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 .
Solution:
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, αCu= 1.70 × 10–5 K–1
For co-efficient of superficial expansion β,and change in temperature ΔT, we have the relation:
\(\frac {Change \,in\,area (\triangle A)}{Original\, area} = β\triangle T\)
Change in diameter = d2 – d1 = 4.2544 – 4.24 = 0.0144 cm
Hence, the diameter increases by 1.44 × 10–2 cm.
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.
Solution:
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.
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).
Solution:
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.
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?
Solution:
Coefficient of volume expansion of glycerin, αV = 49 × 10–5 K–1
Rise in temperature, ΔT = 30°C
Fractional change in its volume = \(\frac {ΔV}V\)
This change is related with the change in temperature as:
Where, m = Mass of glycerine ρT1= Initial density at T1
ρT2= Final density at T2
\(\frac {ρT_1 -ρT_2}{ρT_2} =a_v\triangle T\)
Where,
\(\frac {ρT_1 -ρT_2}{ρT_2} \) = Fractional change in density
∴ Fractional change in the density of glycerin = 49 ×10–5 × 30 = 1.47 × 10–2
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 .
Solution:
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.
Useful energy, ΔQ = \(\frac {50}{100} \times 1.5 \times 10^6 = 7.5 \times 10^5 J\)
But ΔQ = meΔT
Therefore, in 2.5 minutes of drilling, the rise in the temperature of the block is 103°C.
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).
Solution:
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
∴ M1 = \(\frac QL = \frac {487500}{335} = 1455.22g\)
Hence, the maximum amount of ice that can melt is 1.45 kg.
