Chapter 11.10 - Solved Examples on Various Temperature Scales

In the previous section we completed a discussion on thermal properties of matter. In this section, we will see some solved examples related to temperature scales

• The basics about the three scales for measuring temperature and their interrelations can be seen here and here

Solved example 11.33
The triple points of neon and carbon dioxide are 24.57 K and 216.55 K respectively. Express these temperatures on the Celsius and Fahrenheit scales
Solution:
1. To convert the temperature from Kelvin scale to Celsius scale, we use the relation: C = K - 273.15
2. So triple point of neon in Celsius scale
= (K − 273.15) = (24.57 − 273.15) = −248.58 °C
• Once we get the temperature in Celsius scale, we can easily convert it further into Fahrenheit scale using the relation: $\mathbf\small{\rm{F=32+\frac{9}{5}C}}$
• Thus we get:
Triple point of neon in Fahrenheit scale = $\mathbf\small{\rm{32+\frac{9}{5} \times -248.58}}$ = −415.44 °F  
3. Similarly, triple point of carbon dioxide in Celsius scale
= (K − 273.15) = (216.55 − 273.15) = −56.6 °C
• Thus we get:
Triple point of carbon dioxide in Fahrenheit scale = $\mathbf\small{\rm{32+\frac{9}{5} \times -56.6}}$ = −69.88 °F  

Solved example 11.34
Two absolute scales A and B have triple points of water defined to be 200 A and 350 B. What is the relation between T_A and T_B ?
Solution:
1. A and B are names of the scales, just like C and F
• So we can write:
    ♦ When measured with thermometer A, the triple point of water is 200 °A
    ♦ When measured with thermometer B, the triple point of water is 350 °B
2. It is given that, both A and B are absolute scales
• So in both cases:
The zero reading will be the lowest possible temperature
• In other words, in both cases:
There will not be any negative markings
3. So we can write:
• When the bulb of A is placed at triple point of water, the mercury rises 200 units above zero
• When the bulb of B is placed at triple point of water, the mercury rises 350 units above zero
4. We can compare the heat quantities:
(i) Let H_A be the heat required to raise mercury by one unit in A
    ♦ Then the heat required to raise 200 units in A = 200H_A
(ii) Let H_B be the heat required to raise mercury by one unit in B
    ♦ Then the heat required to raise 350 units in B = 350H_B
5. When placed at the triple point of water, both the thermometers will absorb the same heat
• So we can write: 200H_A = 350H_B
• From this, we get two results:
(i) $\mathbf\small{\rm{H_A=\frac{7}{4}H_B}}$
(ii) $\mathbf\small{\rm{H_B=\frac{4}{7}H_A}}$
6. Suppose that, the thermometer A is placed at a hot body
• Let the reading be 'T_A'
    ♦ That means, the mercury rises by 'T_A' units above zero
    ♦ That means, a heat of (T_A × H_A) is absorbed
    ♦ That means, a heat of (T_A × H_A) is available in that hot body
7. But from 5(i), we have: $\mathbf\small{\rm{H_A=\frac{7}{4}H_B}}$
• So the result in (6) becomes: $\mathbf\small{\rm{T_A \times H_A=T_A \times \frac{7}{4}H_B}}$
• That means, a heat of $\mathbf\small{\rm{T_A \times \frac{7}{4}H_B}}$ is available in the hot body
8. So when the thermometer B is placed in the hot body, it will absorb the heat $\mathbf\small{\rm{T_A \times \frac{7}{4}H_B}}$
• But H_B is the heat required to raise mercury by one unit in B
    ♦ So the mercury will rise by ($\mathbf\small{\rm{T_A \times \frac{7}{4}}}$) units
    ♦ So the reading (T_B) in thermometer B will equal to $\mathbf\small{\rm{T_A \times \frac{7}{4}}}$
• That is: $\mathbf\small{\rm{T_B=T_A \times \frac{7}{4}}}$
■ Thus we get: $\mathbf\small{\rm{T_A=\frac{4}{7}T_B}}$

