lost by the boy = heat gained by the evaporated water

∆Q’ = ∆Q

580 m1 = 50000

m 1= 86.2 g

Rate of extra evaporation caused by the drug is m1/t

m1/t = 86.2/20

=4.31 g/ minute

20) A ‘thermacole’ icebox is a cheap and efficient 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 450C, and coefficient of thermal

conductivity of thermacole is 0.01 J s-1

m-1

K-1

. [Heat of fusion of water = 335 x 103

J kg-1

]

Solution:

Given data is:

Coefficient of thermal conductivity of thermacole, K= 0.01 J s-1

m-1

K-1

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

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

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

Time after which amount of ice left is to be calculated,

t = 6 h 6 x 60 x 60 s = 21600 s

Outside temperature, T = 450C

Heat of fusion of water, L = 335 x 103 J kg

-1

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

The amount of heat conducted by the walls of the container is given by:

Q = KA (T – 0)t/l

Where,

A = surface area of the box = 6s2 = 6 x (0.3)

2 = 0.54 m

2

Q = 0.01 x 0.54 x (45) x 6 x 60 x 60/0.05

If ‘m’ is the mass of ice that melts on absorbing this heat, then Q = m’L

m’ = Q/L

We are equating the heat supplied by the container with the product of mass of ice melted

and latent heat of fusion because only that heat will be used to melt the ice.

Mass of ice left = 4 – 0.313 = 3.687 kg

21) A brass boiler has a base area of 0.15 m2 and thickness 1.0 cm. it boils water at

the rate of 6.0 kg/min when placed on a gas stove. Estimate the temperature of the

part of the flame in contact with the boiler. Thermal conductivity of brass = 109 Js-1

m-1

K-1

; heat of vaporization of water = 2256 x 103 J kg

-1.

Solution:

Given data is:

Rate of boiling of water, dm/dt = 6.0 kg/min

dm/dt = 6/60 kg/s = 0.1 kg s-1

Base area of the boiler, A = 0.15 m2

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

Time, t = 1 min = 60 s

Thermal conductivity of brass, K = 109 J s-1

m-1

K-1

Heat of vaporization, L= 2256 x 103 J kg

-1

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

dQ/dt = KA(T1 – T2)/1 ----------(1)

Where,

T1 = temperature of the flame

T2 = boiling point of water = 1000C

The rate at which the heat is to be supplied to boil the water at the rate dm/dt

dQ/dt = dm/dt L ------------(2)

dm/dt L = KA(T1 – T2)/1

0.1 x 2256 x 103 = 109 x 0.15 x (T1 – 100)/0.01

T1- 100 = 0.1 x 2256 x 103 x 0.01/109 x 0.15

T1 – 100 = 138

T1 = 2380C

So, the temperature of the part of the flame in contact with the boiler is 2380C.

22) Explain why:

(a) A body with large reflectivity 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 efficient in warming a

building than those based on circulation of hot water

Solution:

(a) A body with a large reflectivity is a poor emitter because it is a poor absorber of

radiations. A poor absorber will in turn be a poor emitter of radiations.

(b) Brass is a good conductor of heat, so heat is conducted from the body to the brass

tumbler easily. Thus, the temperature of the body reduces to a lower value and one feels

cooler, but a wooden tray, being a bad conductor of heat, does not allow significant heat to

flow through it, on significant drop in temperature takes place.

Thus, a brass tumbler feels colder than a wooden tray on a chilly day.

(c) Optical pyrometer works on the principle of comparing the brightness of the hot object

to the brightness of a calibrated lamp filament by a human eye. Radiation from the target

surface, whose temperature is to be measured, is focused on to a screen. We measure the

temperature of the body by comparing the brightness of the screen with the brightness of a

calibrated tungsten lamp.

Due to the large temperature difference between the red hot iron and the open

surroundings, the brightness of the screen of the pyrometer illuminated by the red hot iron

placed in open is much less than the brightness of the screen of the pyrometer illuminated

by the red hot iron placed in furnace.

(d) Earth’s atmosphere acts as a blanket for the reflected radiations from the surface of the

earth, so in the absence of atmospheric gases, no extra heat will be trapped and its climate

will be very cold. The atmospheric gases like water vapour, carbon dioxide and also

ozone, commonly known as green house gases, absorb and emit infrared radiation. This is

known as green house effect.

(e) A steam contains a surplus of heat in the form of latent heat (540 cal/g), a heating

system based on the circulation of steam is more efficient in warming a building than that

based on the circulation of hot water, because steam can emit 540 cal/g when it condenses

to water at 1000C.

23) A body cools from 800C to 500C in 5 minutes; calculate the time it takes to cool

from 600C to 30

0C. The temperature of the surroundings is 20

0C.

Solution:

From Newton’s law of cooling, we have:

-dT/dt = K (T – T0)

dT/(K(T-T0) = -kdt ------------(1)

Where,

T is mean excess temperature of the body

T0 is temperature of the surroundings

K is a constant

In the first case the temperature of the body falls from 800C to 50

0C in time,

t = 5 min = 300 s

We get, 80 – 50/5x60 = K [(80 +50/2) – 20]

K = 1/450

Let ‘t’ be the time in which the temperature of the body falls from 600C to 30

0C, then

60-30/t = 1/450[(60+30/2) – 20]

t = 540 s = 9 min

Thus, the body will cool from 600C to 30

0C in 9 minutes.