In the previous section we saw some problems in thermal expansion. In this section we will see anomalous behaviour of water. We will also see thermal stress
• But first we will see some special features of the volume expansion in gases. It can be written in 9 steps:
1. We have the ideal gas equation: PV = nRT
2. Imagine that, temperature is increased while keeping the pressure constant
• Then we get two changes:
♦ a change in volume, ΔV
✰ This is given by: ΔV = (V2 - V1)
♦ a change in temperature ΔT
✰ This is given by: ΔT = (T2 - T1)
3. Forming two equations:
• R is already a constant
• In our present case, n and P are also constants
• So we get two equations:
(i) PV1 = nRT1
(ii) PV2 = nRT2
4. Let us subtract 'PV2' from both sides of (i)
• We get: (PV1 - PV2) = (nRT1- PV2)
5. But from (ii), we have PV2 = nRT2
• So we will change the PV2 on the right side of (4)
• We get: (PV1 - PV2) = (nRT1-nRT2)
⇒ P(V1 - V2) = nR(T1-T2)
⇒ PΔV = nRΔT
6. Dividing both sides by the original volume V1, we get:
$\mathbf\small{\rm{\frac{P \Delta V}{V_1}=\frac{nR \Delta T}{V_1}}}$
⇒ $\mathbf\small{\rm{\frac{\Delta V}{V_1}=\frac{nR \Delta T}{PV_1}}}$
7. But from 3(i), we have: PV1 = nRT1
• So we can change the denominator in the right side. We get:
$\mathbf\small{\rm{\frac{\Delta V}{V_1}=\frac{nR \Delta T}{nRT_1}}}$
⇒ $\mathbf\small{\rm{\frac{\Delta V}{V_1}=\frac{\Delta T}{T_1}}}$
8. But from Eq.11.3, we have: $\mathbf\small{\rm{\frac{\Delta V}{V_1}=\alpha_V \Delta T}}$
• So the result in (7) becomes:
$\mathbf\small{\rm{\alpha_V \Delta T=\frac{\Delta T}{T_1}}}$
⇒ $\mathbf\small{\rm{\alpha_V=\frac{1}{T_1}}}$
9. So we can write:
• For an ideal gas, 𝛼V will depend on the initial temperature at which we begin to heat the gas (at constant pressure)
♦ If the initial temperature is high, 𝛼V will be low
♦ If the initial temperature is low, 𝛼V will be high
Anomalous behaviour of water
• Anomalous means: deviating from what is standard, normal, or expected
• Such a behaviour of water can be explained in 15 steps:
1. We know that, when temperature increases, volume of most substances increases
• But in the case of water, the opposite happens between 0 oC and 4 oC
2. In the graph in fig.11.5 below,
• Temperature is plotted along the x-axis
• Volume of 1 kg of water is plotted along the y-axis
 |
| Fig.11.5 |
3. Imagine that, starting from the origin, we are moving towards the right along the x-axis
• So the temperature of the 1 kg water will be increasing from 0 oC
4. When the temperature increases, we expect the volume also to increase
• But here, the volume decreases
5. The decreasing trend continues upto 4 oC
• So the volume will be minimum at 4 oC
• So the red segment of the graph denotes the decreasing trend
6. Once the temperature increases above 4 oC, the volume begins to increase
• This increasing trend is denoted by the yellow segment of the graph
7. We can write in the reverse direction also:
• Imagine that, starting from the room temperature (say 31 oC), we are moving towards the left along the x-axis
• So the temperature of the 1 kg water will be decreasing from 31 oC
8. When the temperature decreases, we expect the volume also to decrease
• Here, the volume indeed decreases
• This is denoted by the yellow segment of the graph. The decreasing trend continues upto 4 oC
• So the volume will be minimum at 4 oC
10. Once the temperature decreases below 4 oC, the volume begins to increase
• This increasing trend is denoted by the red segment of the graph
11. So in either case, the least possible volume of water occurs at 4 oC
12. Now, we consider density
• We know that, density is inversely proportional to volume
• So we can write:
♦ When temperature of water increases from 0 oC,
♦ the density goes on increasing
♦ the maximum density is attained at 4 oC
♦ This is denoted by the red segment in fig.b
13. After 4 oC, the density decreases
• This is denoted by the yellow segment in fig.b
14. When the temperature of water decreases (from say 31 oC)
♦ The density goes on increasing