Solved example 11.35
Using the method used in the above solved example 11.34, obtain a relation between Celsius and Fahrenheit scales
Solution:
1. When comparing Celsius scale and Fahrenheit scale, we can write:
    ♦ Freezing point of water: 0 °C → 32 °F
    ♦ Boiling point of water: 100 °C → 212 °F
2. Let H be the heat required to raise the temperature of water from freezing point to boiling point
• Then this H is capable of:
    ♦ raising (100 - 0 = 100) units in the C thermometer
    ♦ raising (212 - 32 = 180) units in the F thermometer
3. We can compare the heat quantities:
(i) Let H_C be the heat required to raise mercury by one unit in C
    ♦ Then the heat required to raise 100 units in C = 100H_C
(ii) Let H_F be the heat required to raise mercury by one unit in F
    ♦ Then the heat required to raise 180 units in F = 180H_F
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100H_C = 180H_F
• From this, we get two results:
(i) $\mathbf\small{\rm{H_C=\frac{9}{5}H_F}}$
(ii) $\mathbf\small{\rm{H_F=\frac{5}{9}H_C}}$
5. Suppose that, the thermometer C is placed at a hot body
• Let the reading be 'T_C'
    ♦ That means, the mercury rises by 'T_C' units above zero
    ♦ That means, a heat of (T_C × H_C) is absorbed
    ♦ That means, a heat of (T_C × H_C) is available in that hot body
6. But from 4(i), we have: $\mathbf\small{\rm{H_C=\frac{9}{5}H_F}}$
• So the result in (5) becomes: $\mathbf\small{\rm{T_C \times H_C=T_C \times \frac{9}{5}H_F}}$
• That means, a heat of $\mathbf\small{\rm{T_C \times \frac{9}{5}H_F}}$ is available in the hot body
7. So when the thermometer F is placed in the hot body, it will absorb the heat $\mathbf\small{\rm{T_C \times \frac{9}{5}H_F}}$
• But H_F is the heat required to raise mercury by one unit in F
    ♦ So the mercury will rise by ($\mathbf\small{\rm{T_C \times \frac{9}{5}}}$) units
          ✰ Recall that, the heat causing this rise is above the freezing point of water
          ✰ At the freezing point of water, the thermometer F will be already showing 32 F
          ✰ This is shown in fig.11.21(a) below
    ♦ So the reading (T_F) in thermometer F will equal to $\mathbf\small{\rm{32+T_C \times \frac{9}{5}}}$

Fig.11.21

■ Thus we get: $\mathbf\small{\rm{T_F=32+\frac{9}{5}T_C}}$
8. The above expression helps us to quickly convert C to F
• We can do the reverse also. That is., we can find an expression to convert F to C
• The following steps from (9) to (11) will give us the expression
9. Suppose that, the thermometer F is placed at a hot body
• Let the reading be 'T_F'
    ♦ That means, the mercury rises by 'T_F' units above zero
    ♦ That means, a heat of (T_F × H_F) is absorbed
    ♦ That means, a heat of (T_F × H_F) is available in that hot body
• It is necessary to keep in mind that, this heat (T_F × H_F) is inclusive of the freezing point of water
• So the heat available above the freezing point of water is (T_F - 32) × H_F]
10. But from 4(i), we have: $\mathbf\small{\rm{H_F=\frac{5}{9}H_C}}$
• So the result in (9) becomes: $\mathbf\small{\rm{(T_F - 32) \times H_F=(T_F - 32) \times \frac{5}{9}H_C}}$
• That means, a heat of $\mathbf\small{\rm{(T_F - 32) \times \frac{5}{9}H_C}}$ is available in the hot body
11. So when the thermometer C is placed in the hot body, it will absorb the heat $\mathbf\small{\rm{(T_F - 32) \times \frac{5}{9}H_C}}$
• But H_C is the heat required to raise mercury by one unit in C
    ♦ So the mercury will rise by ($\mathbf\small{\rm{(T_F - 32) \times \frac{5}{9}}}$) units
          ✰ This is shown in fig.11.21(b) above
    ♦ So the reading (T_C) in thermometer C will equal to $\mathbf\small{\rm{(T_F - 32) \times \frac{5}{9}}}$
■ Thus we get: $\mathbf\small{\rm{T_C=\frac{5}{9}(T_F - 32)}}$

