Chapter 11.9 - Newton's Law of Cooling

In the previous section we saw heat transfer by convection and radiation. In this section, we will see Newton's Law of cooling

Cooling is same as 'loss of heat'. Some basics about loss of heat can be written in 4 steps:
1. Consider a cup of tea placed on the table
• We know that, as time passes, the tea will become lesser and lesser hot
    ♦ This is because, heat flows from the tea into the surroundings
    ♦ That means, the tea loses heat
2. We want to know the rate at which heat is lost
‘Rate at which heat is lost’ can be explained in 4 steps:
(i) Let us record the temperature of the tea at various instances using a thermometer and a stop-watch
• Let the temperature of the tea be:
    ♦ T₁ when the reading in the stop-watch is t₁
    ♦ T₂ when the reading in the stop-watch is t₂
(ii) Then the differences can be calculated as:
    ♦ difference in temperature = (T₁ - T₂)
    ♦ difference in time = (t₂-t₁)
(iii) The difference in temperature (T₁ - T₂) can be related to ‘difference in heat contents’ using the specific heat capacity of the tea
• The specific heat capacity is a constant. So the difference (T₁- T₂) will give an indication about the ‘heat loss’ during the time interval (t₂- t₁)
(iv) So the rate at which heat is lost will be equal to $\mathbf\small{\rm{\frac{T_1-T_2}{t_2-t_1}}}$
3. But to confirm the result in 2(iv), we have to repeat the experiment
• We will repeat the experiment on the same tea
• The tea is continuously loosing heat. We will take one more set of readings:
(i) Let the temperature of the tea be:
    ♦ T₃ when the reading in the stop-watch is t₃
    ♦ T₄ when the reading in the stop-watch is t₄
(ii) Then the differences can be calculated as:
    ♦ difference in temperature = (T₃ - T₄)
    ♦ difference in time = (t₄ - t₃)
(iii) As before, the difference (T₃- T₄) will give an indication about the ‘heat loss’ during the time interval (t₄ - t₃)
(iv) So the rate at which heat is lost will be equal to $\mathbf\small{\rm{\frac{T_3-T_4}{t_4-t_3}}}$
4. Here we see a problem:
• The result in 2(iv) will not be equal to result in 3(iv)
• That means:
The tea is losing heat at different rates
• Such a situation needs further investigation. So we will do an experiment in the lab

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The experiment can be explained in 22 steps:
1. Take 300 mL water in a calorimeter
• Cover it with a two holed lid
    ♦ Put a stirrer through one hole
    ♦ Put a thermometer through the other hole
          ✰ Make sure that, the bulb of the thermometer is immersed in water
2. Note the reading of the thermometer
    ♦ This reading will be the temperature of the surroundings
    ♦ We will denote it as T_s
3. Heat the water until it’s temperature becomes about (T_s + 40) °C
• Stop heating by removing the heat source
4. Start the stop-watch
• The initial reading in the stop-watch will be 0 s
    ♦ We will denote this instant as t₍₀₎
• At that instant t₍₀₎, take the reading in the thermometer
    ♦ Let it be T₍₀₎
• So at the instant t₍₀₎, the water is at a temperature of (T₍₀₎-T_s) higher than the surrounding temperature
5. Stir the water gently using the stirrer
Continue stirring until the reading in the stop-watch is 60 s
    ♦ We will denote this instant as t₍₆₀₎
• At that instant t₍₆₀₎, take the reading in the thermometer
    ♦ Let it be T₍₆₀₎
• So at the instant t₍₆₀₎, the water is at a temperature of (T₍₆₀₎-T_s) higher than the surrounding temperature
6. Continue stirring until the reading in the stop-watch is 120 s
    ♦ We will denote this instant as t₍₁₂₀₎
• At that instant t₍₁₂₀₎, take the reading in the thermometer
    ♦ Let it be T₍₁₂₀₎
• So at the instant t₍₁₂₀₎, the water is at a temperature of (T₍₁₂₀₎-T_s) higher than the surrounding temperature
7. Continue stirring until the reading in the stop-watch is 180 s
    ♦ We will denote this instant as t₍₁₈₀₎
• At that instant t₍₁₈₀₎, take the reading in the thermometer
    ♦ Let it be T₍₁₈₀₎
• So at the instant t₍₁₈₀₎, the water is at a temperature of (T₍₁₈₀₎-T_s) higher than the surrounding temperature
8. As the water is losing heat, T₍₀₎, T₍₆₀₎, T₍₁₂₀₎ . . . etc., will be progressively decreasing
• So (T₍₀₎-T_s), (T₍₆₀₎-T_s), (T₍₁₂₀₎-T_s) . . . etc., will be progressively decreasing
9. Continue the procedure a total of n times until (T_(n)-T_s) is about 5 °C
• That means, the last reading is taken at the instant when the water is about 5 °C above the surrounding temperature
    ♦ Recall that, at the first instant, the difference (T₍₀₎-T_s) was about 40 °C
10. We can tabulate the readings as shown below:

