In the previous section we saw the boiling point and the latent heat of vaporisation. In this section we will discuss about Evaporation.
Let us do an experiment:
1. Take some spirit in a watch glass. Take some vinegar in another watch glass. Place them at farther ends of a table.
2. We can see that the quantity of both spirit and vinegar becomes lesser and lesser. That means both are changing from liquid state to gaseous state.
3. But this change in spirit is at a faster rate than in vinegar
4. If we heat the vinegar, it will also change into gaseous form at a faster rate
5. But for spirit, no heating is required
Another experiment:
1. Wrap some cotton around the bulb of a thermometer.
2. Record the initial temperature shown by the thermometer.
3. Wet the cotton using some spirit. Observe the new temperature.
4. We can see that the temperature decreases rapidly.
• This is because, the spirit absorbs heat from it's surroundings. Here, the surroundings is the bulb of the thermometer.
• The heat thus absorbed is used by the spirit to change from liquid state to gaseous state.
• Since the heat is taken up from the bulb, the bulb becomes cold. So the thermometer shows a lesser temperature.
• We can directly feel this if we place a piece of cotton dipped in spirit on our hands. We feel cold where our hand is in contact with the cotton. This is because, heat at that portion is taken up by the spirit. This heat is used by the spirit to change from liquid state to gaseous state.
1. Water molecules comes out to the outer surface of the pot through the minute pores in the walls of the pot
2. Those molecules are exposed to the atmosphere. They take up heat from the walls of the pot and change into vapours.
3. Thus the walls and the water inside gets cooled
■ A sweating person feels more cold if he is sitting under a fan
1. Sweating produces lot of water particles on the surface of the body.
2. The molecules in these water particles absorb heat from the body and change to vapour. So the body gets cooled.
3. When the fan is switched on, the newly formed vapours are quickly removed. So more and more water evaporates. That is., evaporation takes place at a faster rate.
4. So the sweating person feels more cold.
• In the same way, if we wet our hands and wave it in the air, we feel more cold.
■ Spreading out wet cloths
1. After washing, and squeezing out water from clothes, we need to spread them out in open air
2. If we keep it in the squeezed form, they will not become dry. This is because evaporation will not take place from the interior portions
Following are the factors which affect evaporation:
• Nature of substance
• Atmospheric temperature
• Atmospheric pressure
• Presence of wind
• Humidity
• Surface area
The following table 6.6 shows a comparison between Evaporation and Vaporisation:
Table 6.6
Global warming
• Global warming is the increase in temperature of the earth's surface (both land and water) and also of the atmosphere.
• During the past few decades, global warming has been increasing at an alarming rate.
Let us see what causes this phenomenon:
1. The earth receives heat from the sun. This heat is radiated back into the atmosphere.
2. This heat which is radiated back, gets trapped in the green house gases (carbon dioxide, methane etc.,) .
3. This trapped heat keeps the earth warm enough to sustain life.
4. But now, human activities like deforestation, burning of fossil fuels etc., creates large quantities of green house gases. Artificial green house gases like chloro fluoro carbon (CFC) are also produced by human activities.
5. They accumulate in the atmosphere and traps more heat than what is necessary.
6. This causes increase in temperature of the earth.
7. As a result:
• Ice and snow will melt and cause rise in sea level. Coastal areas will be submerged. Low lying islands will vanish.
• The climate in many parts of the world will change. Dry areas will become even more drier. There will be shortage for drinking water.
• The cyclones will become stronger and will cause heavier damages.
• In short, the earth will become uninhabitable.
8. So we must all try to prevent global warming.
• Reducing the use of fossil fuels
• Using public transport facilities instead of individual cars
• Avoiding the use of CFC for refrigeration
• preventing deforestation
• Planting more plants and trees etc., are some efficient methods which will reduce global warming
2 kg ice at -10 oC is continuously heated to melt it completely and then to vapourise it. Calculate the quantity of heat required.