Solved example 11.36
On a new scale of temperature called the W scale, the freezing and boiling points of water are 39 °W and 239 °W respectively. What will be the temperature on the new scale, corresponding to a temperature of 39 °C on the Celsius scale ?
Solution:
1. When comparing Celsius scale and W scale, we can write:
    ♦ Freezing point of water: 0 °C → 39 °W
    ♦ Boiling point of water: 100 °C → 239 °W
2. Let H be the heat required to raise the temperature of water from freezing point to boiling point
• Then this H is capable of:
    ♦ raising (100 - 0 = 100) units in the C thermometer
    ♦ raising (239 - 39 = 200) units in the W thermometer
3. We can compare the heat quantities:
(i) Let H_C be the heat required to raise mercury by one unit in C
    ♦ Then the heat required to raise 100 units in C = 100H_C
(ii) Let H_W be the heat required to raise mercury by one unit in W
    ♦ Then the heat required to raise 200 units in W = 200H_W
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100H_C = 200H_W
• From this, we get two results:
(i) $\mathbf\small{\rm{H_C=2H_W}}$
(ii) $\mathbf\small{\rm{H_W=0.5H_C}}$
5. Suppose that, the thermometer C is placed at a hot body
• Let the reading be 'T_C'
    ♦ That means, the mercury rises by 'T_C' units above zero
    ♦ That means, a heat of (T_C × H_C) is absorbed
    ♦ That means, a heat of (T_C × H_C) is available in that hot body
6. But from 4(i), we have: $\mathbf\small{\rm{H_C=2H_W}}$
• So the result in (5) becomes: $\mathbf\small{\rm{T_C \times H_C=T_C \times 2H_W}}$
• That means, a heat of $\mathbf\small{\rm{T_C \times 2H_W}}$ is available in the hot body
7. So when the thermometer W is placed in the hot body, it will absorb the heat $\mathbf\small{\rm{T_C \times 2H_W}}$
• But H_W is the heat required to raise mercury by one unit in W
    ♦ So the mercury will rise by ($\mathbf\small{\rm{T_C \times 2}}$) units
          ✰ Recall that, the heat causing this rise is above the freezing point of water
          ✰ At the freezing point of water, the thermometer W will be already showing 39 °W
          ✰ This is shown in fig.11.22(a) below
    ♦ So the reading (T_W) in thermometer W will equal to $\mathbf\small{\rm{39+T_C \times 2}}$

Fig.11.22

■ Thus we get: $\mathbf\small{\rm{T_W=39+2T_C}}$
8. We are given that, T_C = 39 °C
So the corresponding T_W = (39 + 2 × 39) = 117 °W

