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:
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| 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 Cp
• If the experiment is done as in 16(ii), the result is called molar specific heat capacity at constant volume
♦ It’s symbol is Cv
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.,
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| 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
• 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
In the next section, we will see Change of phase
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