In the previous section we discussed the change of state from solid to liquid. In this section we will discuss the next higher level. That is., change from liquid state to gaseous state.
• Consider some water in a vessel. It's initial temperature is ordinary room temperature (say 30 oC). It is continuously heated. Then it's temperature will go on increasing. It will become 31, 32, 33, . . . so on. This is shown by the segment AB in the graph below:
• But when the temperature reaches 100 oC, the rise in temperature will stop. This is indicated by the point B
• The heat energy supplied after this point will be used to break the bonds between the water molecules.
• At 100 oC, the water will begin to boil. It will boil at the constant temperature of 100 oC, until all the bonds are broken. This is indicated by the horizontal line BC.
• But when all the bonds are broken, there will not be any water left. All the water will be converted to steam.
• If we can confine this newly produced steam in a vessel (with all necessary safety precautions), then the further heat energy supplied will cause an increase in temperature of that steam. This is indicated by the rising line after C
• So we find that water boils at a constant temperature of 100 oC
• At a temperature higher than 100 oC, there will not be any water but only steam
■ The boiling point of a liquid is the constant temperature at which a liquid boils and changes into it's gaseous state at normal atmospheric pressure.
■ The conversion of a liquid into it's gaseous state at it's boiling point is called vaporisation
■ Latent heat of vaporisation (Lv) of a substance is the quantity of heat absorbed by 1 kg of the liquid to change into it's gaseous state at it's boiling point with out change in temperature.
• The following table 6.5 shows the Lv values of some common substances
Table 6.5
• We have seen that, the quantity of heat required for converting a liquid of mass 1 kg into gaseous form is it's Lv
• So the quantity of heat required for converting a liquid of mass m kg into gaseous form is mLv
Let us analyse the values in the above table:
A. Methanol:
1. Consider some methanol kept in a vessel. Let it's temperature be 30 oC.
2. Let this methanol be heated. Once we begin heating, the temperature will begin to rise.
• That is., it will become 31, 32, 33, . . . so on. The graph indicating this rise in temperature will be a rising line.
♦ During this rise in temperature, the methanol do not boil
3. But once it reaches 64 oC, the temperature will not rise.
• The heat supplied just after reaching 64 oC, will be used to break the strong bonds between the methanol molecules.
• Each 1 kg of methanol in that vessel will require 112×104 joules of heat energy to break all the bonds in it.
4. If we supply the required energy with in a short span of time, like by using a large burner, then the bonds will break within that short span of time. We will find that the methanol began to boil sooner.
• For our present discussion, the time in which 112×104 J kg-1 is supplied is not important.
• What is important is that, 112×104 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the methanol begins to boil.
• But the temperature of the newly formed gaseous methanol will remain at 64 oC.
• This is because, all the energy being received is used up for breaking more and more bonds. There is no energy available to increase the kinetic energy of the newly formed vapour molecules.
• So the temperature do not rise above 64 oC during this process. The graph indicating this constant temperature will be a horizontal line.
• When all the bonds are broken, we will get 'all gaseous methanol' (at 64 oC) and 'no liquid methanol'.
6. Further energy supply will be used to increase the temperature of the gaseous methanol from 64 oC up wards. So the graph will again be a rising line.
■ The reader may write the steps for the other substances in table 6.5 in the same way
• In the above table 6.5, water has the highest Lv value. This means that a lot of heat energy will be required to make water boil and escape into gaseous form.
• This high value is a blessing. The ponds and rivers do not dry up easily, and thus the loss of water from the surface of the earth is reduced.
Some applications of high Lv value of water in our daily life:
1. Cooking food in steam is faster.
• This is because, steam contains more heat energy
♦ The heat which brought it up to the boiling point + the latent heat
• So more heat is transferred to the food and it gets cooked faster.
• But 'boiling water' contains only that heat energy which brought it upto the boiling point
2. Steam is used in thermal power plants
• In thermal power plants, water is first boiled to make steam.
♦ This boiling can be done using fossil fuels like coal or diesel. Nuclear energy can also be used.
• The steam thus created is used to rotate turbines. The turbines are attached to generators. Thus electricity is produced.
• Using steam have some advantages:
♦ After steam has done it's job of rotating the turbines, it is condensed to give back the water. This water can be used again to make steam
♦ While condensing, the steam will release latent heat. Using special devices, this heat can be utilized to heat more water, thus reducing the use of fossil fuels
3. Burns (injuries caused by exposure to heat or flame) caused by steam:
• Consider some boiling water.
♦ The maximum temperature that the boiling water can attain is 100 oC
• Steam will be continuously produced from that boiling water.
♦ The freshly formed steam will be at 100 oC.
