In the previous section we saw the principle of method of mixtures. In this section we will see change of state.
• We know that there are three states of matter: Solid, Liquid and Gas.
• The change from one state to another occurs by either one of the two:
♦ the absorption of heat
♦ release of heat.
• This can be shown in the form of flow charts:
♦ SOLID (absorbs heat) ⟶ LIQUID (absorbs heat) ⟶ GAS
♦ GAS (releases heat) ⟶ LIQUID (releases heat) ⟶ SOLID
1. Fill half a beaker with ice pieces. Insert a thermometer into it. Fix the thermometer in a stand in such a way that the thermometer does not touch the walls of the beaker.
2. Record the initial temperature.
3. Slowly heat the beaker in a water bath. At the same time start a stop watch.
4. Record the readings in the thermometer every 30 seconds. Also record the changes taking place in the state of the ice pieces. The table 6.3 shown below can be conveniently used for this purpose.
Table 6.3
5. Using the temperatures draw a time – temperature graph. The graph should be drawn with time in the X axis and temperature in the Y axis. The following table shows the recordings made while conducting this experiment in a lab.
The graph based on the table is shown below:
1. The scale along X axis is: 1 unit = 30 seconds
The scale along Y axis is: 1 unit = 20 oC
2. The initial temperature (when time = 0 s) is 0 oC. So the first point in the graph is (0,0)
• We can see that, from the first point itself, the graph is a straight horizontal line upto 120 seconds
• That means., there is no rise in temperature upto 120 s
3. After 120 s, the temperature increases
• We have to see the details during the first 120 s.
• At this time the ice is melting. But it takes 120 s for all the ice to melt. And during this melting process, the temperature does not rise.
4. So where does the heat energy supplied from the water bath go?
Ans: During the 120 seconds, the applied heat energy is used up to break up the bonds between the molecules.
• When the bonds are broken, the molecules become free to flow, and the 'solid ice' becomes 'liquid water'.
• During the process of breaking up of bonds, all the heat energy is used up for that purpose. So there is no increase in kinetic energy. And hence no increase in temperature.
• Thus we get a horizontal line from (0,0) to (120,0).
5. At (120,0), all the ice have melted. The heat energy received from that point onwards will be used to increase the kinetic energy. So the temperature will rise after (120,0)
■ So 'melting point' is a 'constant temperature' at which a solid changes into it's liquid at normal atmospheric pressure.
'Normal atmospheric pressure' is mentioned because, if the solid is not under such a pressure, the melting point will be a different temperature.
We can consider the reverse also:
1. If a liquid substance continues to release heat energy, it's temperature will continuously decrease.
2. But at a certain temperature, it will stop decreasing. Even if it continues to release heat energy, the temperature will not decrease.
3. This is because, the heat released during this time is the energy which kept the molecules apart. When that energy is lost, the molecules bind together to form a solid.
■ The constant temperature at which the liquid changes into it's solid form is called freezing point.
• We can represent the two points as follows:
♦ Melting point ⟹ SOLID (absorbs heat) ⟶ LIQUID
♦ Freezing point ⟹ LIQUID (releases heat) ⟶ SOLID
For most substances, melting point and freezing point is the same temperature
• It may be noted that, the time 120 s is not constant.
♦ If we take a larger quantity of ice, it may take more than 120 s for all the ice to become water
♦ If we take a smaller quantity of ice, it may take less than 120 s for all the ice to become water
• Also, if we supply a larger quantity of heat, within a short span of time, by say using a larger burner, all the ice will become water in a short span of time.
• So the 'time required to break all the bonds' is not constant.
• But the 'quantity of energy required by 1 kg' of the substance to break all the bonds will be a constant. This quantity is called the Latent heat of fusion. ('fusion' is another term for 'melting'). So we can write the definition:
■ Latent heat of fusion (Lf) of a solid is the quantity of heat absorbed by 1 kg of that solid to change into it's liquid state at it's melting point, without change in temperature
• Since it is the energy per 1 kg, it's unit is J⁄kg. That is., J kg-1.
The following table 6.4 shows the Lf values of some common substances
Table 6.4
Let us analyse the values in the above table:
A. Ice:
1. Consider some ice kept in a vessel. Let it's temperature be -10 oC.
2. Let this ice be heated. Once we begin heating, the temperature will begin to rise.
• That is., it will become -9, -8, -7, . . . so on. The graph indicating this rise in temperature will be a rising line.
♦ During this rise in temperature, the ice do not melt
3. But once it reaches 0 oC, the temperature will not rise.
• Also, the ice do not completely melt just when it becomes 0 oC.
