In the previous section we saw that the freezing point of water is marked as 0 C in the Celsius scale and 32 F in the Fahrenheit scale. In this section we will continue to closely follow the works of the scientists.
13. The scientists continued their work:
• The first group of scientists marked a point B at some distance to the right of A. At that point B, they wrote: 100o C. This is shown in fig.6.6 below:
• They said: 'The heat content at that point is just high enough to make water boil'
 |
| Fig.6.6 |
• 'Just high enough' means that, if we place water any where to the left of B, it will not boil'
14. Now, the second group of scientists wrote 212o F at the point B. This is shown in fig.6.7 below:
• They also said the same: 'The heat content at that point is just high enough to make water boil'
 |
| Fig.6.7 |
15. The 'quantity of heat' at the point B is the same. But the two groups of scientists used different systems of measurements. That is why we have two different values : 100 and 212 at the point B
• Now we want to find a relation between the two systems. We will find the relation in steps:
1. Note the peculiarities of the two points:
• At A, the heat is just low enough to make the water freeze
• At B, the heat is just high enough to make the water boil
2. So we are looking at a specific region to the right side of A and to the left side of B.
• That is., we are interested in the region between A and B
3. Let us first consider the Celsius scale.
• We have zero at A and 100 at B
• Let us divide the region between A and B into 100 equal parts. So each part will be equivalent to 1o C. This is shown in the fig.6.8 below:
• In the fig., it is divided into 10 equal parts. This is for clarity. If we divide each of these parts further into 10 equal parts, we will get exact 100. But then the fig. will become crowded. For our present discussion, it is sufficient to base our discussion on 10 equal parts
■ Consider any two consecutive marks. Say 30 and 40
• We come to an important question:
■ What is the difference in the heat content between these two marks?
The question can be elaborated:
■ We know that 40 will be having a higher heat content than 30. But how much more heat is available at 40?
Let us find the answer:
• Consider the graph shown in fig.6.9 below. The magenta line is the graph showing uniform increase in the heat.
• Consider the red vertical lines corresponding to the points 10 and 20. The two red lines are of the same height.
• The height of these two red lines will give the heat content at point 10. But at 20, the heat content will be more than that at 10. So the red line is not enough for the heat at 20
♦ That is., when compared to 10, some 'extra heat' is available at 20.
• A small white vertical line is drawn above the red line at 20. This white line is the 'extra heat' available at 20.
• In the same way, the small white line at 40 is the extra heat available at 40, when compared to the heat at 30.
• The small white line at 80 is the extra heat available at 80, when compared to the heat at 70.
• We can see that, all the small white lines are of the same height. This can be proved mathematically using some very basic properties of triangles. The reader may try it himself/herself.
♦ We can prove it for any two consecutive points.
■ So we get an important result: The 'difference in heat content' between any two consecutive points is the same.
• The 'same difference' is obtained because two conditions are satisfied:
♦ The magenta coloured graph is a straight line without any bends or curves
♦ The 10 divisions between zero and 100 are all equal
• If both these conditions are not satisfied, we will not get the 'same difference'.
Based on this information, we can see the principle of a thermometer.
Consider the thermometer (Celsius scale) shown in fig.6.10 below:
• The bulb at the lower end is a reservoir for mercury.
• This bulb is brought into contact with the point at which temperature is to be measured.
• The heat is then transferred from that point to the mercury.
• The mercury then expands and rises up through the inner tube. Thus we will get the readings
• Suppose that the bulb is brought in contact with the '30 mark' in our rod in fig.6.9 above. Then the mercury in fig.6.10 will rise upto the 30o mark.
• Next, the bulb is brought into contact with the '40 mark' in our rod in fig.6.9.
• Here the heat content is more. So more heat will flow into the mercury. The mercury will expand further and reach the 40o mark.
• Next, the bulb is brought into contact with the '50 mark' in our rod in fig.6.9.
• Here the heat content is more. So more heat will flow into the mercury. The mercury will expand further and reach the 50o mark.
