Chapter 11.4 - Change of State

In the previous section we saw specific heat capacity and calorimetry. In this section, we will see change of state

For the discussion on change of state, we have to learn about Latent heat of fusion (L_f), Latent heat of vaporisation (L_v), Boiling point, and evaporation. The links to detailed notes about these topics are given below:

Details about Latent heat of fusion

Details about Latent heat of vaporisation

Details about Latent heat of evaporation

Details about boiling point

Based on the above detailed notes, we can see some solved examples. Link to four such solved examples is given below:

Solved examples 11.17 to 11.20
 

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In the above discussions, we see that, the temperature-time graph helps us to obtain valuable information about specific heats and latent heats. Let us see some more properties of this graph. It can be explained in 15 steps:

1. Fig.11.10 below, shows a temperature-time graph
    ♦ time is plotted along the x-axis
    ♦ temperature is plotted along the y-axis

Fig.11.10

2. Heat is supplied from a burner. This heat-supply is at a constant rate
■ That means:
The ‘quantity of heat supplied per second’ remains the same
• Let the ‘quantity of heat supplied per second’ be k
3. Mark two pairs of points: P, Q and U, V  on the x-axis
    ♦ Let the length PQ = Δt
    ♦ Then the heat supplied during the interval from A to B = k Δt
4. Let length PQ = length UV
• Then we get:
    ♦ heat supplied during the interval from P to Q
    ♦ = heat supplied during the interval from U to V
    ♦ = k Δt
■ That means, the same heat is supplied in both intervals
5. Now consider the two triangles: P’Q’R and U’V’W
    ♦ The bases P’Q’ and U’V’ are equal
    ♦ But the altitudes Q’R and V’W are different
6. The altitudes denote the change in temperatures
• Let Q’R = ΔT and V’W = ΔT’
• In our present case, we can see that, ΔT’ is greater than ΔT
• We know that
    ♦ From O to A, the substance is in the solid state
    ♦ From B to C, the substance is in the liquid state
• So we can write:
    ♦ When an energy of (k Δt) is supplied
          ✰ the substance in the solid state
          ✰ attains a temperature change of ΔT
    ♦ When the same energy of (k Δt) is supplied
          ✰ the same substance in the liquid state
          ✰ attains a greater temperature change of ΔT'
7. We know the two general equations:
(i) Energy supplied to the solid = m_Solid × s_Solid × rise in temperature of solid
(ii) Energy supplied to the liquid = m_Liquid × s_Liquid × rise in temperature of liquid
8. In our present case:
    ♦ Energy supplied to the solid = Energy supplied to the liquid = k Δt
    ♦ m_Solid = m_Liquid = m
• So the two equations in (7) become:
(i) k Δt = m × s_Solid × ΔT
(ii) k Δt = m × s_Liquid × ΔT'
9. Since the two energies are equal, we can equate them:
m × s_Solid × ΔT = m × s_Liquid × ΔT'
⇒ $\mathbf\small{\rm{s_{Solid}=s_{Liquid} \times \frac{\Delta T'}{\Delta T}}}$
10. In our present case, ΔT' is greater than ΔT
    ♦ So $\mathbf\small{\rm{\frac{\Delta T'}{\Delta T}}}$ will be greater than '1'
    ♦ So s_Solid will be greater than s_Liquid
11. Let us consider the slopes of the two lines OA and BC. We see that:
• OA is flatter
    ♦ In other words, OA has a small slope
• BC is steeper
    ♦ In other words, BC has a large slope
12. To compare the slopes of two lines, we can make use of two triangles like those we see in fig.11.10
■ If their bases are equal, we get:
• The triangle with greater altitude
    ♦ will indicate
    ♦ a greater slope
• The triangle with lesser altitude
    ♦ will indicate
    ♦ a lesser slope
13. Now we can write in terms of slope:
• We know that
    ♦ From O to A, the substance is in the solid state
    ♦ From B to C, the substance is in the liquid state
■ So we can write:
If OA (solid portion) is flatter and BC (liquid portion) is steeper, then:
s_Solid will be greater than s_Liquid
14. Now the reader may try the following:
(i) Draw a temperature-time graph in such a way that:
OA is steeper and BC is flatter
(ii) Write all the steps from (1) to (13)
(iii) Hence prove that:
If OA is steeper and BC is flatter, then:
s_Solid will be lesser than s_Liquid
15. Thus we can write:
Specific heat capacity at the flatter portion will be larger