• The heat supplied just after reaching 0 oC, will be used to break the strong bonds between the water molecules in the ice.
• Each 1 kg of ice in that vessel will require 335×103 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 ice began to melt sooner.
• For our present discussion, the time in which 335×103 J kg-1 is supplied is not important.
• What is important is that, 335×103 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the ice begins to melt.
• But the temperature of the newly formed water will remain at 0 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 water.
• So the temperature do not rise above 0 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 water' (at 0 oC) and 'no ice'.
6. Further energy supply will be used to increase the temperature from 0 oC up wards. So the graph will again be a rising line.
B. Silver:
1. Consider some silver kept at room temperature say 30 oC.
2. Let this silver 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 silver do not melt
3. But once it reaches 962 oC, the temperature will not rise.
• Also, the silver do not completely melt just when it becomes 962.
• The heat supplied just after reaching 962 oC, will be used to break the strong bonds between the molecules in the silver.
• Each 1 kg of solid silver will require 88×103 joules of heat energy to break all the bonds it 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 silver began to melt sooner.
• For our present discussion, the time in which 88×103 J kg-1 is supplied is not important.
• What is important is that, 88×103 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the silver begins to melt.
• But the temperature of the newly formed liquid silver will remain at 962 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 liquid silver.
• So the temperature do not rise above 962 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 liquid silver' (at 962 oC) and 'no solid silver'.
6. Further energy supply will be used to increase the temperature from 962 oC up wards. So the graph will again be a rising line
C. Copper:
The reader may write the steps for copper in the same way
Now we can explain some common phenomenon that we see around us:
1. Large glaciers do not melt all of a sudden:
• Glaciers are large masses of ice.
• During winter seasons, the temperature of the surroundings will most probably be below zero. So there is no chance for them to melt.
• But when summer approaches, the temperature may become zero or above. Then the ice should melt.
• But the glaciers do not melt all of a sudden. This is because, latent heat of fusion of ice is large even for 1 kg.
• So very large heat will be required for huge masses of ice.
• But the phenomenon of Global warming is supplying this required quantity at an alarming rate.
2. It feels much colder when an ice piece at 0 oC is placed in the mouth than when drinking water which is at 0 oC:
• When we place ice in our mouth we feel cold because heat flows from the mouth into the ice. The mouth loses heat and we feel cold
♦ When we place cold water also, the same thing happens: Heat flows from the mouth into the water. The mouth loses heat and we feel cold
• But the ice takes up more heat from the mouth. This is because, the ice first has to become water at 0 oC, by taking up the latent heat of fusion. Once it becomes water at 0 oC, it takes additional heat to increase it's temperature from 0 oC upwards.
• So in the case of ice, there is an additional latent heat, being taken from the mouth. So we feel much colder
Solved example 6.8
(a) A 1.5 kg sample of copper is at 1083 oC and is continuously heated to melt it into it's liquid state. How much heat will be required to make the sample completely melt into liquid copper at 1083 oC?
(b) A 1.0 kg sample of ice is at 0 oC and is continuously heated to melt it into it's liquid state. How much heat will be required to make the sample completely melt into water at 0 oC?
Solution:
(a) • The melting point of copper is 1083 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid copper at 1083 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 1.5 kg of the solid copper is 1083 oC
♦ Point B at which all the solid is melted into 1.5 kg of liquid copper which is at 1083 oC
• Beyond point B, the heat energy will be used to increase the temperature of the liquid. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of solid copper at 1083 oC to 1 kg of liquid copper at 1083 oC = Lf of copper = 180 × 103 J kg-1
2. So Energy required for 1.5 kg = 1.5 × 180 × 103 = 270 × 103 J
(b) • The melting point of ice is 0 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid water at 0 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 1.0 kg of the ice is 0 oC
♦ Point B at which all the ice is melted into 1.0 kg of water which is at 0 oC
• Beyond point B, the heat energy will be used to increase the temperature of the water. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 335 × 103 J kg-1
• This is the required value because, the given sample is also of 1 kg.