11. Now we can plot a graph
    ♦ The values in the first column of the table are plotted along the x-axis
    ♦ The values in the third column of the table are plotted along the y-axis
• The result is the red curve shown in fig.11.20(a) below:

Fig.11.20

• The following steps from (12) to (22) will give us some interesting information about the above graph:
12. On the x-axis, mark any four points P, Q, U and V in such a way that, PQ = UV
    ♦ This is shown in fig.11.20(b)
    ♦ Since PQ is equal to UV, we are taking two ‘equal time duration’
13. Draw vertical green dashed lines through those four points
    ♦ The vertical dashed lines will intersect the red curve
14. Through those points of intersection, draw horizontal green dashed lines
    ♦ The horizontal dashed lines will intersect the y-axis at P’, Q’, U’ and V’
15. Consider the distance between P’ and Q’
• P’ is the temperature: T_(P) – T_s
    ♦ It is the temperature of the water (above surrounding temperature) when time is P
• Q’ is the temperature: T_(Q) – T_s
    ♦ It is the temperature of the water (above surrounding temperature) when time is Q
• So we get:
Distance between P’ and Q’
= [(T_(P) – T_s) - (T_(Q) – T_s)] = [T_(P) – T_(Q)]
• So P’Q’ indicates the heat lost in the duration PQ
16. Similarly, we can prove:
U’V’ indicates the heat lost in the duration UV
17. The duration PQ and duration UV are the same
• But we see that, the heat loss U’V’ is far lass than the heat loss P’Q’
18. We can write in the terms of rate at which heat is lost:
• Rate at which heat is lost during PQ = $\mathbf\small{\rm{\frac{P'Q'}{PQ}}}$
• 'Rate at which heat is lost' is same as 'Rate of cooling'
• So we get:
Rate of cooling during PQ = $\mathbf\small{\rm{\frac{P'Q'}{PQ}}}$
19. Similarly, we get:
Rate of cooling during UV = $\mathbf\small{\rm{\frac{U'V'}{UV}}}$
20. The denominators in (18) and (19) are the same
• The numerator in (18) is larger than the numerator in (19)
    ♦ So the result in (18) will be larger than the result in (19)
• So we can write:
    ♦ Rate of cooling during PQ
    ♦ is greater than
    ♦ Rate of cooling during UV
21. We can write this in two steps:
(i) In the initial stages, when the temperature is high,
    ♦ More heat is lost in any given duration
(ii) In the later stages, when the temper is low,
    ♦ Less heat is lost in that same duration
22. In other words:
• In the initial stages, when the 'temperature difference' is high,
    ♦ Rate of cooling is high
• In the later stages, when the 'temperature difference' is low,
    ♦ Rate of cooling is low

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Now we can write about Newton's Law of cooling. It can be written in 6 steps:
1. We see that, ‘rate of cooling’ is not a constant. It changes continuously
2. Sir Isaac Newton discovered that, the rate depends on the difference between two temperature values:
(i) The temperature of the body
    ♦ We can denote it as T
(ii) The temperature of the surroundings
    ♦ We can denote it as T_s
3. The rate is directly proportional to the difference (T - T_s)
• That means:
    ♦ If (T - T_s) is large, the rate will be large

          ✰ More heat will be lost in any given time duration

    ♦ If (T - T_s) is small, the rate will be small

          ✰ Less heat will be lost in that same given time duration

4. It is obvious that:
A large (T - T_s) indicates a hot body
• So it is obvious that:
A body will lose more heat when it is ‘more hot’ than when it is ‘less hot’ (even when the time intervals are the same)
5. In addition to the above information, Newton discovered that:
• The rate of cooling depends on two more factors:
    ♦ The nature of surface of the body
    ♦ The area of the surface of the body
• The above two items are constants. So they can be represented by a single constant K
• Thus we get Eq.11.14:
Rate of cooling of a body = K (T - T_s)
6. Each body will have a unique value for K
• If we can find K, we will be able to calculate the rate of cooling
• If we get the rate of cooling, we will be able to calculate the time required for a body to cool down to any required temperature
• The following two solved examples will demonstrate such a procedure