• Now, the 'extra heat' between 30 and 40 in fig 6.9 caused the mercury to rise up from 30o to 40o in fig.6.10
• We have seen that in fig.6.9, 'extra heat' between 30 and 40 will be same as the 'extra heat' between 40 and 50
• So, the following two quantities are the same:
♦ amount of expansion from 30 to 40
♦ amount of expansion from 40 to 50
This is shown in the fig.6.11 below:
So the markings in the thermometer should be at equal intervals
■ Now we will continue with the main discussion: To find the relation between Celsius scale and Fahrenheit scale.
4. We will show the Fahrenheit scale also along with the Celsius scale. So the fig.6.8 above has to be modified. The modified fig.6.12 is shown below:
5. Note the peculiar way in which the markings are made in the Fahrenheit scale.
• The region between 32 and 212 is divided into 10 equal parts.
• But the divisions are marked as 50, 68, 86, . . . Instead of 50, 60, 70, . . .
• Why is this so?
We have:
• In the Celsius scale, (boiling point - freezing point) = (100 - 0) = 100
• In the Fahrenheit scale, (boiling point - freezing point) = (212 - 32) = 180
• So the region between 0 and 100 in the Celsius scale is divided into 100 equal parts
• In the same way, the region between 32 and 212 in the Fahrenheit scale must be divided into 180 equal parts
♦ For getting 180 equal parts, the region is first divided into 10
♦ Then each of these 10 is further divided into 18 equal parts
• So the first mark after 32 is (32+18) = 50. The mark after 50 is (50+18) = 68
and so on . . .
• The two thermometers can be shown together as below:
This fig. is taken from wikimedia commons. The original can be seen here.
6. Let Qc be the quantity of heat required for the rise in each small unit in the Celsius scale
• Then total heat quantity from 0o C to 100o C = Qc ×100 = 100 Qc
7. Let Qf be the quantity of heat required for the rise in each small unit in the Fahrenheit scale
• Then total heat quantity from 32o F to 212o F = Qf ×180 = 180 Qf
8. The quantities in (6) and (7) are the same. So we can write:
Eq.6.1:
100 Qc = 180 Qf
⇒ 5 Qc = 9 Qf ⇒ Qc = (9⁄5) Qf
9. From this result, we can note some interesting points:
• In the Celsius scale, the region is divided into 100 equal units. But in the Fahrenheit scale, the same region is divided into 180 equal units
• So the units in the Fahrenheit scale will be smaller than the units in the Celsius scale.
• We find that this is true when we see the result in (8): Qf is to be multiplied by a factor which is 'greater than 1' to get Qc
10. The result in (8) is the basic relation between the two scales. We can use it to convert from one scale to another. Let us see an example:
• The Celsius thermometer in a room indicates that the temperature in that room is 28o C. How much is it in Fahrenheit?
Solution:
1. The reading is 28o C. So it is 28 units above zero in the Celsius scale
2. The value corresponding to 0o C in the Fahrenheit scale is 32o F.
3. Since 28 is above zero, the reading corresponding to 28 in the Fahrenheit scale will be above 32
• We want to know 'how many units' above 32
4. We have seen that each unit in Fahrenheit scale is smaller than the units in Celsius scale.
• So the heat producing a rise of one unit in Celsius scale will produce a rise of more than one unit in Fahrenheit scale. We have seen that it is (9⁄5) times
• So the rise of 28 units in the Celsius scale will cause a rise of [28 × (9⁄5)] = 50.4 units in the Fahrenheit scale
5. The '28 units' is above zero in the Celsius scale
• So the '50.4 units' is above 32 in the Fahrenheit scale
• Thus the reading in Fahrenheit scale is (32+50.4) = 82.4 F
Based on this example we can write a general form:
If C is the reading in Celsius scale, the corresponding reading in Fahrenheit scale will be [32 + (9⁄5) × C]
So if F is the reading corresponding to C, we can write:
Eq.6.2:
Next we will see how to convert Fahrenheit to Celsius. Here also we will use an example
• The Fahrenheit thermometer in a room indicates that the temperature in that room is 95o F. How much is it in Celsius?
