Thermal Expansion
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Thermal Expansion
When a body gets heat, the amplitude of vibration of its particles increases. As a result, its length, area and volume may increase.
Coefficient of Linear Expansion (Linear Expansivity), α
Consider a rod of initial length l1 at temperature θ1. When its temperature increases to θ2, its length becomes l2.
Change in length:
Δl = l2 − l1
Change in temperature:
Δθ = θ2 − θ1
Experimentally:
Δl ∝ l1 … (i)
Δl ∝ Δθ … (ii)
Combining (i) and (ii):
Δl = αl1Δθ … (iii)
Therefore:
α = Δl/(l1Δθ)
From Δl = αl1(θ2 − θ1):
l2 − l1 = αl1(θ2 − θ1)
l2 = l1[1 + α(θ2 − θ1)]
Coefficient of Superficial Expansion (Superficial Expansivity), β
Consider a solid square of initial area A1 at temperature θ1. When its temperature increases to θ2, its area becomes A2.
Change in area:
ΔA = A2 − A1
Change in temperature:
Δθ = θ2 − θ1
Experimentally:
ΔA ∝ A1 … (i)
ΔA ∝ Δθ … (ii)
Combining:
ΔA = βA1Δθ … (iii)
β = ΔA/(A1Δθ)
Coefficient of Cubical Expansion (Cubical Expansivity), γ
Consider a solid cube of initial volume V1 at temperature θ1. When its temperature rises to θ2, its volume becomes V2.
Change in volume:
ΔV = V2 − V1
Change in temperature:
Δθ = θ2 − θ1
Experimentally:
ΔV ∝ V1 … (i)
ΔV ∝ Δθ … (ii)
Combining:
ΔV = γV1Δθ … (iii)
γ = ΔV/(V1Δθ)
Relation Between α, β and γ
Relation Between α and β
For a square of side l:
A1 = l12 … (i)
A2 = l22 … (ii)
From superficial expansion:
A2 = A1(1 + βΔθ) … (iii)
From linear expansion:
l2 = l1(1 + αΔθ) … (iv)
Using (iv) in (ii):
A2 = [l1(1 + αΔθ)]2
A2 = l12[1 + 2αΔθ + α2Δθ2]
Since α is small, the higher-order term is neglected:
A2 = A1(1 + 2αΔθ)
Comparing with A2 = A1(1 + βΔθ):
β = 2α
Relation Between α and γ
For a cube:
V1 = l13 … (i)
V2 = l23 … (ii)
From cubical expansion:
V2 = V1(1 + γΔθ) … (iii)
From linear expansion:
l2 = l1(1 + αΔθ) … (iv)
Using (iv):
V2 = l13(1 + αΔθ)3
V2 = l13[1 + 3αΔθ + 3α2Δθ2 + α3Δθ3]
Neglecting higher-order terms:
V2 = V1(1 + 3αΔθ)
Comparing with (iii):
γ = 3α
Expansion of Liquid
Consider a liquid kept in a vessel up to level A. When the vessel is heated initially, the vessel expands and the level of liquid falls to B. When heating is continued, the liquid expands and rises to level C.
According to the source figure:
Coefficient of Real Expansion (Real Expansivity), γr
Coefficient of Apparent Expansion (Apparent Expansivity), γa
Cubical Expansivity of Vessel, γv
Since:
ΔVr = ΔVa + ΔVv
γrVΔθ = γaVΔθ + γvVΔθ
γr = γa + γv
For a solid vessel, γv = 3αv, so:
γr = γa + 3αv
Variation of Density with Temperature
Let a body have mass M and volume V1 at temperature θ1. Its density is:
When heated to θ2, its volume becomes V2, while its mass remains the same:
Dividing (ii) by (i):
ρ2/ρ1 = V1/V2 … (iii)
If γ is the cubical expansivity:
V2 = V1(1 + γΔθ) … (iv)
Using (iv) in (iii):
ρ2 = ρ1/(1 + γΔθ)
or:
ρ2 = ρ1(1 + γΔθ)−1
Using binomial expansion and neglecting higher powers:
ρ2 = ρ1(1 − γΔθ)
Dulong and Petit’s Method
The experiment described in the scanned notes is used for measuring the real expansivity of a liquid and is based on the principle of hydrostatics.
A U-shaped glass tube is filled with liquid. One side is kept in ice and salt, and the other side is kept in a steam arrangement. The horizontal part is wrapped with wet cloth to reduce exchange of heat with the surroundings. Two thermometers measure the temperatures on the two sides.
Let h1, ρ1, T1 and h2, ρ2, T2 be the heights, densities and temperatures of the liquid on the two sides.
From hydrostatic equilibrium:
h2ρ2g = h1ρ1g
h1/h2 = ρ2/ρ1 … (i)
Also:
ρ2/ρ1 = 1/(1 + γΔθ) … (ii)
Using (ii) in (i):
h1/h2 = 1/(1 + γΔθ)
h2 = h1(1 + γΔθ)
γΔθ = (h2 − h1)/h1
γ = (h2 − h1)/(h1Δθ)
Solved Numericals from the Scanned Notes
Numerical 1 – 400 cm³ Glass Flask Filled with Mercury
Question: A glass flask of volume 400 cm³ is just filled with mercury at 0°C. How much mercury will overflow when the temperature rises to 80°C?
