Heat and Temperature
Exercises - Q21 to Q34
30. Two steel rods and an aluminum rod of equal length lₒ and equal cross-section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficients of linear expansion of aluminum and steel are αₐ and αₛ respectively. Young's modulus of aluminum is Yₐ and of steel is Yₛ.
For Aluminum =lₒ(1+αₐθ)
Exercises - Q21 to Q34
21. A glass window is to be fit in an aluminum frame. The temperature on a working day is 40°C and the glass window measures exactly 20 cm x 30 cm. What should be the size of the aluminum frame so that there is no stress on the glass in winter even if the temperature drops to 0°C? Coefficients of linear expansion for glass and aluminum are 9.0x10⁻⁶/°C and 24x10⁻⁶/°C respectively.
Answer: The size of the frame and the glass should be equal at 0°C. For glass length L at 40°C, its length at 0°C,
Lₒ = L{1 - (40-0)*9.0x10⁻⁶}
=L*0.99964 cm
For this length of the aluminum frame at 0°C, length at 40°C should be
L' = Lₒ(1+40*24x10⁻⁶)
=L*0.99964(1.00096)
=L*1.0006
So for L = 20 cm, L' = 20.012 cm, and
for L = 30 cm, L' = 30.018 cm.
Hence the required size of the aluminum frame is 20.012cmx30.018 cm.
22. The volume of a glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? Coefficients of cubical expansion of mercury and glass are 1.8x10⁻⁴/°C and 9.0x10⁻⁶/°C respectively.
Answer: Let the required volume of mercury be = A cc. So the volume of the remaining space = (1000 - A) cc. At temperature T, the volume of the glass vessel = 1000{1+9.0x10⁻⁶(T-20)} and the volume of the mercury
=A{1+1.8x10⁻⁴(T-20)}
The difference in the volume at this temperature
=1000{1+9.0x10⁻⁶(T-20)}-A{1+1.8x10⁻⁴(T-20)}
=(1000-A)+9000x10⁻⁶(T-20)-1.8x10⁻⁴(T-20)A
But it is constant and equal to 1000-A.
Equating we get,
(1000-A)+9000x10⁻⁶(T-20)-1.8x10⁻⁴(T-20)A =1000-A
→1.8x10⁻⁴(T-20)A=9000x10⁻⁶(T-20)
→A = 90/1.8 =900/18 =50 cc.
23. An aluminum can of cylindrical shape contains 500 cm³ of water. The area of the inner cross-section of the can is 125 cm². All measurements refer to 10°C. Find the rise in water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminum = 23x10⁻⁶/°C and the average coefficient of volume expansion of water = 3.2x10⁻⁴/°C respectively.
Answer: At 10°C, let the radius of the cross-section of the can = r. Hence,
πr² = 125
→r² = 125/π,
→r = 6.31 cm.
The height of the water level in the can = 500/125 cm = 4 cm.
When the temperature rises to 80°C, the difference in the temperature =80-10 =70°C.
Now the radius =6.31(1+70*23x10⁻⁶)
=6.32 cm
The area of cross-section now
=π*6.32²
=125.48 cm²
The volume of water
= 500(1+70*3.2x10⁻⁴)
= 511.2 cm³
Hence the height of the water level
=511.2/125.48
=4.074 cm
Hence the rise in water level =4.074-4.000 =0.074 cm
If the increase in cross-section is neglected then the height of water level =511.2/125 =4.089 cm
The rise in water level =4.089-4.000
=0.089 cm.
24. A glass vessel measures exactly 10 cm x10 cm x10 cm at 0°C. It is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm³ of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5x10⁻⁶/°C.
Answer: The length of the edge of the glass at 10°C = 10(1+10*6.5x10⁻⁶) cm
=10.00065 cm
Hence the volume of the vessel at 10°C
=(10.00065)³ cm³
=1000.195 cm³
The original volume of mercury at 0°C Vₒ=10³ cm³ =1000 cm³.
=1000 cm³
The volume of mercury at 10°
V=1000.195+1.6 =1001.795 cm³.
If ɣ be the coefficient of volume expansion then, V=Vₒ(1+ɣT)
→1001.795=1000(1+ɣ*10)
→1+10*ɣ = 1.0018
→ɣ =0.0018/10
→ɣ =1.8x10⁻⁴/°C.
25. The densities of wood and benzene at 0°C are 880 kg/m³ and 900 kg/m³ respectively. The coefficients of volume expansion are 1.2x10⁻³/°C for wood and 1.5x10⁻³/°C for benzene. At what temperature will a piece of wood just sink in benzene?
Answer: Consider 1 m³ of wood. It's weight = mg = 880g N.