♦ But if we have the required safety equipments, we can collect that steam, and increase it's temperature further above 100 oC
• Consider the two:
(i) Boiling water which is at 100 oC
(ii) Freshly formed steam which is at 100 oC
Which one of them will cause a more severe burn?
• Ans: The burn from the steam will be more severe.
• This is because, stem has more heat
♦ The heat which brought it up to the boiling point + the latent heat
• So more heat will be transferred to the body, and the burn will be more severe
Solved example 6.10
1 kg of steam at 100 oC is condensed back to water at the same temperature. It is then cooled to 30 oC. Calculate the heat given out during this process.
Solution:
1. Heat given out when 1 kg of stem at 100 is condensed back to water at the same temperature
= Latent heat of steam = 226×104 = 2260000 J
2. Heat given out when this newly condensed water is cooled to 30 = mcθ = mc(θ2-θ1)
= 1×4200×(30 -100) = 1×4200×-70 = 294000 J
3. So total heat given out = 2260000 + 294000 = 2554000 J
1. Consider some water being heated in a vessel.
• Under normal atmospheric pressure, when the heat acquired by it is enough to break the internal bonds, it will begin to boil.
• At that time it's temperature will show 100 oC
2. But in some cases, the following things happen:
• The heat acquired by it is enough to break the internal bonds
• The heat acquired by it is enough to show 100 oC
■ But the water does not boil.
3. When does this happen?
• This happens when the pressure acting on the surface of the water is above normal atmospheric pressure.
• In that case, the molecules of water are pressed down by the external pressure.
• So they cannot escape from the liquid.
4. If this happens, we will have to continue supplying heat.
• The temperature will rise above 100 oC. Then the molecules will acquire enough energy and will be able to escape. Thus the water will boil.
5. This property of water is used effectively in a pressure cooker.
Let us see the basics about it's working:
(i) Inside the pressure cooker, high pressure builds up above the surface of the water
(ii) So the water will boil only at a higher temperature.
(iii) That means, water does not escape as steam and is available to cook the food.
(iv) Also this water is at a temperature higher than 100 oC. So it contains a lot of heat.
(v) Thus the food gets cooked faster. But the Pressure cooker should have all the required safety features.
• That means, larger heat energy will be required to boil that water.
• It can be used as coolants in motor vehicles.
• This is because, such water will be able to absorb more heat (produced by the engines) and even then, remain in liquid state with out boiling
• Consider some water in a vessel. It's initial temperature is ordinary room temperature (say 30 oC). It is continuously heated. Then it's temperature will go on increasing. It will become 31, 32, 33, . . . so on. This is shown by the segment AB in the graph below:
• The heat energy supplied after this point will be used to break the bonds between the water molecules.
• At 100 oC, the water will begin to boil. It will boil at the constant temperature of 100 oC, until all the bonds are broken. This is indicated by the horizontal line BC.
• But when all the bonds are broken, there will not be any water left. All the water will be converted to steam.
• If we can confine this newly produced steam in a vessel (with all necessary safety precautions), then the further heat energy supplied will cause an increase in temperature of that steam. This is indicated by the rising line after C
• At a temperature higher than 100 oC, there will not be any water but only steam
■ The boiling point of a liquid is the constant temperature at which a liquid boils and changes into it's gaseous state at normal atmospheric pressure.
■ The conversion of a liquid into it's gaseous state at it's boiling point is called vaporisation
■ Latent heat of vaporisation (Lv) of a substance is the quantity of heat absorbed by 1 kg of the liquid to change into it's gaseous state at it's boiling point with out change in temperature.
• The following table 6.5 shows the Lv values of some common substances
Table 6.5
| Substance | Boiling point (oC) | Lv (J kg-1) |
|---|---|---|
| Methanol | 64 | 112 × 104 |
| Ethanol | 79 | 85 × 104 |
| Mercury | 357 | 27 × 104 |
| Water | 100 | 226 × 104 |
• We have seen that, the quantity of heat required for converting a liquid of mass 1 kg into gaseous form is it's Lv
• So the quantity of heat required for converting a liquid of mass m kg into gaseous form is mLv
Let us analyse the values in the above table:
A. Methanol:
1. Consider some methanol kept in a vessel. Let it's temperature be 30 oC.
2. Let this methanol be heated. Once we begin heating, the temperature will begin to rise.
• That is., it will become 31, 32, 33, . . . so on. The graph indicating this rise in temperature will be a rising line.
♦ During this rise in temperature, the methanol do not boil
3. But once it reaches 64 oC, the temperature will not rise.
• The heat supplied just after reaching 64 oC, will be used to break the strong bonds between the methanol molecules.
• Each 1 kg of methanol in that vessel will require 112×104 joules of heat energy to break all the bonds in it.
4. If we supply the required energy with in a short span of time, like by using a large burner, then the bonds will break within that short span of time. We will find that the methanol began to boil sooner.
