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 TA and TB ?
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 HA be the heat required to raise mercury by one unit in A
♦ Then the heat required to raise 200 units in A = 200HA
(ii) Let HB be the heat required to raise mercury by one unit in B
♦ Then the heat required to raise 350 units in B = 350HB
5. When placed at the triple point of water, both the thermometers will absorb the same heat
• So we can write: 200HA = 350HB
• 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 'TA'
♦ That means, the mercury rises by 'TA' units above zero
♦ That means, a heat of (TA × HA) is absorbed
♦ That means, a heat of (TA × HA) 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 HB 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 (TB) 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 HC be the heat required to raise mercury by one unit in C
♦ Then the heat required to raise 100 units in C = 100HC
(ii) Let HF be the heat required to raise mercury by one unit in F
♦ Then the heat required to raise 180 units in F = 180HF
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100HC = 180HF
• 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 'TC'
♦ That means, the mercury rises by 'TC' units above zero
♦ That means, a heat of (TC × HC) is absorbed
♦ That means, a heat of (TC × HC) 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 HF 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 (TF) 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 'TF'
♦ That means, the mercury rises by 'TF' units above zero
♦ That means, a heat of (TF × HF) is absorbed
♦ That means, a heat of (TF × HF) is available in that hot body
• It is necessary to keep in mind that, this heat (TF × HF) is inclusive of the freezing point of water
• So the heat available above the freezing point of water is (TF - 32) × HF]
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 HC 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 (TC) 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 HC be the heat required to raise mercury by one unit in C
♦ Then the heat required to raise 100 units in C = 100HC
(ii) Let HW be the heat required to raise mercury by one unit in W
♦ Then the heat required to raise 200 units in W = 200HW
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100HC = 200HW
• 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 'TC'
♦ That means, the mercury rises by 'TC' units above zero
♦ That means, a heat of (TC × HC) is absorbed
♦ That means, a heat of (TC × HC) 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 HW 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 (TW) 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, TC = 39 °C
So the corresponding TW = (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 HC be the heat required to raise mercury by one unit in C
♦ Then the heat required to raise 100 units in C = 100HC
(ii) Let HU be the heat required to raise mercury by one unit in U
♦ Then the heat required to raise 70 units in U = 70HU
4. Which ever be the thermometer, the heat mentioned in (2) will be the same H
• So we can write: H = 100HC = 70HU
• 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 'TU'
♦ That means, the mercury rises by 'TU' units above zero
♦ That means, a heat of (TU × HU) is absorbed
♦ That means, a heat of (TU × HU) is available in that hot body
[It is necessary to keep in mind that, this heat (TU × HU) is inclusive of the freezing point of water. So the heat available above the freezing point of water is (TU - 10) × HU]
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 HC 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 (TC) 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: TU = 59 °U
So we can write: $\mathbf\small{\rm{T_C=\frac{10}{7}(59 - 10)}}$
Thus we get: TC = 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 × 105 Pa
♦ At normal melting point of sulphur, the pressure is 1.797 × 105 Pa
Thermometer B:
♦ At triple-point of water, the pressure is 0.200 × 105 Pa
♦ At normal melting point of sulphur, the pressure is 0.287 × 105 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
In the next chapter, we will see Thermodynamics
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