♦ the maximum density is attained at 4 oC
♦ After 4 oC, the density decreases
15. So in either case, the maximum possible density of water occurs at 4 oC
1. Consider the top surface of a lake
• In cold weather, the temperature of this top surface goes on decreasing from say 31 oC
2. The volume also goes on decreasing
• So density goes on increasing
3. The denser water at the top layer sinks to the bottom
♦ The layer which was just below the top layer rises to the top
♦ This layer also cools down and sinks to the bottom
♦ This process continues and eventually, the whole lake will become cold
4. But even after becoming cold, the temperature goes on decreasing
• A stage will be reached where the temperature of the top most layer falls just below 4 oC
• At that stage, that top layer will begin to expand
5. Due to expansion, density becomes low
• So the top most layer will not sink down
6. It remains at the top and gets colder and colder to become ice
• Thus a layer of ice will be formed at the top of the lake
7. Below the ice layer, liquid water will be present
• Fish and other aquatic life can survive in this water
• Thus the anomalous behaviour of water helps to protect aquatic life in winter
Next, we will see the basics of thermal stress. It can be written in 5 steps:
1. In fig.11.6(a) below, a metal rod is fixed between two rigid supports
 |
| Fig.11.6 |
2. Let us increase the temperature of the rod
• The rod will try to increase in length
3. But such a linear expansion is prevented by the rigid supports
4. The rod applies a force on either supports
• The supports apply equal and opposite force on the rod
• So the free body diagram of the rod will be as shown in fig.b
5. Due to the forces shown in fig.b, the rod will be under stress
• This stress is called thermal stress
Let us see an application of thermal stress. it can be written in 6 steps:
1. Consider two steel rails AB and BC in fig.11.6(c)
• Let each have a length of 5 m and cross sectional area of 40 cm2
2. Suppose that, the surrounding temperature increase by 10 oC
• The length of the rails will increase. So there will be a linear strain
3. We can calculate the linear strain in just 3 steps:
(i) We have Eq.11.1: Δl = 𝛼L l1 Δt
(ii) This can be rearranged as: $\mathbf\small{\rm{\frac{\Delta l}{l}=\alpha_L \Delta t}}$
(iii) So the linear strain, $\mathbf\small{\rm{\frac{\Delta l}{l}}}$ = 1.2 × 10-5 ×10 = 1.2 × 10-4.
4. There will be a large number of rails connected together in a railway track
• So the rails are not free to expand
• Since the expansion is prevented, compressive forces will develop. This is shown in fig.11.6(d)
5. Now we apply Young's modulus of elasticity
• We have: $\mathbf\small{\rm{\frac{Linear\;stress}{Linear \, strain}= Young's \,modulus}}$
⇒ Linear stress = Linear strain × Young's modulus
• Substituting the values, we get:
Linear stress = (1.2 × 10-4) × (2 × 1011) = 2.4 × 107 N m-2
6. But $\mathbf\small{\rm{Stress=\frac{Force}{Area}}}$
• So we get: Force = Stress × area =(2.4 × 107) × (40 × 10-4) = 96 × 103 N
• This much force can easily bend the rails at the joint
• So special provisions have to be made at the joint. We will see those special provisions in higher classes
Now we will see some solved examples
Solved example 11.8
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:
• The brass wire is held taut with out any tension
• That means:
There is no sagging. The wire is pulled from two ends up to the stage when it becomes just horizontal. Further pulling will induce tension. But in our present case, there is no further pulling
• But when the wire is cooled, it will contract
• But the contraction is prevented. Since the contraction is prevented, tensile force will be induced in the wire. We are asked to calculate this tensile force
1. Change in temperature, Δt = (27 - (-39)) = 66 oC
2. We have: $\mathbf\small{\rm{\frac{\Delta l}{l}=\alpha_L \Delta t}}$
• So the linear strain, $\mathbf\small{\rm{\frac{\Delta l}{l}}}$ = 2.0 × 10-5 × 66 = 132 × 10-5.