The wood will sink at that temperature at which the volume of benzene displaced has a weight of more than 880g. Let that temperature = T.
The volume of wood here
V= (1 m³)(1+T*1.2x10⁻³)
= 1+1.2x10⁻³T m³
=volume of the benzene displaced.
Now the volume of 1 m³ of benzene at temperature T =(1 m³)(1+1.5x10⁻³T)
=1+1.5x10⁻³T m³.
Hence density = 900/(1+1.5x10⁻³T) kg/m³.
So the weight of the benzene displaced
=(1+1.2x10⁻³T)*900g/(1+1.5x10⁻³T)
This should be equal to 880g. Equating,
(1+1.2x10⁻³T)*900=880(1+1.5x10⁻³T)
→90+0.108T =88+0.132T
→(0.132-0.108)T = 2
→T = 2/0.024 ≈ 83°C.
26. A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?
Answer: Here the rod rests on the smooth horizontal base. The length will increase due to temperature increase but there will be no longitudinal stress on the rod. Since there is no stress hence there will be no longitudinal strain developed.
27. A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2x10⁻⁵/°C.
Answer: The increase in unclamped rod's length at 50°C, ΔL = L*(50-20)*1.2x10⁻⁵
=36x10⁻⁵L
Hence the clamped rod has a contraction =ΔL
Strain =elongation/original length
=-ΔL/L
=-36x10⁻⁵L/L
=-3.6x10⁻⁴
28. A steel wire of cross-sectional area 0.5 mm² is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. The coefficient of linear expansion of steel is 1.2x10⁻⁵/°C and its Young's modulus is 2.0x10¹¹ N/m².
Answer: Here, let the length of wire at 20°C =L. The change in free length of the wire at 0°C, ΔL =αTL
=1.2x10⁻⁵*20L
The strain of the fixed wire =ΔL/L
=24x10⁻⁵
Area of the wire = 0.5 mm²
=0.5x10⁻⁶ m²
If the tension developed = F,
the stress =F/0.5x10⁻⁶ N/m²
But stress =strain*Elasticity
→F/0.5x10⁻⁶ = 24x10⁻⁵*2.0x10¹¹
→F =24*0.5*2 N = 24 N.
29. A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20°C. If the temperature rises to 100°C what force will the rod exert on one of the clamps? Area of the cross-section of the rod = 2.00 mm². Coefficient of linear expansion of steel = 12.0x10⁻⁶/°C and Young's modulus of steel =2.00x10¹¹ N/m².
Answer: Let the force developed =F.
Area of cross-section =2.00 mm² =2.00x10⁻⁶ m²
Stress = F/2.00x10⁻⁶ N/m²
=5x10⁵F N/m²
Un restrained change in length,
ΔL = αTL
→ΔL/L =αT
→Strain = αT =12x10⁻⁶*(100-20)
=12*80*10⁻⁶
= 9.6x10⁻⁴
Stress/ strain = Elasticity
→5x10⁵F/9.6x10⁻⁴ = 2.00x10¹¹
→F = 2.00x10¹¹*9.6x10⁻⁴/5x10⁵ N
→F = 3.84x10² N =384 N.
30. Two steel rods and an aluminum rod of equal length lₒ and equal cross-section are joined rigidly at their ends as shown in the figure below. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficients of linear expansion of aluminum and steel are αₐ and αₛ respectively. Young's modulus of aluminum is Yₐ and of steel is Yₛ.
Figure for Q-30
Answer: First consider that only one end is joined and other is free to expand.
Free length at temperature θ,
For steel =lₒ(1+αₛθ)
For Aluminum =lₒ(1+αₐθ)
Diagram for Q-30
The free length of both metals will be different at θ, hence in the joined state one metal will be in tension and the other in compression. Such that both forces will be equal in magnitude but opposite in direction (P). Since the area of cross-section of steel-rods is twice the aluminum rod, the stress in aluminum =P/A and in the steel =P/2A. Where A is the area of cross-section of each rod.
So for aluminum, Stress/strain=Yₐ
→Strain = Stress/Yₐ
→ẟₐ/lₒ =P/AYₐ
→ẟₐ = Plₒ/AYₐ = contraction of aluminum.
Similarly elongation of steel
ẟₛ= Pl₀/2AYₛ
The sum of both contractions of aluminum and elongation of steel will be equal to the difference of the free lengths at θ. i.e.
