Page No 295: - Chapter 11 Thermal Properties Of Fluids class 11 ncert solutions Physics - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question 11.5:

Two ideal gas thermometers Aand Buse 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 Aand B?

(b) What do you think is the reason behind the slight difference in Answers of thermometers Aand B? (The thermometers are not faulty). What further procedure is needed in the experiment to reduce the discrepancy between the two readings?

Answer:

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

At this temperature, pressure in thermometer A, P_A = 1.250 × 10⁵ Pa

Let T₁ be the normal melting point of sulphur.

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

According to Charles’ law, we have the relation:

= 392.69 K

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer A is 392.69 K.

At triple point 273.16 K, the pressure in thermometer B, P_B = 0.200 × 10⁵ Pa

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

According to Charles’ law, we can write the relation:

Therefore, the absolute temperature of the normal melting point of sulphur as read by thermometer B is 391.98 K.

(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.

Question 11.6:

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? Coefficient of linear expansion of steel = 1.20 × 10^–5 K^–1.

Answer:

Length of the steel tape at temperature T = 27°C, l = 1 m = 100 cm

At temperature T₁ = 45°C, the length of the steel rod, l₁ = 63 cm

Coefficient of linear expansion of steel, α = 1.20 × 10^–5 K^–1

Let l₂ be the actual length of the steel rod and l‘ be the length of the steel tape at 45°C.

Hence, the actual length of the steel rod measured by the steel tape at 45°C can be calculated as:

 = 63.0136 cm

Therefore, the actual length of the rod at 45.0°C is 63.0136 cm. Its length at 27.0°C is 63.0 cm.

Question 11.7:

A large steel wheel is to be fitted 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 coefficient of linear expansion of the steel to be constant over the required temperature range: α_steel = 1.20 × 10^–5 K^–1.

Answer:

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

27 + 273 = 300 K

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

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

Coefficient of linear expansion of steel, α_steel = 1.20 × 10^–5 K^–1

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

= – 0.01 cm

Temperature T₁, can be calculated from the relation:

Δd = d₁α_steel (T₁ – T)

0.01 = 8.70 × 1.20 × 10^–5 (T₁ – 300)

(T₁ – 300) = 95.78

∴T₁= 204.21 K

= 204.21 – 273.16

= –68.95°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.

Question 11.8:

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? Coefficient of linear expansion of copper = 1.70 × 10^–5 K^–1.

Answer:

Initial temperature, T₁ = 27.0°C

Diameter of the hole at T₁, d₁ = 4.24 cm

Final temperature, T₂ = 227°C

Diameter of the hole at T₂ = d₂

Co-efficient of linear expansion of copper, α_Cu= 1.70 × 10^–5 K^–1

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

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

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

Question 11.9:

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-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1; Young’s modulus of brass = 0.91 × 10¹¹ Pa.

Answer:

Initial temperature, T₁ = 27°C

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

Final temperature, T₂ = –39°C

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

Tension developed in the wire = F

Coefficient of linear expansion of brass, α= 2.0 × 10^–5 K^–1

Young’s modulus of brass, Y = 0.91 × 10¹¹ Pa

Young’s modulus is given by the relation:

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 = αL(T₂ – T₁) … (ii)

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

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

Hence, the tension developed in the wire is 3.8 ×10² N.

Question 11.10:

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-efficient of linear expansion of brass = 2.0 × 10^–5 K^–1, steel = 1.2 × 10^–5 K^–1).

Answer:

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 = 50 cm

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

Length of the steel rod at T₂, l₂ = 50 cm

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

Coefficient of linear expansion of brass, α₁ = 2.0 × 10^–5K^–1

Coefficient of linear expansion of steel, α₂ = 1.2 × 10^–5K^–1

For the expansion in the brass rod, we have:

For the expansion in the steel rod, we have:

Total change in the lengths of brass and steel,

Δl = Δl₁ + Δl₂

= 0.2205 + 0.126

= 0.346 cm

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

Since the rod expands freely from both ends, no thermal stress is developed at the junction.

Question 11.11:

The coefficient of volume expansion of glycerin is 49 × 10^–5 K^–1. What is the fractional change in its density for a 30 °C rise in temperature?

Answer:

Coefficient of volume expansion of glycerin, αV = 49 × 10^–5 K^–1

Rise in temperature, ΔT = 30°C

Fractional change in its volume =

This change is related with the change in temperature as:

Where,

m = Mass of glycerine

= Initial density at T₁

= Final density at T₂

Where,

= Fractional change in density

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

Question 11.12:

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.

Answer:

Power of the drilling machine, P = 10 kW = 10 × 10³ W

Mass of the aluminum block, m = 8.0 kg = 8 × 10³ g

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

Specific heat of aluminium, c = 0.91 J g^–1 K^–1

Rise in the temperature of the block after drilling = δT

Total energy of the drilling machine = Pt

= 10 × 10³ × 150

= 1.5 × 10⁶ J

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

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

Question 11.13:

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

Answer:

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

Rise in the temperature of the copper block, Δθ = 500°C

Specific heat of copper, C = 0.39 J g^–1 °C^–1

Heat of fusion of water, L = 335 J g^–1

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

= 2500 × 0.39 × 500

= 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 melted ice, Q = m₁L

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

Question 11.14:

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 150 cm³ 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?

Answer:

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

Initial temperature of the metal, T₁ = 150°C

Final temperature of the metal, T₂ = 40°C

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

Volume of water, V = 150 cm³

Mass (M) of water at temperature T = 27°C:

150 × 1 = 150 g

Fall in the temperature of the metal:

ΔT = T₁ – T₂ = 150 – 40 = 110°C

Specific heat of water, C_w = 4.186 J/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 C_wΔT^’

= (M + m′) C_w ΔT^’ … (ii)

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

mCΔT = (M + m^’) C_w ΔT^’

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

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