Solved example 11.37
The steam point and ice point of a mercury thermometer U are marked as 80 °U and 10 °U respectively. At what temperature on the Celsius scale, the reading of this thermometer will be 59 °U ?
Solution:
1. When comparing Celsius scale and U scale, we can write:
    ♦ Freezing point of water: 0 °C → 10 °U
    ♦ Boiling point of water: 100 °C → 80 °U
2. Let H be the heat required to raise the temperature of water from freezing point to boiling point
• Then this H is capable of:
    ♦ raising (100 - 0 = 100) units in the C thermometer
    ♦ raising (80 - 10 = 70) units in the U thermometer
3. We can compare the heat quantities:
(i) Let H_C be the heat required to raise mercury by one unit in C
    ♦ Then the heat required to raise 100 units in C = 100H_C
(ii) Let H_U be the heat required to raise mercury by one unit in U
    ♦ Then the heat required to raise 70 units in U = 70H_U
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100H_C = 70H_U
• From this, we get two results:
(i) $\mathbf\small{\rm{H_C=\frac{7}{10}H_U}}$
(ii) $\mathbf\small{\rm{H_U=\frac{10}{7}H_C}}$
5. Suppose that, the thermometer U is placed at a hot body
• Let the reading be 'T_U'
    ♦ That means, the mercury rises by 'T_U' units above zero
    ♦ That means, a heat of (T_U × H_U) is absorbed
    ♦ That means, a heat of (T_U × H_U) is available in that hot body
[It is necessary to keep in mind that, this heat (T_U × H_U) is inclusive of the freezing point of water. So the heat available above the freezing point of water is (T_U - 10) × H_U]
6. But from 4(i), we have: $\mathbf\small{\rm{H_U=\frac{10}{7}H_C}}$
• So  we get: $\mathbf\small{\rm{(T_U - 10) \times H_U=(T_U - 10) \times \frac{10}{7}H_C}}$
• That means, a heat of $\mathbf\small{\rm{(T_U - 10) \times \frac{10}{7}H_C}}$ is available in the hot body
7. So when the thermometer C is placed in the hot body, it will absorb the heat $\mathbf\small{\rm{(T_U - 10) \times \frac{10}{7}H_C}}$
• But H_C is the heat required to raise mercury by one unit in C
    ♦ So the mercury will rise by ($\mathbf\small{\rm{(T_U - 10) \times \frac{10}{7}}}$) units
          ✰ This is shown in fig.11.22(b) above
    ♦ So the reading (T_C) in thermometer C will equal to $\mathbf\small{\rm{(T_U - 10) \times \frac{10}{7}}}$
■ Thus we get: $\mathbf\small{\rm{T_C=\frac{10}{7}(T_U - 10)}}$
8. We are given that: T_U = 59 °U
So we can write: $\mathbf\small{\rm{T_C=\frac{10}{7}(59 - 10)}}$
Thus we get: T_C = 70 °C

Solved example 11.38
We have learnt about linear expansion caused by increase in temperature. Explain how it can be used to make a Celsius scale thermometer
Solution:
1. First step is to obtain two fixed points:
• Let l₀ be the length of the thermometer when the surrounding temperature is 0 °C
• Let l₁₀₀ be the length of the thermometer when the surrounding temperature is 100 °C
2. Change in length:
• Change in length caused by an increase of 100 °C = (l₁₀₀ - l₀)
• So change in length caused by an increase of 1 °C = $\mathbf\small{\rm{\frac{l_{100} - l_0}{100}}}$
3. Let l be the length when the surrounding temperature is T °C
• Then the change in length suffered by the thermometer due to the increase from 0 °C to T °C = (l - l₀)
4. Increase by 1 °C will cause an increase of $\mathbf\small{\rm{\frac{l_{100} - l_0}{100}}}$
• So increase of T °C will cause an increase in length of $\mathbf\small{\rm{\frac{l_{100} - l_0}{100}T}}$
5. But the increase in length due to T °C is equal to (l - l₀)
• So the result in (4) is equal to (l - l₀)
• We can write: $\mathbf\small{\rm{\frac{l_{100} - l_0}{100}T=l-l_0}}$
• Thus we get: $\mathbf\small{\rm{T=\left (\frac{l - l_0}{l_{100} - l_0} \right )\times 100}}$
■ l₀ and l₁₀₀ will be known quantities. To determine the temperature T, all we need is the length l

