Chapter 11.3 - Specific Heat Capacity

In the previous section we saw thermal stress. In this section we will see specific heat capacity

To learn the basics about specific heat capacity, we do three experiments. They can be explained in 18 steps:
1. A schematic representation of the first experiment is shown in fig.11.8 below:

Fig.11.8

• Based on the first fig., we can write 3 steps:
(i) Heat m kg of water so that, it’s temperature rises by 20 _^oC
    ♦ Note down the time required for this 20 _^oC rise in temperature. Let it be t
(ii) Heat another m kg of water so that it’s temperature rises by 40 _^oC
    ♦ Use the same source of heat
    ♦ Note down the time required for this 40 _^oC rise in temperature
          ✰ This time will be about 2t
(iii) Remember that, we used the same source of heat
• So, since the time is doubled, we can write:
If Q units of heat energy is supplied in the first case, 2Q units of heat energy is supplied in the second case
■ It is clear that, the quantity of heat energy depends on the rise in temperature, ΔT
2. A schematic representation of the second experiment is also shown in fig.11.8 above
Based on the second fig., we can write 3 steps:
(i) Heat m kg of water so that, it’s temperature rises by 20 _^oC
    ♦ Note down the time required for this 20 _^oC rise in temperature. Let it be t
(ii) Heat 2m kg of water so that it’s temperature rises by 20 _^oC
    ♦ Use the same source of heat
    ♦ Note down the time required for this 20 _^oC rise in temperature
          ✰ This time will be about 2t
(iii) Remember that, we used the same source of heat
• So, since the time is doubled, we can write:
If Q units of heat energy is supplied in the first case, 2Q units of heat energy is supplied in the second case
■ It is clear that, the quantity of heat energy depends on the mass of the sample
3. A schematic representation of the third experiment is also shown in fig.11.8 above
• Based on the third fig., we can write 3 steps:
(i) Heat m kg of water so that, it’s temperature rises by 20 _^oC
    ♦ Note down the time required for this 20 _^oC rise in temperature. Let it be t
(ii) Heat m kg of oil so that it’s temperature rises by 20 _^oC
    ♦ Use the same source of heat
    ♦ Note down the time required for this 20 _^oC rise in temperature
          ✰ This time will be different from t
(iii) Remember that, we used the same source of heat
• So, since the time is different, we can write:
If Q units of heat energy is supplied in the first case, Q’ units of heat energy is supplied in the second case
■ It is clear that, the quantity of heat energy depends on the nature of the sample
4. So the ‘quantity of heat required’ to warm a substance depends on three items:
    ♦ Mass of the substance, m
    ♦ Change in temperature, ΔT
    ♦ Nature of the substance
5. To show the 'dependence on the three items mathematically', we introduce two quantities: Heat capacity and Specific heat capacity
6. Heat capacity can be described in 3 steps
(i) Q units of heat is supplied to a sample
(ii) The temperature of the sample increases by ΔT _^oC
(iii) So the heat required to increase the temperature by 1 _^oC will be equal to $\mathbf\small{\rm{\frac{Q}{\Delta T}}}$
• This heat is called heat capacity of the sample. It’s symbol is S
• So we get Eq.11.6: $\mathbf\small{\rm{S=\frac{Q}{\Delta T}}}$
7. Let us see the practical application of heat capacity. It can be written in 4 steps:
(i) We take various samples containing various substances
(ii) Heat each sample to increase the temperature by ΔT
• This ΔT need not be same for the samples
• If it is difficult to increase the temperature of a sample, a smaller ΔT can be adopted
(iii) Note down the Q required for achieving the ΔT for each sample
(iv) Then all we need to do is: Apply Eq.11.6 for each sample
• Thus we can obtain heat capacity (S value) of any sample
8. In calculating the S values in this way,
    ♦ we are not considering mass of substance
    ♦ we are not considering nature of substance
• Various samples will be having different S values
• The difference in S values may be due to either one or both of the two factors below:
    ♦ Difference in masses of the samples
    ♦ Difference in nature of the samples
9. So we need to be more accurate. For that, we introduce the quantity: Specific heat capacity
It can be explained in steps
(i) Q units of heat is supplied to a sample
(ii) The temperature of the sample increases by ΔT _^oC
(iii) So the heat required to increase the temperature by 1 _^oC will be equal to $\mathbf\small{\rm{\frac{Q}{\Delta T}}}$