V1m = V1f = 400 cm³
γm = 1.8 × 10−4 K−1
γf = 5.4 × 10−6 K−1
Δθ = 80°C
V2m = 400[1 + (1.8 × 10−4 × 80)] = 405.76 cm³
V2f = 400[1 + (5.4 × 10−6 × 80)] = 400.1728 cm³
Overflow = V2m − V2f
Overflow = 5.5872 cm³
Numerical 2 – 50 cm³ Glass Vessel Filled with Mercury
Question: A glass vessel of volume 50 cm³ is filled with mercury and heated from 20°C to 60°C. What volume of mercury will overflow?
Δθ = 40°C
V2m = 50[1 + (1.8 × 10−4 × 40)] = 50.36 cm³
V2v = 50[1 + (5.4 × 10−6 × 40)] = 50.0108 cm³
Overflow = 50.36 − 50.0108 = 0.3492 cm³
Numerical 3 – 200 cm³ Flask Filled with Mercury
Question: A glass flask of volume 200 cm³ is filled to the brim with mercury at 20°C. How much mercury overflows when the temperature is raised to 100°C?
Δθ = 80°C
V2m = 200[1 + (1.8 × 10−4 × 80)] = 202.88 cm³
V2f = 200[1 + (5.4 × 10−6 × 80)] = 200.0864 cm³
Overflow = 2.7936 cm³
Numerical 4 – Copper Vessel Filled with Glycerin
Question: A copper vessel of volume 100 cm³ at 15°C is filled with glycerin. If the temperature rises to 25°C, find the amount of glycerin spilled out.
Δθ = 10°C
The source uses γglycerin = 49 × 10−5 K−1 and γcopper vessel = 5.1 × 10−5 K−1.
Expanded glycerin volume = 100[1 + (49 × 10−5 × 10)] = 100.49 cm³
Expanded vessel volume = 100[1 + (5.1 × 10−5 × 10)] = 100.051 cm³
Spilled glycerin = 0.439 cm³
Numerical 5 – 500 cm³ Glass Flask Filled with Mercury
Question: A glass flask of volume 500 cm³ is just filled with mercury at 0°C. How much mercury overflows when the temperature is raised to 80°C?
V2m = 500[1 + (1.8 × 10−4 × 80)] = 507.2 cm³
V2f = 500[1 + (5.4 × 10−6 × 80)] = 500.216 cm³
Overflow = 6.984 cm³
Numerical 6 – Steel Wire Fixed Between Rigid Supports
Question: A steel wire 8 m long and 4 mm in diameter is fixed between two rigid supports. Calculate the increase in tension when its temperature falls by 10°C.
Y = 2 × 1011 N m−2
α = 1.2 × 10−5 K−1
d = 4 mm = 0.004 m, Δθ = 10°C
A = πd²/4
Since the wire is restrained, thermal strain = αΔθ.
F = YAαΔθ
F = 96π N ≈ 301.6 N
Numerical 7 – Steel Wire of Radius 2 mm
Question: The two ends of a steel wire of length 8 m and radius 2 mm are fixed to rigid supports. Calculate the increase in tension when its temperature falls by 10°C.
Y = 2 × 1011 N m−2
r = 0.002 m, α = 1.2 × 10−5 K−1, Δθ = 10°C
F = Yπr²αΔθ
F = 96π N ≈ 301.6 N
Numerical 8 – Error in Measurement with an Iron Rod
Question: An iron rod of length 100 m at 10°C is used to measure a distance of 2 km on a day when the temperature is 40°C. Calculate the error in measuring the distance.
α = 1.6 × 10−6 °C−1
Δθ = 30°C
Length of rod at 40°C:
l = 100[1 + (1.6 × 10−6 × 30)] = 100.0048 m
The source then scales the 2 km measurement:
Measured actual distance = 1.000048 × 2000 = 2000.096 m
Error = 0.096 m
Numerical 9 – Aluminium and Brass Rulers
Question: The markings on an aluminium ruler and a brass ruler are perfectly aligned at 0°C. How far apart will their 20.0 cm marks be at 100°C if the left-hand ends remain precisely aligned?
The source uses αAl = 2.4 × 10−5 °C−1 and αbrass = 2.0 × 10−5 °C−1.
Aluminium mark: 20[1 + (2.4 × 10−5 × 100)] = 20.048 cm
Brass mark: 20[1 + (2.0 × 10−5 × 100)] = 20.04 cm
Difference = 0.008 cm
Numerical 10 – Iron Rod Measured by a Brass Scale
Question: The length of an iron rod is measured by a brass scale. When both are at 10°C, the measured length is 50 cm. What is the length of the rod at 40°C when measured by the brass scale at 40°C?