Solution:
1. The reading is 95o F. So it is (95-32) = 63 units above 32 in the Fahrenheit scale
2. The value corresponding to 32o F in the Celsius scale is 0o C.
3. Since 95 is above 32, the reading corresponding to 95 in the Celsius scale will be above 0
• We want to know 'how many units' above 0
4. Let it be 'C' units above zero
5. We have seen that each unit in Fahrenheit scale is smaller than the units in Celsius scale.
• So the heat producing a rise of one unit in Celsius scale will produce a rise of more than one unit in Fahrenheit scale. We have seen that it is (9⁄5) times
• So the rise of C units in the Celsius scale will cause a rise of [C × (9⁄5)] units in the Fahrenheit scale
6. In this example, [C × (9⁄5)] = 63 = (95-32) = (F-32)
So C = 63 × (5⁄9) = 35
7. In the above step, 63 = (95-32) = [(given value of F)-32]
So we can write a general form:
C = (F-32)× (5⁄9)
• This can be written as:
Eq.6.3:
Note that, the above equation can be obtained by rearranging Eq.6.2 also
• After reading the basics about Fahrenheit scale, the reader may naturally wonder:
■ If 32 is the freezing point of water, what is zero?
• To find the answer, we have to look into some history:
• The Fahrenheit scale was invented by a German scientist Daniel Gabriel Fahrenheit., who lived and worked in Amsterdam, Netherlands.
• He invented the scale in the 1720s. At that time, the lowest temperature achievable in his laboratory was the temperature of a 'special mixture'.
• It was a mixture of salt, water and ammonium chloride. It had a heat content much lower than the freezing point of ordinary water.
• As it was the lowest possible temperature, he assigned the number 'zero' to it.
• At A, the heat is just low enough to make the water freeze
• At B, the heat is just high enough to make the water boil
2. So we are looking at a specific region to the right side of A and to the left side of B.
• That is., we are interested in the region between A and B
3. Let us first consider the Celsius scale.
• We have zero at A and 100 at B
• Let us divide the region between A and B into 100 equal parts. So each part will be equivalent to 1o C. This is shown in the fig.6.8 below:
 |
| Fig.6.8 |
■ Consider any two consecutive marks. Say 30 and 40
• We come to an important question:
■ What is the difference in the heat content between these two marks?
The question can be elaborated:
■ We know that 40 will be having a higher heat content than 30. But how much more heat is available at 40?
Let us find the answer:
• Consider the graph shown in fig.6.9 below. The magenta line is the graph showing uniform increase in the heat.
 |
| Fig.6.9 |
• The height of these two red lines will give the heat content at point 10. But at 20, the heat content will be more than that at 10. So the red line is not enough for the heat at 20
♦ That is., when compared to 10, some 'extra heat' is available at 20.
• A small white vertical line is drawn above the red line at 20. This white line is the 'extra heat' available at 20.
• In the same way, the small white line at 40 is the extra heat available at 40, when compared to the heat at 30.
• The small white line at 80 is the extra heat available at 80, when compared to the heat at 70.
• We can see that, all the small white lines are of the same height. This can be proved mathematically using some very basic properties of triangles. The reader may try it himself/herself.
♦ We can prove it for any two consecutive points.
■ So we get an important result: The 'difference in heat content' between any two consecutive points is the same.
• The 'same difference' is obtained because two conditions are satisfied:
♦ The magenta coloured graph is a straight line without any bends or curves
♦ The 10 divisions between zero and 100 are all equal
• If both these conditions are not satisfied, we will not get the 'same difference'.
Consider the thermometer (Celsius scale) shown in fig.6.10 below:
 |
| Fig.6.10 |
• This bulb is brought into contact with the point at which temperature is to be measured.
• The heat is then transferred from that point to the mercury.
• The mercury then expands and rises up through the inner tube. Thus we will get the readings
• Suppose that the bulb is brought in contact with the '30 mark' in our rod in fig.6.9 above. Then the mercury in fig.6.10 will rise upto the 30o mark.
• Next, the bulb is brought into contact with the '40 mark' in our rod in fig.6.9.
• Here the heat content is more. So more heat will flow into the mercury. The mercury will expand further and reach the 40o mark.
• Next, the bulb is brought into contact with the '50 mark' in our rod in fig.6.9.