(iv) So the heat required to increase the temperature of 1 kg of the substance by 1 _^oC will be equal to $\mathbf\small{\rm{\frac{Q}{m \Delta T}}}$
(m is the mass of the sample. It should be measured in kg) 
• This heat is called specific heat capacity of the sample. It’s symbol is s
• So we get Eq.11.7: $\mathbf\small{\rm{s=\frac{Q}{m \Delta T}}}$
10. Let us see the practical application of specific heat capacity. It can be written in 4 steps:
(i) We take various samples containing various substances
(ii) Heat each sample to increase the temperature by ΔT
• This ΔT need not be same for the samples
• If it is difficult to increase the temperature of a sample, a smaller ΔT can be adopted
(iii) Note down the Q required for achieving the ΔT for each sample
(iv) Then all we need to do is: Apply Eq.11.7 for each sample
• While reporting the results, write the name of the corresponding substances also
• By looking at such a report, a person can know this:
    ♦ Heat energy required
    ♦ to raise the temperature by 1 _^oC
    ♦ for 1 kg mass
    ♦ of a particular substance
• Thus, all required details are included
11. An important note can be written in 3 steps:
(i) In the above discussion, we did some experiments
    ♦ Each of those experiments start when we begin the supply of heat
    ♦ Each of those experiments end when we stop the supply of heat
(ii) It is compulsory that, from the start to the end, there must be no phase change
■ That means:
• If at the start, the sample is in solid state, the experiment should end before melting point is reached
• If at the start, the sample is in liquid state, the experiment should end before boiling point is reached
(iii) In later sections, we will see the reason for making this compulsory
12. Now we can write the definition of specific heat capacity of a substance
    ♦ It is the amount of heat absorbed or rejected
    ♦ by one kg of the substance
    ♦ while it’s temperature changes by 1 _^oC
    ♦ and not undergoing any phase change
13. Let us write the unit of s:
• From Eq.11.7, we have:
Unit of s = $\mathbf\small{\rm{\frac{(J)}{(kg)(K)}=J \,kg^{-1} \,K^{-1}}}$
14. We can derive another relation also:
• Dividing Eq.11.6 by 11.7, we get: $\mathbf\small{\rm{\frac{S}{s}= \frac{\frac{Q}{\Delta T}}{\frac{Q}{m \Delta T}}=m}}$
• Thus we get Eq.11.8: $\mathbf\small{\rm{s=\frac{S}{m}}}$
15. We have seen that, to find the specific heat capacity, we need to know the mass of the substance in the sample. Sometimes, instead of mass, we use ‘number of moles’, n
• In such cases, we divide $\mathbf\small{\rm{\frac{Q}{\Delta T}}}$ by n
(Recall that, earlier, we divided $\mathbf\small{\rm{\frac{Q}{\Delta T}}}$ by m)
• So instead of ‘per kg’, we get ‘per moles’
• The result is called molar specific heat capacity. It’s symbol is C
• So we can write Eq.11.9: $\mathbf\small{\rm{C=\frac{S}{n}=\frac{Q}{n \;\Delta T}}}$
16. Now we will see the C values of gases. The present step (16) and the next step (17) will help us to understand them
• To find the specific heat capacity, we heat the substance
• But if the substance is in the gaseous state, heating can be done in two ways:
(i) The gas is allowed to expand freely
    ♦ Then, the pressure experienced by the gas will be constant
    ♦ The volume will change
(ii) The gas is not allowed to expand freely. That is., volume is kept constant
    ♦ Then, the pressure experienced by the gas will change continuously
    ♦ Volume will remain constant
17. Two types of molar specific heat capacity for gases:
• If the experiment is done as in 16(i), the result is called molar specific heat capacity at constant pressure
    ♦ It’s symbol is C_p
• If the experiment is done as in 16(ii), the result is called molar specific heat capacity at constant volume
    ♦ It’s symbol is C_v
18. Comparison between various substances:
• We can obtain the specific heat capacity of various substances from the data book
    ♦ We see that, water has a very high specific heat capacity
• s for water is 4186 J kg_^-1 K_^-1
    ♦ This is much larger than that of iron or aluminum
          ✰ s of iron is 450 J kg_^-1 K_^-1
          ✰ s of aluminum is 900 J kg_^-1 K_^-1
■ Let us see two advantages of water having such a high specific heat capacity:
(i) Water can be used as a coolant in automobile radiators
• In the radiators, water absorb a large amount of heat. It’s temperature will rise only slowly. Because, 1 kg of water will need 4186 J of heat energy to increase temperature even by 1 _^oC
• So water will reach the boiling point only slowly
(ii) Water can be used as a heater in hot water bags
• This is because, water cools down only slowly
• 4186 J will have to be rejected to lower the temperature even by 1 _^oC

