# Question-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:**

Triple point of neon = T

_{1}= 24.57 K

∴ C = T

_{1}K - 273.15

= 24.57 - 273.15

= -248.58°C

=

F =

= (-248.58) + 32 = - 447.44 + 32

Temperature of neon in Fahrenheit = -415.44°F.

(b) Triple point of CO

_{2}= T

_{2}= 216.55 K

∴ C = T

_{2}K - 273.15

= 216.55 - 273.15

= -56.60°C

F =

= (-56.60) +32 = -101.88 + 32

Temperature of CO

_{2}in Fahrenheit = -69.88°F.

# Question-2

**A constant volume gas thermometer using helium records a pressure of 20.0 k Pa at the triple-point of water and pressure of 14.3 k Pa at the temperature of 'dry ice' (solid CO**

_{2}). What is the temperature of 'dry ice'?**Solution:**

Triple point water = T

_{tr}= 273.16 K

Pressure at the triple point of water P

_{tr}= 20 × 10

^{3}Pa

Pressure at the temperature of dry ice P = 14.3 × 10

^{3}Pa

We know that

Therefore temperature of dry ice T = T

_{tr}

= × 273.16 K

T = 195.31 K.

# Question-3

**Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T**

_{A}and T_{B}?**Solution:**

We know that absolute temperature can be written as

T = P × (T

_{tr}/P

_{tr})

We have the triple points given at two different scales A & B. Let it be T

_{A}and T

_{B}respectively.

T

_{A}= (P/P

_{tr})(200) →1 (in scale A)

T

_{B}= (P/P

_{tr})(350) →2 (in scale B)

Dividing 1 by 2 we get

T

_{A}/T

_{B}= 200/350

= 4/7

∴ T

_{A}= (4 /7) T

_{B}

This is the relation between T

_{A}& T

_{B}.

# Question-4

**The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law:**

R = R

The resistance is 101.6 ohms, at the triple point of water and 165.5 ohms at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 ohms?

R = R

_{0}[1+5 ×10^{-3}(T - T_{0})]The resistance is 101.6 ohms, at the triple point of water and 165.5 ohms at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 ohms?

**Solution:**

R = R

_{0}[1 + 5 × 10

^{-3}(T - T0)]

R = R

_{0}[1 + α (T - T

_{0})]

When T

_{0}= 273.16 K, R

_{0}= 101.6 ohm

and T = 600.5 K, R = 165.5 ohm, then

165.5 = 101.6 [1 + α (600.5 - 273.16)]

= 101.6 + 101.6α × 327.34

63.9 = 101.6 × α × 327.34 --------(1)

When R = 123.4 ohm, then

123.4 = 101.6 [1 + α (T - 273.16)]

= 101.6 + 101.6a (T - 273.16)

21.8 = 101.6 × α (T - 273.16) --------(2)

Dividing eqn. (1) and (2)

T - 273.16 = = 111.67

T = 111.67 + 273.16 = 384.83 K.

# Question-5

**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 was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0 °C and 100 °C respectively. On the absolute scale, one of the fixed point is the triple point of water which on the Kelvin absolute scale is assigned the number 273.16K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature T

(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 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 was originally done in the Celsius scale)?

(b) There were two fixed points in the original Celsius scale as mentioned above which were assigned the numbers 0 °C and 100 °C respectively. On the absolute scale, one of the fixed point is the triple point of water which on the Kelvin absolute scale is assigned the number 273.16K. What is the other fixed point on this (Kelvin) scale?

(c) The absolute temperature (Kelvin scale) T is related to the temperature T

_{c}on the Celsius scale by T_{c}= T - 273.15 K. 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:**

(a) The triple point of water occurs only at one pressure and also all thermometers agree at this temperature and hence unique. The melting point and boiling point changes with pressure and also they may change with the purity of the substance.

(b) One fixed point on the Kelvin scale is the triple point of water. The other fixed point is the absolute zero.

(c) If we let 't' to represent the Celsius temperature, then we have t = t - 273.15°C. We see that the triple point of water (= 273.16° K by differentiation) corresponds to 0.01°C. If we have t = T - 273.16°C then, triple point of water is 0°C. In order to get the triple point as 0.01°C, we should have t = T - 273.15°C.

(d) Temperature of the triple point of water on an absolute scale whose unit interval size is equal to that of Fahrenheit scale is (9/5)T

_{c}= 9 /5(273.16) = 491.69.

# Question-6

**(a) What is the absolute temperature of normal melting point of sulphur as read by thermometers A & B?**

(b) What do you think is the reason for slightly different answers from A and B? (The Thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?(b) What do you think is the reason for slightly different answers from 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) The absolute temperature can be calculated using the formula

T(p) =

The absolute melting point of sulphur as read by Thermometer A and B can be calculated as following

Thermometer A:

T(P) = = 392.69K

Thermometer B:

T(P) = = 391.98K

(b) We could see that there is a discrepancy in the values measured by thermometer A and B. This arises because the gases are not perfectly ideal. This can be avoided if the readings are taken for lower and lower pressures and the plot between temperature measured versus absolute pressure of the gas at triple point should be extrapolated to obtain temperature in the limit pressure tends to zero, when the gases approach ideal gas behaviour.