ẟₐ+ẟₛ = lₒαₐθ-lₒαₛθ
→Plₒ/AYₐ +Plₒ/2AYₛ = lₒαₐθ-lₒαₛθ
→P/A =2YₐYₛ(αₐθ-αₛθ)/(Yₐ+2Yₛ)
Hence ẟₛ =2lₒYₐYₛ(αₐθ-αₛθ)/{2Yₛ(Yₐ+2Yₛ)}
=lₒYₐ(αₐθ-αₛθ)/(Yₐ+2Yₛ)
Free length of steel at θ=lₒ+lₒαₛθ
Hence final joined length of steel
=Free length+ẟₛ
=lₒ+lₒαₛθ+lₒYₐ(αₐθ-αₛθ)/(Yₐ+2Yₛ)
=lₒ[1+(αₛθYₐ+2αₛθYₛ+αₐθYₐ-αₛθYₐ)/(Yₐ+2Yₛ)]
=lₒ[1+(αₐYₐ+2αₛYₛ)θ/(Yₐ+2Yₛ)]
31. A steel ball initially at a pressure of 1.0x10⁵ Pa is heated from 20°C to 120°C keeping its volume constant. Find the pressure inside the ball. Coefficient of linear expansion of steel = 12x10⁻⁶/°C and bulk modulus of steel = 1.6x10¹¹ N/m².
Answer: Bulk modulus, B =p/(ΔV/V)
→ΔV =pV/B
{p= pressure, V=volume, ΔV=change in volume}
α=Coefficient of linear expansion of steel=12x10⁻⁶/°C
ɣ=Coefficient of volumetric expansion =3α =36x10⁻⁶/°C
B=1.6x10¹¹ N/m² (Given)
So, ΔV = ɣV(120°-20°)
→pV/B=ɣV*100
→p=ɣB*100
→p=36x10⁻⁶*1.6x10¹¹*100
→p=5.76x10⁸ Pa.
32. Show the moment of inertia of a solid body of any shape changes with temperature as I = Iₒ(1+2αθ), where Iₒ = is the moment of inertia at 0°C and α is the coefficient of linear expansion of the solid.
Answer: The moment of inertia of a solid body, Iₒ =Σmᵢrᵢ², where rᵢ is the distance of a particle from the axis at 0°C.
At temperature θ, the distance now becomes rᵢ'.
rᵢ' =rᵢ(1+αθ)
Now moment of inertia at temperature θ, I =Σmᵢrᵢ'² =Σmᵢrᵢ²(1+αθ)²
→I =(Σmᵢrᵢ²)(1+2αθ+α²θ²)
{since α² is a very very small quantity, the term α²θ² may be neglected}
→I = Iₒ(1+2αθ). Proved.
33. A torsional pendulum consists of a solid disc connected to a thin wire (α=2.4x10⁻⁵/°C) at its center. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).
Answer: If the torsional constant of the wire be = k, the time period of the pendulum T at 0°C = 2π√(Iₒ/k)
where Iₒ is the moment of inertia at 0°C.
At 5°C, the moment of inertia, I'=Iₒ(1+2αθ) =Iₒ(1+2*5α) =Iₒ(1+10α)
At 45°C, the moment of inertia,
I" = Iₒ(1+2α*45) =Iₒ(1+90α)
At 5°C time period, T=2π√(I'/k)
At 45°C time period,
T' =2π√(I"/k)
The difference in time period =T'-T
Percentage change in the time period
=(T'-T)*100/T
=(T'/T - 1)*100
={(√I"/k)/√(I'/k)-1}*100
={√(I"/I')-1}*100
={√Iₒ√(1+90α)/√Iₒ√(1+10α) -1}*100
=√(1+90*2.4x10⁻⁵)/√(1+10*2.4x10⁻⁵ -1)*100
=100*{(√(1.00216/1.00024)-1}
=100*{1.00096 - 1}
=0.096
=9.6x10⁻²
34. A circular disk made of iron is rotated about its axis at a constant velocity ⍵. Calculate the percentage change in the linear speed of a particle of the rim as the disk is slowly heated from 20°C to 50°C keeping the angular velocity constant. Coefficient of linear expansion of iron = 1.2x10⁻⁵/°C.
Answer: Given, α = 1.2x10⁻⁵/°C
Temperature difference, θ =50°-20° =30°C.
The linear speed of the particle,
v =⍵r, where r is the distance of the particle from the axis of rotation.
When the temperature changes, the distance r changes due to linear thermal expansion. The changed distance,
r' = r+rαθ
Now the linear speed v'=⍵r'
→v' = ⍵r+⍵rαθ
Now the percentage change in the linear speed = (v'-v)*100/v
=100(v'/v - 1)
=100{(⍵r+⍵rαθ)/⍵r - 1}
=100{1 +αθ -1}
=100αθ
=100*1.2x10⁻⁵*30
=3600x10⁻⁵
=3.6x10⁻².
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CHAPTER- 21 - Speed of Light
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CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
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