Solved example 11.10
400 J heat is given to 0.1 kg of a metal. Temperature of the metal increases by 20 _^oC. What is the specific heat capacity of the metal?
Solution:
1. 400 J of heat is required for a ΔT of 20 _^oC
• So, for a ΔT of 1 _^oC, (⁴⁰⁰⁄₂₀) = 20 J will be required
2. 20 J is required for a ΔT of 1 _^oC for a mass of 0.1 kg
• So, for a mass of 1 kg, (²⁰⁄_0.1)= 200 J will be required
3. s of the metal is 200 J kg_^-1 K_^-1
4. Alternatively, we can use Eq.11.7: $\mathbf\small{\rm{s=\frac{Q}{m \Delta T}}}$
• Substituting the known values, we get: $\mathbf\small{\rm{s=\frac{(400)}{(0.1)(20)}}}$ = 200 J kg_^-1 K_^-1.  

Calorimetry

• The word calorimetry means: heat measurement
    ♦ Calorie means heat energy
    ♦ Metry means measurement
• The basics of calorimetry can be written in 6 steps
1. Fig.11.9(a) shows a 3D space. It is colored green
• It is not a plane surface. It is a 3D space. It can be a class room, a lab room, a stadium etc.,

Fig.11.9

2. In this 3D space, we place a box made of insulating material
• This is denoted by the brown rectangle in fig.b
    ♦ Though it is shown as a rectangle, it is a 3D box
    ♦ The insulating material is present on all six sides
3. In fig.b, the green space can be considered as:
• The surroundings of the box
• Due to the presence of the insulating material on all the six sides,
    ♦ There is no passage of heat from the box into the green space (surroundings)
    ♦ There is no passage of heat from the surroundings into the box
4. Inside the box, there are only two items:
    ♦ A hot body shown in red color
    ♦ A cold body shown in white color
■ When the two bodies are brought into contact with each other, heat flows from the hot body to the cold body
■ That means:
    ♦ The hot body loses some heat
    ♦ The cold body gains some heat
5. Since there is no loss into the surroundings, we can write:
    ♦ Heat lost by the hot body
    ♦ is equal to
    ♦ Heat gained by the cold body
■ This is the principle of calorimetry
6. The hot body and the cold body are together known as a system
• We call it a ‘system’ because, there is more than one component
    ♦ The components work together to give the desired results
(For example, the computer system consists of four components: The monitor, CPU, keyboard and UPS. We get desired results only when the four components work together. If there is only one component, we need not call it a system. We can call it by it’s individual name)
• In our present case, the system consists of two components: The hot body and the cold body
■ Since there is no loss or gain of heat from the surroundings, we call it: an isolated system
• The ‘sides of the box’ forms the boundary of the system