• Lf of ice = 336× 103 J kg-1 • Lv of water = 226 × 104 J kg-1
• c of ice = 2100 J kg-1 K-1
• c of water = 4200 J kg-1 K-1
Solution:
The given sample of ice is at -10 oC. We have to go through the following stages:
Stage 1: Heat the ice to bring it's temperature up to it's melting point, which is 0 oC
Stage 2: Heat the ice at oC to melt it into water at 0 oC
Stage 3: Heat the water at 0 oC to bring it's temperature upto it's boiling point, which is 100 oC
Stage 4: Heat the water at 100 oC to change it into vapour at 100 oC
Now we can write the steps:
1. Heat for stage 1 = mcθ = 2×2100×[0-(-10)] = 2×2100×10= 42000 J
2. Heat for stage 2:
• Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 336 × 103 J kg-1
• So Energy required for 2 kg = 2 × 336 × 103 = 672 × 103 J
3. Heat for stage 3 = mcθ = 2×4200×(100-0) = 2×4200×100= 840000 J
4. Heat for stage 4:
• Energy required to vapourise 1 kg of water at 100 oC to 1 kg of steam at 100 oC = Lv of water = 226 × 104 J kg-1
• So Energy required for 2 kg = 2 × 226 × 104 = 452 × 104 J
5. So total heat energy required = 42000 + 672000 + 840000 + 2260000 = 3814000 J
Solved example 6.12
An ice block is at 273 K. A hole was drilled in it. When the hole was filled with water at 373 K, 2 kg of ice melted and changed into water at 273 K. If so, what is the mass of the water that was used?
Solution:
1. The ice block was already at it's melting point (273 K). So any heat supplied at this stage will be used to break the internal bonds of ice. There will be no rice in temperature until all ice has melted.
2. It is given that water at 273 K was obtained. So indeed there is no rise in temperature.
3. Mass of water obtained is 2 kg. So it is clear that, 2 kg of ice (at 273 K) melted to give 2 kg of water (at 273 K)
4. The heat required for this melting can be found out as follows:
• Energy required to melt 1 kg of ice at 0 oC (273 K) to 1 kg of water at 0 oC = Lf of ice = 336 × 103 J kg-1
• So Energy required for 2 kg = 2 × 336 × 103 = 672 × 103 J
5. This much energy is supplied by the hot water which was at 100 oC (373 K)
• Energy lost by the hot water was used to melt the ice
6. Energy lost by the hot water = mcθ = m×4200×[0-(-100)] = m×4200×(-100)= -420000m J (The negative sign indicates that heat is lost)
7. Equating the results in (4) and (6) we get:
670 × 103 = 420000 m same as m = 672000⁄420000 = 1.6 kg
Solved example 6.13
The graphic representation of cooling of 500 grams of a liquid is given below:
Answer the following questions:
(i) What is the melting point?
(ii) Is there any change in temperature from 50 s to 60 s?
(iii) Where does the decrease in kinetic energy of the molecules begin? Where does the decrease in potential energy of the molecules begin?
(iv) What is the state of the matter after 60 s?
(v) How much heat energy is given out during the first 50 s?
• c of the liquid = 3000 J kg-1 K-1
• Lf of the liquid = 150 × 103 J kg-1
Solution:
Part (i):
1. The liquid cools down from point A to point B
During this process, the temperature continuously decreases with the passage of time
2. But after point B, the temperature is constant. So B corresponds to the melting point (freezing point)
The temperature at B is 90 oC. So the melting point is 90 oC.
Part (ii)
There is no change in temperature during the time span between 50 s and 60 s
Part (iii)
1. The decrease in kinetic energy of the molecules occurs between A and B and also between C and D
2. The decrease in potential energy of the molecules occurs between B and C
Part (iv)
After 60 s, the matter will be in the solid state
Part (v)
Heat given out = mcθ = 0.5×3000×(90-120) = 0.5×3000×(-30)= -45000 J (The negative sign indicates that heat is lost)
Let us do an experiment:
1. Take some spirit in a watch glass. Take some vinegar in another watch glass. Place them at farther ends of a table.
2. We can see that the quantity of both spirit and vinegar becomes lesser and lesser. That means both are changing from liquid state to gaseous state.