• Here the heat content is more. So more heat will flow into the mercury. The mercury will expand further and reach the 50o mark.
• Now, the 'extra heat' between 30 and 40 in fig 6.9 caused the mercury to rise up from 30o to 40o in fig.6.10
• We have seen that in fig.6.9, 'extra heat' between 30 and 40 will be same as the 'extra heat' between 40 and 50
• So, the following two quantities are the same:
♦ amount of expansion from 30 to 40
♦ amount of expansion from 40 to 50
This is shown in the fig.6.11 below:
 |
| Fig.6.11 |
4. We will show the Fahrenheit scale also along with the Celsius scale. So the fig.6.8 above has to be modified. The modified fig.6.12 is shown below:
 |
| Fig.6.12 |
• The region between 32 and 212 is divided into 10 equal parts.
• But the divisions are marked as 50, 68, 86, . . . Instead of 50, 60, 70, . . .
• Why is this so?
We have:
• In the Celsius scale, (boiling point - freezing point) = (100 - 0) = 100
• In the Fahrenheit scale, (boiling point - freezing point) = (212 - 32) = 180
• So the region between 0 and 100 in the Celsius scale is divided into 100 equal parts
• In the same way, the region between 32 and 212 in the Fahrenheit scale must be divided into 180 equal parts
♦ For getting 180 equal parts, the region is first divided into 10
♦ Then each of these 10 is further divided into 18 equal parts
• So the first mark after 32 is (32+18) = 50. The mark after 50 is (50+18) = 68
and so on . . .
• The two thermometers can be shown together as below:
6. Let Qc be the quantity of heat required for the rise in each small unit in the Celsius scale
• Then total heat quantity from 0o C to 100o C = Qc ×100 = 100 Qc
7. Let Qf be the quantity of heat required for the rise in each small unit in the Fahrenheit scale
• Then total heat quantity from 32o F to 212o F = Qf ×180 = 180 Qf
8. The quantities in (6) and (7) are the same. So we can write:
Eq.6.1:
100 Qc = 180 Qf
⇒ 5 Qc = 9 Qf ⇒ Qc = (9⁄5) Qf
9. From this result, we can note some interesting points:
• In the Celsius scale, the region is divided into 100 equal units. But in the Fahrenheit scale, the same region is divided into 180 equal units
• So the units in the Fahrenheit scale will be smaller than the units in the Celsius scale.
• We find that this is true when we see the result in (8): Qf is to be multiplied by a factor which is 'greater than 1' to get Qc
10. The result in (8) is the basic relation between the two scales. We can use it to convert from one scale to another. Let us see an example:
• The Celsius thermometer in a room indicates that the temperature in that room is 28o C. How much is it in Fahrenheit?
Solution:
1. The reading is 28o C. So it is 28 units above zero in the Celsius scale
2. The value corresponding to 0o C in the Fahrenheit scale is 32o F.
3. Since 28 is above zero, the reading corresponding to 28 in the Fahrenheit scale will be above 32
• We want to know 'how many units' above 32
4. We have seen that each unit in Fahrenheit scale is smaller than the units in Celsius scale.
• So the heat producing a rise of one unit in Celsius scale will produce a rise of more than one unit in Fahrenheit scale. We have seen that it is (9⁄5) times
• So the rise of 28 units in the Celsius scale will cause a rise of [28 × (9⁄5)] = 50.4 units in the Fahrenheit scale
5. The '28 units' is above zero in the Celsius scale
• So the '50.4 units' is above 32 in the Fahrenheit scale
• Thus the reading in Fahrenheit scale is (32+50.4) = 82.4 F
If C is the reading in Celsius scale, the corresponding reading in Fahrenheit scale will be [32 + (9⁄5) × C]
So if F is the reading corresponding to C, we can write:
Eq.6.2:
Next we will see how to convert Fahrenheit to Celsius. Here also we will use an example
• The Fahrenheit thermometer in a room indicates that the temperature in that room is 95o F. How much is it in Celsius?