• For our present discussion, the time in which 112×104 J kg-1 is supplied is not important.
• What is important is that, 112×104 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the methanol begins to boil.
• But the temperature of the newly formed gaseous methanol will remain at 64 oC.
• This is because, all the energy being received is used up for breaking more and more bonds. There is no energy available to increase the kinetic energy of the newly formed vapour molecules.
• So the temperature do not rise above 64 oC during this process. The graph indicating this constant temperature will be a horizontal line.
• When all the bonds are broken, we will get 'all gaseous methanol' (at 64 oC) and 'no liquid methanol'.
6. Further energy supply will be used to increase the temperature of the gaseous methanol from 64 oC up wards. So the graph will again be a rising line.
■ The reader may write the steps for the other substances in table 6.5 in the same way
• In the above table 6.5, water has the highest Lv value. This means that a lot of heat energy will be required to make water boil and escape into gaseous form.
• This high value is a blessing. The ponds and rivers do not dry up easily, and thus the loss of water from the surface of the earth is reduced.
Some applications of high Lv value of water in our daily life:
1. Cooking food in steam is faster.
• This is because, steam contains more heat energy
♦ The heat which brought it up to the boiling point + the latent heat
• So more heat is transferred to the food and it gets cooked faster.
• But 'boiling water' contains only that heat energy which brought it upto the boiling point
2. Steam is used in thermal power plants
• In thermal power plants, water is first boiled to make steam.
♦ This boiling can be done using fossil fuels like coal or diesel. Nuclear energy can also be used.
• The steam thus created is used to rotate turbines. The turbines are attached to generators. Thus electricity is produced.
• Using steam have some advantages:
♦ After steam has done it's job of rotating the turbines, it is condensed to give back the water. This water can be used again to make steam
♦ While condensing, the steam will release latent heat. Using special devices, this heat can be utilized to heat more water, thus reducing the use of fossil fuels
3. Burns (injuries caused by exposure to heat or flame) caused by steam:
• Consider some boiling water.
♦ The maximum temperature that the boiling water can attain is 100 oC
• Steam will be continuously produced from that boiling water.
♦ The freshly formed steam will be at 100 oC.
♦ But if we have the required safety equipments, we can collect that steam, and increase it's temperature further above 100 oC
• Consider the two:
(i) Boiling water which is at 100 oC
(ii) Freshly formed steam which is at 100 oC
Which one of them will cause a more severe burn?
• Ans: The burn from the steam will be more severe.
• This is because, stem has more heat
♦ The heat which brought it up to the boiling point + the latent heat
• So more heat will be transferred to the body, and the burn will be more severe
Solved example 6.10
1 kg of steam at 100 oC is condensed back to water at the same temperature. It is then cooled to 30 oC. Calculate the heat given out during this process.
Solution:
1. Heat given out when 1 kg of stem at 100 is condensed back to water at the same temperature
= Latent heat of steam = 226×104 = 2260000 J
2. Heat given out when this newly condensed water is cooled to 30 = mcθ = mc(θ2-θ1)
= 1×4200×(30 -100) = 1×4200×-70 = 294000 J
3. So total heat given out = 2260000 + 294000 = 2554000 J
Changes of boiling point
Let us discuss the circumstances under which the boiling point changes1. Consider some water being heated in a vessel.
• Under normal atmospheric pressure, when the heat acquired by it is enough to break the internal bonds, it will begin to boil.
• At that time it's temperature will show 100 oC
2. But in some cases, the following things happen:
• The heat acquired by it is enough to break the internal bonds
• The heat acquired by it is enough to show 100 oC
■ But the water does not boil.
3. When does this happen?
• This happens when the pressure acting on the surface of the water is above normal atmospheric pressure.
• In that case, the molecules of water are pressed down by the external pressure.
• So they cannot escape from the liquid.
4. If this happens, we will have to continue supplying heat.
• The temperature will rise above 100 oC. Then the molecules will acquire enough energy and will be able to escape. Thus the water will boil.
5. This property of water is used effectively in a pressure cooker.
Let us see the basics about it's working:
(i) Inside the pressure cooker, high pressure builds up above the surface of the water
(ii) So the water will boil only at a higher temperature.
(iii) That means, water does not escape as steam and is available to cook the food.
(iv) Also this water is at a temperature higher than 100 oC. So it contains a lot of heat.
(v) Thus the food gets cooked faster. But the Pressure cooker should have all the required safety features.
Coolants
• When some substances are dissolved in water, the boiling point will increase.• That means, larger heat energy will be required to boil that water.
• It can be used as coolants in motor vehicles.
• This is because, such water will be able to absorb more heat (produced by the engines) and even then, remain in liquid state with out boiling
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