■ We can see that, though the given ice has less mass than the given copper, that ice requires more energy
Solved example 6.9
Calculate the quantity of heat required to convert 5 kg of ice at 0 oC into water at the same temperature
Solution:
The melting point of ice is 0 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid water at 0 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 5.0 kg of the ice is 0 oC
♦ Point B at which all the ice is melted into 5.0 kg of water which is at 0 oC
• Beyond point B, the heat energy will be used to increase the temperature of the water. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 335 × 103 J kg-1
2. So Energy required for 5 kg = 5 × 335 × 103 = 1675 × 103 J
Solved example 6.10
A solid is at an initial temperature of 0 oC. It is continuously heated. The graph showing the change in temperature according to the heat energy absorbed is given below:
Based on the graph, calculate the following:
(a) Mass of the solid
(b) Latent heat of fusion (Lf) of the substance
Given: Specific heat capacity of the substance = 500 J kg-1 K-1
Solution:
Part (a):
1. Let 'm' be the mass of the solid
2. Consider the first sloping line in the graph. The end point of this sloping line is (1000, 80). From this, we get two information:
• The total energy absorbed by the substance upto that point is 1000 J
• This 1000 J causes an increase in temperature from 0 oC to 80 oC
So we can write:
• Q = 1000 J
• rise in temperature (θ) = (θ2 - θ1) = (80 - 0) = 80
3. We have: Q = mcθ.
substituting the known values, we get:
1000 = m×500×80 ⟹1000 = m×40000 ⟹ m = 0.025 kg
Part (b):
1. From the graph we can see that, the end points of the horizontal line are: (1000,80) and (2000,80). From this we get two information:
• When the temperature becomes 80 oC, it becomes constant. That means, it has reached the melting point.
• It continues to melt at the constant temperature of 80 oC until it's whole mass becomes liquid.
• The total energy absorbed until the beginning of the melting process is 1000 J
• The total energy absorbed until the end of the melting process is 2000 J
• So the total energy absorbed during the melting process is (2000 - 1000) = 1000 J
• Once it completely becomes liquid, the temperature will begin to rise again
2. So 0.025 kg of that substance requires 1000 J to completely melt into liquid state
• So energy required by 1 kg = 1000⁄0.025 = 40000 J
3. The energy required by 1 kg is the latent heat of fusion. So we can write:
• Lf of the substance = 40000 J kg-1
• We know that there are three states of matter: Solid, Liquid and Gas.
• The change from one state to another occurs by either one of the two:
♦ the absorption of heat
♦ release of heat.
• This can be shown in the form of flow charts:
♦ SOLID (absorbs heat) ⟶ LIQUID (absorbs heat) ⟶ GAS
♦ GAS (releases heat) ⟶ LIQUID (releases heat) ⟶ SOLID
■ How do a change in state occur when heat is either absorbed or released?
To find the answer, we will do an experiment:1. Fill half a beaker with ice pieces. Insert a thermometer into it. Fix the thermometer in a stand in such a way that the thermometer does not touch the walls of the beaker.
2. Record the initial temperature.
3. Slowly heat the beaker in a water bath. At the same time start a stop watch.
4. Record the readings in the thermometer every 30 seconds. Also record the changes taking place in the state of the ice pieces. The table 6.3 shown below can be conveniently used for this purpose.
Table 6.3
| Time (s) | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| Temperature | ||||||||||
| Change in state of ice pieces | ||||||||||
| Time (s) | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|
| 30 | 60 | 90 | 120 | 150 | 180 | 210 | 240 | 270 | 300 | |
| Temperature | 0 | 0 | 0 | 0 | 30 | 60 | 90 | 120 | 150 | 180 |
| Change in state of ice pieces | Ice continues to melt | No ice piece left | ||||||||
1. The scale along X axis is: 1 unit = 30 seconds
The scale along Y axis is: 1 unit = 20 oC
2. The initial temperature (when time = 0 s) is 0 oC. So the first point in the graph is (0,0)
• We can see that, from the first point itself, the graph is a straight horizontal line upto 120 seconds
• That means., there is no rise in temperature upto 120 s
3. After 120 s, the temperature increases
• We have to see the details during the first 120 s.
• At this time the ice is melting. But it takes 120 s for all the ice to melt. And during this melting process, the temperature does not rise.
4. So where does the heat energy supplied from the water bath go?
Ans: During the 120 seconds, the applied heat energy is used up to break up the bonds between the molecules.
• When the bonds are broken, the molecules become free to flow, and the 'solid ice' becomes 'liquid water'.
• During the process of breaking up of bonds, all the heat energy is used up for that purpose. So there is no increase in kinetic energy. And hence no increase in temperature.
• Thus we get a horizontal line from (0,0) to (120,0).
5. At (120,0), all the ice have melted. The heat energy received from that point onwards will be used to increase the kinetic energy. So the temperature will rise after (120,0)
■ So 'melting point' is a 'constant temperature' at which a solid changes into it's liquid at normal atmospheric pressure.
'Normal atmospheric pressure' is mentioned because, if the solid is not under such a pressure, the melting point will be a different temperature.