Solution:
1. The reading is 95o F. So it is (95-32) = 63 units above 32 in the Fahrenheit scale
2. The value corresponding to 32o F in the Celsius scale is 0o C.
3. Since 95 is above 32, the reading corresponding to 95 in the Celsius scale will be above 0
• We want to know 'how many units' above 0
4. Let it be 'C' units above zero
5. We have seen that each unit in Fahrenheit scale is smaller than the units in Celsius scale.
• So the heat producing a rise of one unit in Celsius scale will produce a rise of more than one unit in Fahrenheit scale. We have seen that it is (9⁄5) times
• So the rise of C units in the Celsius scale will cause a rise of [C × (9⁄5)] units in the Fahrenheit scale
6. In this example, [C × (9⁄5)] = 63 = (95-32) = (F-32)
So C = 63 × (5⁄9) = 35
7. In the above step, 63 = (95-32) = [(given value of F)-32]
So we can write a general form:
C = (F-32)× (5⁄9)
• This can be written as:
Eq.6.3:
Note that, the above equation can be obtained by rearranging Eq.6.2 also
Let us see a few solved examples:
Solved example 6.1
The normal human body temperature is 98.6o F. How much is this in the Celsius scale?
Solution:
1. In this problem, we have to convert from Fahrenheit to Celsius. We can use Eq.6.3:
C = (5⁄9)×[F-32]
2. Substituting the given value of F, we get:
C = (5⁄9)×[98.6-32] = (5⁄9)×[66.6] = 37o C
Solved example 6.2
If the average temperature of a day is 30o C, how much will it be in Fahrenheit scale?
Solution:
1. In this problem, we have to convert from Fahrenheit to Celsius. We can use Eq.6.2:
F = (9⁄5)×C + 32
2. Substituting the given value of C, we get:
F = (9⁄5)×30 + 32 = 54 + 32 = 86o F
Graphical method to convert from Celsius to Fahrenheit and vice versa
• Conversion between the two scales can be done using a graph also. Let us see the method:
• In fig.6.13 below, the Celsius scale is marked along the X axis and the Fahrenheit scale is marked along the Y axis.
 |
| Fig.6.13 |
• We can take any convenient scale. In the fig.6.13, the following scales are used:
♦ Each unit along the X axis represents 10o C
♦ Each unit along the Y axis represents 20o F
• The first step is to mark the known points. We need at least two points to draw a line. We do have two such points: (0,32) and (100,212). These are shown with blue coloured 'x' marks
• The blue line joining those two points is our graph.
• Let us do some conversions. We will see the same solved examples 6.1 and 6.2
• 6.1: We want to convert 98.6o F to Celsius
• For that, draw a horizontal line through 98.6 on the Y axis. In the fig.6.13, it is the horizontal white colored dashed line (very close to 100 on the Y axis)
• This line intersects the blue graph at a point.
• From the point of intersection, drop a vertical line. In the fig.6.13, it is the vertical white colored dashed line
• This line intersects the X axis at a point. This point of intersection is our required point. If we zoom in, we will find that it is (37,0). So 37o C is equivalent to 98.6o F
• 6.2: We want to convert 30o C to Fahrenheit
• For that, draw a vertical line through 30 on the X axis. In the fig.6.13, it is the vertical mgenta colored dashed line (through 30 on the X axis)
• This line intersects the blue graph at a point.
• From the point of intersection, draw a horizontal line to the left. In the fig.6.13, it is the horizontal magenta colored dashed line
• This line intersects the Y axis at a point. This point of intersection is our required point. If we zoom in, we will find that it is (0,86). So 86o F is equivalent to 30o C
■ If 32 is the freezing point of water, what is zero?
• To find the answer, we have to look into some history:
• The Fahrenheit scale was invented by a German scientist Daniel Gabriel Fahrenheit., who lived and worked in Amsterdam, Netherlands.
• He invented the scale in the 1720s. At that time, the lowest temperature achievable in his laboratory was the temperature of a 'special mixture'.
• It was a mixture of salt, water and ammonium chloride. It had a heat content much lower than the freezing point of ordinary water.
• As it was the lowest possible temperature, he assigned the number 'zero' to it.
In the next section, we will see the Kelvin scale.
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