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• Next we will see the application of calorimetry
• Using calorimetry, we can determine unknown s values (specific heat capacity values) of various materials
• Before doing such calculations, we will first see how the arrangement shown in fig.11.9(b) can be achieved in practice. It can be written in 5 steps
1. The box shown in fig.11.9(b) is made of wood
    ♦ The inside of the box is covered with a layer of insulating material
    ♦ Usually glass wool is used as the insulating material
    ♦ Some images can be seen here
2. The cold body consists of three items:
    ♦ A metallic vessel
    ♦ A liquid (usually water) contained inside the metallic vessel
    ♦ A stirrer (made of the same metal as that of the vessel)
          ✰ When water is stirred, heat will be uniformly distributed in the system
3. The hot body can be arranged by any one of the two methods given below:
(i) Heat a body outside the box, measure it’s temperature and quickly place it inside the water in the  metallic vessel
(ii) Place combustible material in a stainless steel chamber, place the chamber in the water in the metallic vessel and ignite using an electric fuse
4. A thermometer can be inserted through a hole in the wooden box
• In this way, the final temperature of the water can be measured
5. Thus the isolated system can be obtained in practice

■ Some solved examples will help us to get a good understanding about the procedure. Link to five such solved examples is given below:

Solved examples 11.11 to 11.16

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In the next section, we will see Change of phase

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Chapter 11.2 - Thermal stress

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, 𝛼_V will depend on the initial temperature at which we begin to heat the gas (at constant pressure)
    ♦ If the initial temperature is high, 𝛼_V will be low
    ♦ If the initial temperature is low, 𝛼_V 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

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Let us see an effect of this anomalous behaviour. It can be written in 7 steps:

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

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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

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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 cm_²
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 = 𝛼_L 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 × 10_¹¹) = 2.4 × 10_⁷ N m_^-2
6. But $\mathbf\small{\rm{Stress=\frac{Force}{Area}}}$
• So we get: Force = Stress × area =(2.4 × 10_⁷) × (40 × 10_^-4) = 96 × 10_³ 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

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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 × 10_¹¹ 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 × 10_¹¹) = 120.12 × 10_⁶ N m-2
6. But $\mathbf\small{\rm{Stress=\frac{Force}{Area}}}$
• So we get: Force = Stress × area =(120.12 × 10_⁶) × (π × 1_² × 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 = 𝛼_L l₁ Δt
• So we can write:
Δl_B = 𝛼_L(B) l_1(B) Δt = (2.0 × 10_^-5) × (50) × (210) = 21000 × 10_^-5 cm
Δl_S = 𝛼_L(S) l_1(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: Δl_S = 𝛼_L(S) l Δt
    ♦ The copper strip will increase in length by: Δl_C = 𝛼_L(C) l Δt
5. We see that, the only difference is in the 𝛼_L values
• We have:
    ♦ 𝛼_L(S) = 1.2 × 10^_-5 K^_-1
    ♦ 𝛼_L(C) = 1.7 × 10^_-5 K^_-1
6. 𝛼_L(C) is larger. So it is clear that, Δl_C will be larger
• The final lengths are: (l + Δl_S) and (l + Δl_C)
• 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 𝛼_L is on the convex side
10. Let us decrease the surrounding temperature
    ♦ The steel strip will decrease in length by: Δl_S = 𝛼_L(S) l Δt
    ♦ The copper strip will decrease in length by: Δl_C = 𝛼_L(C) l Δt
11. We see that, the only difference is in the 𝛼_L values
• We have:
    ♦ 𝛼_L(S) = 1.2 × 10^_-5 K^_-1
    ♦ 𝛼_L(C) = 1.7 × 10^_-5 K^_-1
12. 𝛼_L(C) is larger. So it is clear that, Δl_C will be larger
• The final lengths are: (l - Δl_S) and (l - Δl_C)
• 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 𝛼_L is on the concave side

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In the next section, we will see specific heat



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