3. But this change in spirit is at a faster rate than in vinegar
4. If we heat the vinegar, it will also change into gaseous form at a faster rate
5. But for spirit, no heating is required
Another experiment:
1. Wrap some cotton around the bulb of a thermometer.
2. Record the initial temperature shown by the thermometer.
3. Wet the cotton using some spirit. Observe the new temperature.
4. We can see that the temperature decreases rapidly.
• This is because, the spirit absorbs heat from it's surroundings. Here, the surroundings is the bulb of the thermometer.
• The heat thus absorbed is used by the spirit to change from liquid state to gaseous state.
• Since the heat is taken up from the bulb, the bulb becomes cold. So the thermometer shows a lesser temperature.
• We can directly feel this if we place a piece of cotton dipped in spirit on our hands. We feel cold where our hand is in contact with the cotton. This is because, heat at that portion is taken up by the spirit. This heat is used by the spirit to change from liquid state to gaseous state.
• In the above experiments, the molecules at the surface of the liquid escapes into the atmosphere.
• Once they have escaped, there is no binding forces between those molecules. That is., they change from liquid state to gaseous state.
• When this change occurs, some heat is absorbed from the surroundings. So the surroundings gets cooled.
■ Evaporation is the process by which a liquid changes to it's vapor form by absorbing heat from the surroundings. This is a slow and normal process that takes place on the surface of liquids at all temperatures. During evaporation, the substance which supplies the heat gets cooled.
Let us see some day to day situations where evaporation is observed:
■ Water kept in earthen pots gets cooled well1. Water molecules comes out to the outer surface of the pot through the minute pores in the walls of the pot
2. Those molecules are exposed to the atmosphere. They take up heat from the walls of the pot and change into vapours.
3. Thus the walls and the water inside gets cooled
■ A sweating person feels more cold if he is sitting under a fan
1. Sweating produces lot of water particles on the surface of the body.
2. The molecules in these water particles absorb heat from the body and change to vapour. So the body gets cooled.
3. When the fan is switched on, the newly formed vapours are quickly removed. So more and more water evaporates. That is., evaporation takes place at a faster rate.
4. So the sweating person feels more cold.
• In the same way, if we wet our hands and wave it in the air, we feel more cold.
■ Spreading out wet cloths
1. After washing, and squeezing out water from clothes, we need to spread them out in open air
2. If we keep it in the squeezed form, they will not become dry. This is because evaporation will not take place from the interior portions
Following are the factors which affect evaporation:
• Nature of substance
• Atmospheric temperature
• Atmospheric pressure
• Presence of wind
• Humidity
• Surface area
The following table 6.6 shows a comparison between Evaporation and Vaporisation:
Table 6.6
| Evaporation | Vaporisation | |
|---|---|---|
| 1 | Occurs at all temperatures | Occurs only at the boiling point of the liquid |
| 2 | Takes place only from the surface of the liquid | Takes place from all parts of the liquid |
| 3 | Influenced by atmospheric pressure, temperature and humidity | Not at all influenced by atmospheric pressure, temperature or humidity |
| 4 | Evaporation causes cooling | No cooling is experienced |
• Global warming is the increase in temperature of the earth's surface (both land and water) and also of the atmosphere.
• During the past few decades, global warming has been increasing at an alarming rate.
Let us see what causes this phenomenon:
1. The earth receives heat from the sun. This heat is radiated back into the atmosphere.
2. This heat which is radiated back, gets trapped in the green house gases (carbon dioxide, methane etc.,) .
3. This trapped heat keeps the earth warm enough to sustain life.
4. But now, human activities like deforestation, burning of fossil fuels etc., creates large quantities of green house gases. Artificial green house gases like chloro fluoro carbon (CFC) are also produced by human activities.
5. They accumulate in the atmosphere and traps more heat than what is necessary.
6. This causes increase in temperature of the earth.
7. As a result:
• Ice and snow will melt and cause rise in sea level. Coastal areas will be submerged. Low lying islands will vanish.
• The climate in many parts of the world will change. Dry areas will become even more drier. There will be shortage for drinking water.
• The cyclones will become stronger and will cause heavier damages.
• In short, the earth will become uninhabitable.
8. So we must all try to prevent global warming.
• Reducing the use of fossil fuels
• Using public transport facilities instead of individual cars
• Avoiding the use of CFC for refrigeration
• preventing deforestation
• Planting more plants and trees etc., are some efficient methods which will reduce global warming
Now we will see a few more solved examples:
Solved example 6.112 kg ice at -10 oC is continuously heated to melt it completely and then to vapourise it. Calculate the quantity of heat required.