Solved example 11.39
We have learnt that, for an ideal gas, $\mathbf\small{\rm{\frac{PV}{T}}}$ is a constant. Explain how this can be used to make a Celsius scale thermometer
Solution:
1. For an ideal gas, $\mathbf\small{\rm{\frac{PV}{T}=a \; constant \;K}}$
• Using this fact, we can make two types of Celsius scale thermometers:
    ♦ Constant volume thermometer
    ♦ Constant pressure thermometer
2. First we will see constant volume thermometer
• In this type, the ideal gas is enclosed inside a glass container of fixed volume
• So the expression $\mathbf\small{\rm{\frac{PV}{T}=K}}$ becomes: $\mathbf\small{\rm{\frac{P}{T}=K}}$
• This is same as: P = KT
■ So, when temperature increases, pressure also increases
3. The two fixed points:
• When the surrounding temperature is 0 °C, let the pressure of the ideal gas be P₀
• When the surrounding temperature is 100 °C, let the pressure of the ideal gas be P₁₀₀
4. Then we can write:
• For in increase of 100 °C, the increase in pressure is (P₁₀₀ - P₀)
• So, for an increase in 1 °C, the increase in pressure will be $\mathbf\small{\rm{\frac{P_{100}-P_0}{100}}}$
5. Let the surrounding temperature, which is to be determined, be T °C
• Increase by 1 °C will cause an increase of $\mathbf\small{\rm{\frac{P_{100} - P_0}{100}}}$
• Then the increase in pressure due to T °C will be $\mathbf\small{\rm{\frac{P_{100}-P_0}{100}T}}$
6. Let the pressure of the ideal gas be P when the surrounding temperature is this T
• Then the increase in pressure suffered by the ideal gas is: (P - P₀)
• So we can write: $\mathbf\small{\rm{P - P_0=\frac{P_{100}-P_0}{100}T}}$
■ Thus we get: $\mathbf\small{\rm{T=\frac{100 \; (P - P_0)}{P_{100}-P_0}}}$
■ P₀ and P₁₀₀ will be known quantities. To determine the temperature T, all we need is the pressure P
7. In a similar way, we can derive the expression for constant pressure thermometer also. We will get:
$\mathbf\small{\rm{T=\frac{100 \; (V - V_0)}{V_{100}-V_0}}}$ 

Solved example 11.40
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law : R = R₀[1 + ⍺(T - T₀)]. The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
1. The two fixed points are:
    ♦ Triple point of water
    ♦ Normal melting point of lead
2. Applying the fixed points:
• Using the first fixed point in the given expression, we get:
101.6 = R₀[1 +  ⍺(273.16 - 273.16)] = R₀
• Using the second fixed point in the given expression, we get:
165.5 = 101.6[1 +  ⍺(600.5 - 273.16)]
    ♦ From this, we get ⍺ = 0.00192
3. The given resistance is 123.4 Ω
• Using the given expression:
123.4 = 101.6[1 +  ⍺(T - 273.16)]
• From this, we get T = 384.8 K

Solved example 11.41
Two ideal gas thermometers A and B use oxygen and hydrogen respectively. The following observations are made:
Thermometer A:
    ♦ At triple-point of water, the pressure is 1.250 × 10_⁵ Pa
    ♦ At normal melting point of sulphur, the pressure is 1.797 × 10_⁵ Pa
Thermometer B:
    ♦ At triple-point of water, the pressure is 0.200 × 10_⁵ Pa
    ♦ At normal melting point of sulphur, the pressure is 0.287 × 10_⁵ Pa
• What is the absolute temperature of normal melting point of sulphur as read by thermometers A and B ?
Solution:
• For an ideal gas, we have: $\mathbf\small{\rm{\frac{P_1}{T_1}=\frac{P_2}{T_2} \; = \; a \; constant}}$
1. Thermometer A:
• Two fixed points are given
• Applying those two fixed points in the ideal gas equation, we get:
$\mathbf\small{\rm{\frac{1.250 \;\times \; 10^{5}}{273.16}=\frac{1.797 \;\times \; 10^{5}}{T}}}$
• Thus we get: T = 392.69 K
2. Thermometer B:
• Two fixed points are given
• Applying those two fixed points in the ideal gas equation, we get:
$\mathbf\small{\rm{\frac{0.200 \;\times \; 10^{5}}{273.16}=\frac{0.287 \;\times \; 10^{5}}{T}}}$
• Thus we get: T = 391.98 K

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In the next chapter, we will see Thermodynamics

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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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