# Question-7

**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 63.0cm on a hot day when the temperature is 45.0°C. What is the length of the same steel rod on a day. What is the length of the same steel rod on a day when the temperature is 27.0°C? Co-efficient of linear expansion of copper = 1.70 ×10**

^{-5}°C^{-1}.**Solution:**

l

_{c }= l

_{0}[1+ α ×t]

Substituting l

_{0 }= 63cm and Δt = 45°C - 27°C = 18°C and α=11 ×10

^{-6}°C

^{-1}

∴ l

_{c }= 63[1 + 11 ×10

^{-6 }×18]

= 63

The length of the rod at 27°C = 63 cm.

# Question-8

**A hole is drilled in 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? Co-efficient of linear expansion of copper = 1.70 ×10**

^{-5}°C^{-1}?**Solution:**

Diameter of the hole = 4.24 cm at 27°C

Temperature T = 227°C

Difference in temperature dt = 227°C - 27°C = 200°C

Co-efficient of linear expansion of copper α = 1.70 ×10

^{-5}°C

^{-1}

Therefore change in the diameter d

*l*= ∝ ×

*l*× dt = 1.7 × 10

^{-5}× 4.24 × 200 = 1.44 × 10

^{-2}cm

Change in the diameter of the hole = 1.44 × 10

^{-2}cm.

# Question-9

**A brass wire 1.8m 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 2.0mm? Co-efficient of linear expansion of brass = 2.0 × 10**

^{-5}°C^{-1}, Young's modulus of brass = 0.91 × 10^{11}Pa.**Solution:**

Length of the brass wire L = 1.8m

Change in temperature t = [27 - (-39)]°C = 66°C

We know that,

Stress = Y × strain;

Strain = where, d = α × L × t

Therefore Strain = = α × t

Therefore Stress = Y × Strain = Y × ∝ × t = (0.91 × 10

^{11}) × (2.0 × 10

^{-5}) × 66 = 3.8 × 10

^{2}N.

# Question-10

**A brass rod of length 50cm and diameter 3.0mm 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-efficient of linear expansion of brass**

= 2.0 × 10

= 2.0 × 10

^{-5}°C^{-1}, Steel = 1.2 × 10^{-5}°C^{-1}).**Solution:**

Two rods are joined in such a way that their ends are not clamped so that they can expand freely.

**For Brass rod**

Length of the rod = 50cm

Increase in temperature dt = 250°C - 40°C = 210°C

Co-efficient of linear expansion of brass α

_{1}= 2.0 × 10

^{-5}°C

Therefore change in length d

*l*= α

_{1}×

*l*× dt = 2 × 10

^{-5}× 50cm × 210 = 0.21cm

**For Steel rod**

Co-efficient of linear expansion of steel α

_{2}= 1.2 × 10

^{-5}°C

Change in the diameter d

*l*= α

_{2}×

*l*× dt = 50 × 1.2 × 10

^{-5}× 210 = 0.13cm

Total change in length = d of Brass + d of steel = (0.21 + 0.13) cm = 0.34cm

No 'thermal stress' is developed at the junction since the rods freely expand.

# Question-11

**The coefficient of volume expansion of glycerin is 49 × 10**

^{-5}°C^{-1}. What is the fractional change in its density for a 30°C rise in temperature?**Solution:**

We know that dv = 3 α × v × dt

Therefore = 3 a ´ dt

Where, 3α = β = coefficient of volume expansion

is the fractional change in volume when the temperature rises by 30°C

Therefore = β × dt

where β = 49 × 10

^{-5}°C

^{-1}, dt = 30°C

Therefore = 49 × 10

^{-5}× 30 = 0.0147.

# Question-12

**A 10kW 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}°C^{-1}?**Solution:**

Specific heat of aluminium = 0.91 J g

^{-1}°C

^{-1}= 0.91 × 10

^{3}J kg

^{-1}K

^{-1}

Power of machine = 10 kW

Energy obtained from machine in 2.5 minutes

= 10 × 10

^{3}× 2.5 × 60 J = 1500 × 10

^{3}J

50% of this energy = 750 × 10

^{3}J

750 × 10

^{3}J is used in heating the aluminium block.

Heat absorbed by the block = mass × specific heat of aluminium × rise in temperature.

750 × 10

^{3}= 8.0 × 0.91 × 10

^{3}× rise in temperature.

Rise in temperature of aluminium = = 103 °C.

# Question-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? (Specific heat of copper = 0.39 J g**

^{-1}°C^{-1}, heat of fusion of water = 335 Jg^{-1}).**Solution:**

When copper block is kept on ice its temperature will decrease from 500°C to 0°C.