We can consider the reverse also:
1. If a liquid substance continues to release heat energy, it's temperature will continuously decrease.
2. But at a certain temperature, it will stop decreasing. Even if it continues to release heat energy, the temperature will not decrease.
3. This is because, the heat released during this time is the energy which kept the molecules apart. When that energy is lost, the molecules bind together to form a solid.
■ The constant temperature at which the liquid changes into it's solid form is called freezing point.
• We can represent the two points as follows:
♦ Melting point ⟹ SOLID (absorbs heat) ⟶ LIQUID
♦ Freezing point ⟹ LIQUID (releases heat) ⟶ SOLID
For most substances, melting point and freezing point is the same temperature
• It may be noted that, the time 120 s is not constant.
♦ If we take a larger quantity of ice, it may take more than 120 s for all the ice to become water
♦ If we take a smaller quantity of ice, it may take less than 120 s for all the ice to become water
• Also, if we supply a larger quantity of heat, within a short span of time, by say using a larger burner, all the ice will become water in a short span of time.
• So the 'time required to break all the bonds' is not constant.
• But the 'quantity of energy required by 1 kg' of the substance to break all the bonds will be a constant. This quantity is called the Latent heat of fusion. ('fusion' is another term for 'melting'). So we can write the definition:
■ Latent heat of fusion (Lf) of a solid is the quantity of heat absorbed by 1 kg of that solid to change into it's liquid state at it's melting point, without change in temperature
• Since it is the energy per 1 kg, it's unit is J⁄kg. That is., J kg-1.
The following table 6.4 shows the Lf values of some common substances
Table 6.4
| Substance | Melting point (oC) | Lf (J kg-1) |
|---|---|---|
| Ice | 0 | 335 × 103 |
| Silver | 962 | 88 × 103 |
| Copper | 1083 | 180 × 103 |
A. Ice:
1. Consider some ice kept in a vessel. Let it's temperature be -10 oC.
2. Let this ice be heated. Once we begin heating, the temperature will begin to rise.
• That is., it will become -9, -8, -7, . . . so on. The graph indicating this rise in temperature will be a rising line.
♦ During this rise in temperature, the ice do not melt
3. But once it reaches 0 oC, the temperature will not rise.
• Also, the ice do not completely melt just when it becomes 0 oC.
• The heat supplied just after reaching 0 oC, will be used to break the strong bonds between the water molecules in the ice.
• Each 1 kg of ice in that vessel will require 335×103 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 ice began to melt sooner.
• For our present discussion, the time in which 335×103 J kg-1 is supplied is not important.
• What is important is that, 335×103 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the ice begins to melt.
• But the temperature of the newly formed water will remain at 0 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 water.
• So the temperature do not rise above 0 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 water' (at 0 oC) and 'no ice'.
6. Further energy supply will be used to increase the temperature from 0 oC up wards. So the graph will again be a rising line.
B. Silver:
1. Consider some silver kept at room temperature say 30 oC.
2. Let this silver 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 silver do not melt
3. But once it reaches 962 oC, the temperature will not rise.
• Also, the silver do not completely melt just when it becomes 962.
• The heat supplied just after reaching 962 oC, will be used to break the strong bonds between the molecules in the silver.
• Each 1 kg of solid silver will require 88×103 joules of heat energy to break all the bonds it 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 silver began to melt sooner.
• For our present discussion, the time in which 88×103 J kg-1 is supplied is not important.
• What is important is that, 88×103 J kg-1 is required to break all the bonds.
5. While this energy is being absorbed, the silver begins to melt.
• But the temperature of the newly formed liquid silver will remain at 962 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 liquid silver.
• So the temperature do not rise above 962 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 liquid silver' (at 962 oC) and 'no solid silver'.
6. Further energy supply will be used to increase the temperature from 962 oC up wards. So the graph will again be a rising line
C. Copper:
The reader may write the steps for copper in the same way
Now we can explain some common phenomenon that we see around us:
1. Large glaciers do not melt all of a sudden:
• Glaciers are large masses of ice.
• During winter seasons, the temperature of the surroundings will most probably be below zero. So there is no chance for them to melt.
• But when summer approaches, the temperature may become zero or above. Then the ice should melt.
• But the glaciers do not melt all of a sudden. This is because, latent heat of fusion of ice is large even for 1 kg.
• So very large heat will be required for huge masses of ice.
• But the phenomenon of Global warming is supplying this required quantity at an alarming rate.