Solved example 11.22
A solid substance is supplied with heat at a constant rate. The variation of temperature with heat input is shown in fig.11.11 below:

Fig.11.11

Based on the graph, answer the following:
(i) What do the horizontal portions AB and CD represent ?
(ii) If CD = 2AB, what do you infer ?
(iii) What does slope DE represent ?
(iv) The slope of OA > slope of BC. What does this indicate ?
Solution:
Part (i)
1. Imagine vertical lines through A and B
• Those vertical lines will meet the x-axis at Q_A and Q_B
    ♦ Q_A is the heat possessed by the body when the temperature is T_A
    ♦ Q_B is the heat possessed by the body when the temperature is T_B
2. But T_A = T_B
• In other words, there is no temperature change from A to B
• Between A and B, the body absorbs a heat of (Q_B-Q_A)
    ♦ All this heat is used for changing the state from solid to liquid
    ♦ There is no heat available to cause a change in temperature
• So the horizontal portion AB indicates that:
T_A is the melting point
3. The body begins to melt just when the temperature reaches T_A
• At B, the melting is complete
• That is, after TA, when an energy of (Q_B-Q_A) is absorbed, the body completely becomes a liquid at constant temperature T_A
4. In the same way:
After T_C, when an energy of (Q_D-Q_C) is absorbed, the body completely becomes a gas at constant temperature T_C
 

Part (ii)
1. Length AB indicates the quantity of heat energy required for melting
    ♦ We know that, this quantity is equal to (L_f m)
2. Length CD indicates the quantity of heat energy required for vaporisation
    ♦ We know that, this quantity is equal to (L_v m)
3. Given that, CD = 2AB
• So we can write: L_v m = 2 L_f m
⇒ L_v = 2 L_f
4. Thus we get:
Latent heat of vaporisation of the given substance is twice it's latent heat of fusion

Part (iii)
1. Imagine a right triangle below the line DE
• Slope of the line will be equal to ^Altitude⁄_Base
2. Base of the right triangle will be equal to heat (Q) absorbed during a particular time interval
• Altitude will be equal to change in temperature (ΔT) during that time interval
3. So slope = $\mathbf\small{\rm{\frac{\Delta T}{Q}}}$
4. But Q = m s_gas Δt
• So the result in (3) becomes:
slope = $\mathbf\small{\rm{\frac{\Delta T}{m s_{gas} \Delta t}=\frac{1}{m s_{gas}}}}$
5. But (m s_gas) is the heat capacity of the gas sample
• So the result in (5) becomes:
slope of line DE = Reciprocal of the heat capacity of the sample in the gaseous state

Part (iv)
1. OA is steeper and BC is flatter
• That means, OA has a greater slope and BC has a smaller slope
2. From part (iii), we have:
slope = Reciprocal of the heat capacity of the sample
3. So we can write in terms of heat capacity:
(i) slope of OA = $\mathbf\small{\rm{\frac{1}{m \, s_{solid}}}}$
(ii) slope of BC = $\mathbf\small{\rm{\frac{1}{m \, s_{liquid}}}}$
4. Given that 3(i) is greater
• Then the denominator of 3(i) must be smaller than the denominator in 3(ii)
5. 'm' is common in both
• So we can write: s_solid is smaller than s_liquid
Note: In the present problem (fig.11.11), we plot heat along the x-axis. But in the earlier fig.11.10, we plotted time along the x-axis. However, both are related because, rate at which heat is supplied is constant

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In the next section, we will see Phase diagram and triple point

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