Specific heat of copper = = 0.39 × 10

^{3}J/kg°C

Heat given out by copper = mass × specific heat rise in temperature

= 2.5 × 0.39 × 10

^{3}× (500 - 0) J

= 487500 J

Using the above heat let us assume that M kg of ice at 0°C is converted to water at 0°C.

Latent heat of fusion of water = 335 J/g = 335 × 10

^{3}J/kg

Heat absorbed by ice = Heat given out by copper

M × 335 × 10

^{3}= 487500

M =

= 1.4552 kg.

# Question-14

**In an experiment on the specific 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 150cc of water at 27°C. The final temperature is 40°C. 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:**

Mass of the metal block M = 0.20 kg = 200 g.

Fall in temperature of the metal block, = (150 - 40) = 110°C

If c be the specific heat of metal, then heat lost by the metal block = 200 × c × 100

Volume of water = 150 cm

^{3}

Mass of water = 150 g

Increase in temperature of water = (40 - 27) °C = 13°C

Heat gained by water = 150 × 13 cal

Water equivalent of calorimeter w = 0.025 kg = 25 g

Heat gained by calorimeter = w × increase in temperature of calorimeter

= 25 × 13 cal

Heat lost by metal block = Heat gained by water + Heat gained by calorimeter

200 × c × 110 = (150 + 25) 13

∴ c = = 0.1

If heat lost to the surroundings, c will be less than the actual value.

# Question-15

**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 are put in the box, estimate the amount of ice remaining after 6h. The outside temperature is 45°C, and co-efficient of thermal conductivity of thermacole = 0.01 J s**

^{-1}m^{-1}°C^{-1}, (heat of fusion of water =335 × 10^{3}Jkg^{-1})?**Solution:**

Side of the cubical box = 30 cm.

Thickness of box = 5cm.

Therefore area = (30 × 30) cm = 900 cm.

Amount of ice put inside = 4kg.

Heat flow into the box Q = , but Q = mL

Therefore mL =

m = = = 0.31 kg

Therefore remaining amount of ice = 4.0 - 0.31 = 3.69 kg

Hence the amount of ice remaining = 3.7 kg.

# Question-16

**A brass boiler has a base area of 0.15 m**

^{2}and thickness 1.0 cm. It boils water at the rate of 6.0 kg/minute 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 J s^{-1}m^{-1}°C^{-1}(heat of vaporization of water = 2256 Jkg^{-1})?**Solution:**

Heat of vaporization of water = 2256 Jkg

^{-1}

A = 0.15 m

^{2}, x = 0.01 m, Mass of water = m = 6.0 kg, time t = 1 min = 60 secs.

Latent heat of vaporization L = 2256 Jkg

^{-1}= 2256 × 10

^{3 }Jg

^{-1}

T

_{2}= 100°C = 373 K

Q = , but Q = mL

Therefore mL =

θ

_{1 }- θ

_{2}= =

θ

_{1}- 373 = 137.98

θ

_{1}= 137.98 + 373 = 510.98 K.

# Question-17

**Explain why:**

(a) a body with large reflectivity is a poor emitter.

(b) a brass tumbler feels much older than a wooden tray on a chilly day.

(c) an optical pyrometer (for measuring high temperatures) calibration 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 system based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

(a) a body with large reflectivity is a poor emitter.

(b) a brass tumbler feels much older than a wooden tray on a chilly day.

(c) an optical pyrometer (for measuring high temperatures) calibration 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 system based on circulation of steam are more efficient in warming a building than those based on circulation of hot water.

**Solution:**

(a) Large reflectivity means less absorption. When a body is a perfectly black body, it absorbs all the radiations and emits all the radiations. Since in the above case absorption is less, emission is very poor.

(b) Brass is a good conductor of heat. When we touch brass tumbler with our fingers, our body heat is quickly conducted to the brass tumbler and hence temperature at the finger tips is reduced. Thus, the brass tumbler feels colder. On the other hand wood is bad conductor, hence our body is not conducted to the wooded tray.

(c) The temperature of red hot iron in the oven is given by E

_{1}= σT

^{4}. When iron is taken out in the open temperature (T0) then its radiation energy is given Es = σ(T

^{4}- T

_{0}

^{4}). Thus the pyrometer measures low values for the red hot iron in the open.

(d) The atmosphere serves purpose of a blanket over the Earth and it does not allow Earth's heat to be radiated during night.

(e) This is due to fact that steam contains more heat in the form of latent heat (540 cal/g) in water.

# Question-18

**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.**

**Solution:**

If the body cools from 80 °C to 60 °C,

Average temperature of the liquid = = 65 °C

Excess temperature = (65 - 20) °C = 45 °C

Rate of fall in temperature,

or

i.e., -6 = K ×45 -----------(1)

If the body cools from 60 °C to 30 °C,

Average temperature of the liquid =

Excess temperature = (45 - 20) °C = 25 °C

Rate of fall in temperature,

i.e., = K ×25 -----------(2)

Dividing (i) by (2), we get

30t = 30 ×9

t = .