2. It feels much colder when an ice piece at 0 oC is placed in the mouth than when drinking water which is at 0 oC:
• When we place ice in our mouth we feel cold because heat flows from the mouth into the ice. The mouth loses heat and we feel cold
♦ When we place cold water also, the same thing happens: Heat flows from the mouth into the water. The mouth loses heat and we feel cold
• But the ice takes up more heat from the mouth. This is because, the ice first has to become water at 0 oC, by taking up the latent heat of fusion. Once it becomes water at 0 oC, it takes additional heat to increase it's temperature from 0 oC upwards.
• So in the case of ice, there is an additional latent heat, being taken from the mouth. So we feel much colder
Solved example 6.8
(a) A 1.5 kg sample of copper is at 1083 oC and is continuously heated to melt it into it's liquid state. How much heat will be required to make the sample completely melt into liquid copper at 1083 oC?
(b) A 1.0 kg sample of ice is at 0 oC and is continuously heated to melt it into it's liquid state. How much heat will be required to make the sample completely melt into water at 0 oC?
Solution:
(a) • The melting point of copper is 1083 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid copper at 1083 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 1.5 kg of the solid copper is 1083 oC
♦ Point B at which all the solid is melted into 1.5 kg of liquid copper which is at 1083 oC
• Beyond point B, the heat energy will be used to increase the temperature of the liquid. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of solid copper at 1083 oC to 1 kg of liquid copper at 1083 oC = Lf of copper = 180 × 103 J kg-1
2. So Energy required for 1.5 kg = 1.5 × 180 × 103 = 270 × 103 J
(b) • The melting point of ice is 0 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid water at 0 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 1.0 kg of the ice is 0 oC
♦ Point B at which all the ice is melted into 1.0 kg of water which is at 0 oC
• Beyond point B, the heat energy will be used to increase the temperature of the water. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 335 × 103 J kg-1
• This is the required value because, the given sample is also of 1 kg.
■ We can see that, though the given ice has less mass than the given copper, that ice requires more energy
Solved example 6.9
Calculate the quantity of heat required to convert 5 kg of ice at 0 oC into water at the same temperature
Solution:
The melting point of ice is 0 oC. The given sample is at this temperature.
• If we give more heat at this point, the temperature will not increase. That heat will be used to break the bonds between the molecules. When all the bonds are broken we will get liquid water at 0 oC.
• We are asked to find the heat energy to be supplied between the following two points:
♦ Point A at which the temperature of 5.0 kg of the ice is 0 oC
♦ Point B at which all the ice is melted into 5.0 kg of water which is at 0 oC
• Beyond point B, the heat energy will be used to increase the temperature of the water. We need the energy between A and B only
• We can use the following steps:
1. Energy required to melt 1 kg of ice at 0 oC to 1 kg of water at 0 oC = Lf of ice = 335 × 103 J kg-1
2. So Energy required for 5 kg = 5 × 335 × 103 = 1675 × 103 J
Solved example 6.10
A solid is at an initial temperature of 0 oC. It is continuously heated. The graph showing the change in temperature according to the heat energy absorbed is given below:
Based on the graph, calculate the following:
(a) Mass of the solid
(b) Latent heat of fusion (Lf) of the substance
Given: Specific heat capacity of the substance = 500 J kg-1 K-1
Solution:
Part (a):
1. Let 'm' be the mass of the solid
2. Consider the first sloping line in the graph. The end point of this sloping line is (1000, 80). From this, we get two information:
• The total energy absorbed by the substance upto that point is 1000 J
• This 1000 J causes an increase in temperature from 0 oC to 80 oC
So we can write:
• Q = 1000 J
• rise in temperature (θ) = (θ2 - θ1) = (80 - 0) = 80
3. We have: Q = mcθ.
substituting the known values, we get:
1000 = m×500×80 ⟹1000 = m×40000 ⟹ m = 0.025 kg
Part (b):
1. From the graph we can see that, the end points of the horizontal line are: (1000,80) and (2000,80). From this we get two information:
• When the temperature becomes 80 oC, it becomes constant. That means, it has reached the melting point.
• It continues to melt at the constant temperature of 80 oC until it's whole mass becomes liquid.
• The total energy absorbed until the beginning of the melting process is 1000 J
• The total energy absorbed until the end of the melting process is 2000 J
• So the total energy absorbed during the melting process is (2000 - 1000) = 1000 J
• Once it completely becomes liquid, the temperature will begin to rise again
2. So 0.025 kg of that substance requires 1000 J to completely melt into liquid state
• So energy required by 1 kg = 1000⁄0.025 = 40000 J
3. The energy required by 1 kg is the latent heat of fusion. So we can write:
• Lf of the substance = 40000 J kg-1
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