Solved example 11.31
A pan filled with hot food cools from 94 °C to 86 °C in 2 minutes when the room temperature is at 20 °C. How long will it take to cool from 71 °C to 69 °C?
Solution:
Data given is:
    ♦ Temperatures at the initial stage: 94 °C and 86 °C
    ♦ Time for cooling = 2 minutes
    ♦ Temperatures at the final stage: 71 °C and 69 °C
    ♦ Time for cooling = ?
    ♦ Surrounding temperature, T_s = 20 °C
Case 1: The food cools from 94 °C to 86 °C
1. When the food is at 94 °C, it is hot
• When it is hot at 94 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{94-86}{2 \times 60}=\frac{4}{60}}}$
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{94+86}{2}=90 \right )}}$
3. So this rate occurs when the difference in temperature is (90 - 20) = 70 °C
4. Applying Eq.11.14, we get: $\mathbf\small{\rm{\frac{4}{60}}}$ = K  × 70
• So K = $\mathbf\small{\rm{\frac{4}{60 \times 70}}}$
Case 2: The body cools from 71 °C to 69 °C
1. When the body is at 71 °C, it is less hot
• When it is less hot at 71 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{71-69}{t}=\frac{2}{t}}}$
    ♦ Where 't' is the time required to cool from 71 °C to 69 °C
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{71+69}{2}=70 \right )}}$
3. So this rate occurs when the difference in temperature is (70 - 20) = 50 °C
4. Applying Eq.11.14, we get: $\mathbf\small{\rm{\frac{2}{t}}}$ = K  × 50
• But from case 1, we have the value of K
• Substituting that value of K, we get:
$\mathbf\small{\rm{\frac{2}{t}=\frac{4}{60 \times 70} \times 50}}$
• So t = 42 s

Solved example 11.32
A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool
from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.
Solution:
Data given is:
    ♦ Temperatures at the initial stage: 80 °C and 50 °C
    ♦ Time for cooling = 5 minutes
    ♦ Temperatures at the final stage: 60 °C and 30 °C
    ♦ Time for cooling = ?
    ♦ Surrounding temperature, T_s = 20 °C
Case 1: The body cools from 80 °C to 50 °C
1. When the body is at 80 °C, it is hot
• When it is hot at 80 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{80-50}{5 \times 60}}}$ = 0.1
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{80+50}{2}=65 \right )}}$
3. So this rate occurs when the difference in temperature is (65 - 20) = 45 °C
4. Applying Eq.11.14, we get; 0.1 = K  × 45
• So K = $\mathbf\small{\rm{\frac{0.1}{45}}}$
Case 2: The body cools from 60 °C to 30 °C
1. When the body is at 60 °C, it is less hot
• When it is less hot at 60 °C, it will give off heat at a particular rate
• This rate will be given by:
$\mathbf\small{\rm{\frac{Change \, in \,temperature}{Time}=\frac{60-30}{t}=\frac{30}{t}}}$
    ♦ Where 't' is the time required to cool from 60 °C to 30 °C
2. This rate occurs when the body is at an average temperature of $\mathbf\small{\rm{\left ( \frac{60+30}{2}=45 \right )}}$
3. So this rate occurs when the difference in temperature is (45 - 20) = 25 °C
4. Applying Eq.11.14, we get; $\mathbf\small{\rm{\frac{30}{t}}}$ = K  × 25
• But from case 1, we have the value of K
• Substituting that value of K, we get:
$\mathbf\small{\rm{\frac{30}{t}=\frac{0.1}{45} \times 25}}$
• So t = 540 s = 9 min

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■ The following points should be noted:
• When the food in solved example 11.31 is at 94 °C, the ‘rate of cooling’ will have a particular value
• When the temperature falls to 93 °C, the rate of cooling will be having another value
• In this way, the rate continuously changes
• We took the average temperature to find K
• But it is possible to find instantaneous rates by applying calculus
• Such a method will give more accurate results. We will see it after completing the basic course in calculus

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We have completed the discussion in this chapter. In the next section, we will see some solved examples related to temperature scales

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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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