Δθ = 30°C
The source uses αiron = 16 × 10−6 °C−1 and αbrass = 24 × 10−6 °C−1.
Rod length at 40°C = 50[1 + (16 × 10−6 × 30)] = 50.024 cm
Corresponding scale length at 40°C = 50[1 + (24 × 10−6 × 30)] = 50.036 cm
Difference = 0.012 cm
Source answer: 49.988 cm
The final value is retained as written in the supplied handwritten solution.
Numerical 11 – Linear Expansivity of Brass
Question: A brass rod of length 0.40 m and a steel rod of length 0.60 m, both initially at 0°C, are heated to 75°C. If the increase in length is the same for both rods, calculate the linear expansivity of brass. The source gives αsteel = 12 × 10−6 K−1.
(Δl)brass = (Δl)steel
lbαbΔθ = lsαsΔθ
0.4αb = 0.6 × 12 × 10−6
αb = 18 × 10−6 K−1
Numerical 12 – Aluminium Rod Measured with a Steel Scale
Question: An aluminium rod, when measured with a steel scale, both at 25°C, appears to be 1 m long. If the scale is correct at 0°C, find the length of the rod at 0°C. The source gives αAl = 26 × 10−6 K−1 and αsteel = 12 × 10−6 K−1.
Length represented by the 1 m steel scale at 25°C:
l25,s = 1[1 + (12 × 10−6 × 25)] = 1.0003 m
The source then sets:
1.0003 = l0,Al[1 + (26 × 10−6 × 25)]
Source answer: l0,Al = 0.99 m
The source’s written final approximation is preserved exactly rather than silently replaced.
Numerical 13 – Copper Wire and Fall in Temperature
Question: A copper wire of diameter 0.5 mm is stretched between two fixed points at 25°C. Calculate the increase in tension if its temperature falls to 0°C.
Y = 1.2 × 1011 N m−2
α = 18 × 10−6 K−1
d = 0.5 mm, r = 0.25 mm = 0.00025 m
Δθ = 25°C
F = Yπr²αΔθ
F = 10.6 N
Numerical 14 – Temperature at Which Wood Just Sinks in Benzene
Question: Using the data written in the source, determine the temperature at which wood just sinks in benzene.
ρbenzene,0 = 9 × 102 kg m−3
ρwood,0 = 8.8 × 102 kg m−3
The source uses γwood = 1.5 × 10−4 K−1 and γbenzene = 1.2 × 10−3 K−1.
At the temperature of just sinking:
ρwood = ρbenzene
8.8 × 102 / [1 + (1.5 × 10−4)T] = 9 × 102 / [1 + (1.2 × 10−3)T]
T = 21.71°C = 294.71 K
Numerical 15 – Density of Silver at 100°C
Question: The density of silver at 0°C is 10310 kg m−3 and its coefficient of linear expansion is 0.000019 K−1. Calculate its density at 100°C.
ρ0 = 10310 kg m−3
α = 0.000019 K−1
γ = 3α
ρ100 = ρ0/[1 + γΔθ]
ρ100 = 10310/[1 + (3 × 0.000019 × 100)]
ρ100 = 10251.56 kg m−3
Numerical 16 – Brass Seconds Pendulum from 10°C to 35°C
Question: A seconds pendulum made of brass keeps correct time at 10°C. How many seconds will it lose or gain per day when the surrounding temperature rises to 35°C?
For a pendulum, T = 2π√(l/g).
T35/T10 = √(l35/l10)
= √[1 + αΔθ]
The source uses α = 2 × 10−5 K−1 and Δθ = 25°C.
T35/T10 = 1.000249969
Loss per day = (1.000249969 − 1) × 24 × 60 × 60
Loss ≈ 21.6 s per day
Numerical 17 – Brass Pendulum from 15°C to 0°C
Question: A brass pendulum clock keeps correct time at 15°C. How many seconds per day will it lose or gain at 0°C?
T15/T0 = √[1 + α(15)]
The source uses α = 2 × 10−5 K−1.
T15/T0 = 1.000149989
Required time difference = (1.000149989 − 1) × 86400
Gain ≈ 12.96 s per day
Numerical 18 – Brass Clock from 15°C to 20°C
Question: The pendulum of a clock is made of brass. If the clock keeps correct time at 15°C, how many seconds per day will it lose at 20°C?
Δθ = 5°C
The source uses α = 2 × 10−5 K−1.
T20/T15 = √[1 + αΔθ] = 1.00004999
Loss per day = (1.00004999 − 1) × 86400
Loss ≈ 4.32 s per day
Numerical 19 – Brass Pendulum from 30°C to 10°C
Question: A clock having a brass pendulum beats seconds correctly when the room temperature is 30°C. How many seconds will it gain or lose per day when the temperature falls to 10°C?
The source gives α = 0.000018 K−1.
Δθ = 20°C
T30/T10 = √[1 + (0.000018 × 20)] = 1.000179984
Time difference = (1.000179984 − 1) × 86400
Gain ≈ 15.55 s per day
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