• Lf of ice = 336× 103 J kg-1 • Lv of water = 226 × 104 J kg-1
• c of ice = 2100 J kg-1 K-1
• c of water = 4200 J kg-1 K-1
Solution:
The given sample of ice is at -10 oC. We have to go through the following stages:
Stage 1: Heat the ice to bring it's temperature up to it's melting point, which is 0 oC
Stage 2: Heat the ice at oC to melt it into water at 0 oC
Stage 3: Heat the water at 0 oC to bring it's temperature upto it's boiling point, which is 100 oC
Stage 4: Heat the water at 100 oC to change it into vapour at 100 oC
Now we can write the steps:
1. Heat for stage 1 = mcθ = 2×2100×[0-(-10)] = 2×2100×10= 42000 J
2. Heat for stage 2:
• Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 336 × 103 J kg-1
• So Energy required for 2 kg = 2 × 336 × 103 = 672 × 103 J
3. Heat for stage 3 = mcθ = 2×4200×(100-0) = 2×4200×100= 840000 J
4. Heat for stage 4:
• Energy required to vapourise 1 kg of water at 100 oC to 1 kg of steam at 100 oC = Lv of water = 226 × 104 J kg-1
• So Energy required for 2 kg = 2 × 226 × 104 = 452 × 104 J
5. So total heat energy required = 42000 + 672000 + 840000 + 2260000 = 3814000 J
Solved example 6.12
An ice block is at 273 K. A hole was drilled in it. When the hole was filled with water at 373 K, 2 kg of ice melted and changed into water at 273 K. If so, what is the mass of the water that was used?
Solution:
1. The ice block was already at it's melting point (273 K). So any heat supplied at this stage will be used to break the internal bonds of ice. There will be no rice in temperature until all ice has melted.
2. It is given that water at 273 K was obtained. So indeed there is no rise in temperature.
3. Mass of water obtained is 2 kg. So it is clear that, 2 kg of ice (at 273 K) melted to give 2 kg of water (at 273 K)
4. The heat required for this melting can be found out as follows:
• Energy required to melt 1 kg of ice at 0 oC (273 K) to 1 kg of water at 0 oC = Lf of ice = 336 × 103 J kg-1
• So Energy required for 2 kg = 2 × 336 × 103 = 672 × 103 J
5. This much energy is supplied by the hot water which was at 100 oC (373 K)
• Energy lost by the hot water was used to melt the ice
6. Energy lost by the hot water = mcθ = m×4200×[0-(-100)] = m×4200×(-100)= -420000m J (The negative sign indicates that heat is lost)
7. Equating the results in (4) and (6) we get:
670 × 103 = 420000 m same as m = 672000⁄420000 = 1.6 kg
Solved example 6.13
The graphic representation of cooling of 500 grams of a liquid is given below:
Answer the following questions:
(i) What is the melting point?
(ii) Is there any change in temperature from 50 s to 60 s?
(iii) Where does the decrease in kinetic energy of the molecules begin? Where does the decrease in potential energy of the molecules begin?
(iv) What is the state of the matter after 60 s?
(v) How much heat energy is given out during the first 50 s?
• c of the liquid = 3000 J kg-1 K-1
• Lf of the liquid = 150 × 103 J kg-1
Solution:
Part (i):
1. The liquid cools down from point A to point B
During this process, the temperature continuously decreases with the passage of time
2. But after point B, the temperature is constant. So B corresponds to the melting point (freezing point)
The temperature at B is 90 oC. So the melting point is 90 oC.
Part (ii)
There is no change in temperature during the time span between 50 s and 60 s
Part (iii)
1. The decrease in kinetic energy of the molecules occurs between A and B and also between C and D
2. The decrease in potential energy of the molecules occurs between B and C
Part (iv)
After 60 s, the matter will be in the solid state
Part (v)
Heat given out = mcθ = 0.5×3000×(90-120) = 0.5×3000×(-30)= -45000 J (The negative sign indicates that heat is lost)
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