3. Now we apply Young's modulus of elasticity
• We have: $\mathbf\small{\rm{\frac{Linear\;stress}{Linear \, strain}= Young's \,modulus}}$
⇒ Linear stress = Linear strain × Young's modulus
• Substituting the values, we get:
Linear stress = (132 × 10-5) × (0.91 × 1011) = 120.12 × 106 N m-2
6. But $\mathbf\small{\rm{Stress=\frac{Force}{Area}}}$
• So we get: Force = Stress × area =(120.12 × 106) × (π × 12 × 10-6) = 377.52 N
Solved example 11.9
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:
1. Change in temperature, Δt = (250 - 40) = 210 oC
2. We have Eq.11.1: Δl = 𝛼L l1 Δt
• So we can write:
ΔlB = 𝛼L(B) l1(B) Δt = (2.0 × 10-5) × (50) × (210) = 21000 × 10-5 cm
ΔlS = 𝛼L(S) l1(S) Δt = (1.2 × 10-5) × (50) × (210) = 12600 × 10-5 cm
3. Thus total expansion of the combined rod
= (21000 + 12600) × 10-5 = 33600 × 10-5 cm = 0.34 cm
4. The ends are free to expand. So there will not be any thermal stress
Bimetallic strip
The working of a bimetallic strip can be explained in 15 steps:
1. In fig.11.7(a) below, two strips of metal are shown
♦ The upper strip is made of iron
♦ The lower strip is made of copper
 |
| Fig.11.7 |
2. There are two important bonds in the fig.
(i) The bond between the two strips
• This bond is so strong that, there is no slipping between them even at high temperatures
(ii) The bond between the combined strips and the support
This bond is so strong that, the combined strips is held tightly at by the support even at high temperatures
3. When the surrounding temperature changes, the strips bend together
♦ They can bend upwards as in fig.b
♦ They can bend downwards as in fig.c
• We want to know which bending occurs
• The following steps from (4) to (15) will give us the answer
4. In fig.11.7(a), both the strips have the same length l
• Let us increase the surrounding temperature by Δt
♦ The steel strip will increase in length by: ΔlS = 𝛼L(S) l Δt
♦ The copper strip will increase in length by: ΔlC = 𝛼L(C) l Δt
5. We see that, the only difference is in the 𝛼L values
• We have:
♦ 𝛼L(S) = 1.2 × 10-5 K-1
♦ 𝛼L(C) = 1.7 × 10-5 K-1
6. 𝛼L(C) is larger. So it is clear that, ΔlC will be larger
• The final lengths are: (l + ΔlS) and (l + ΔlC)
• It is clear that, the final length of copper will be larger
7. Now consider fig.b
♦ The two strips have bent together
♦ They are part of an arc
♦ The arc is part of a circle
8. O is the center of the circle and the red dashed line is the radius of that circle
• Copper strip is in the outer periphery of the circle
• Applying geometrical principles. we can write:
♦ Strip in the outer periphery
♦ will have a greater length
♦ than the strip in the inner periphery
9. So we can write:
• When the surrounding temperature of a bimetallic strip increases, the strip will bent in such a way that:
The metal having greater 𝛼L is on the convex side
10. Let us decrease the surrounding temperature
♦ The steel strip will decrease in length by: ΔlS = 𝛼L(S) l Δt
♦ The copper strip will decrease in length by: ΔlC = 𝛼L(C) l Δt
11. We see that, the only difference is in the 𝛼L values
• We have:
♦ 𝛼L(S) = 1.2 × 10-5 K-1
♦ 𝛼L(C) = 1.7 × 10-5 K-1
12. 𝛼L(C) is larger. So it is clear that, ΔlC will be larger
• The final lengths are: (l - ΔlS) and (l - ΔlC)
• It is clear that, the final length of copper will be smaller
13. Now consider fig.c
♦ The two strips have bent together
♦ They are part of an arc
♦ The arc is part of a circle
14. O is the center of the circle and the red dashed line is the radius of that circle
• Copper strip is in the inner periphery of the circle
• Applying geometrical principles. we can write:
♦ Strip in the inner periphery
♦ will have a lesser length
♦ than the strip in the outer periphery
15. So we can write:
• When the surrounding temperature of a bimetallic strip decreases, the strip will bent in such a way that:
The metal having greater 𝛼L is on the concave side
In the next section, we will see specific heat
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