WBCHSE Class 11 Physics Change Of State Of Matter Fusion And Solidification Notes

Properties Of Bulk Matter Change Of State Of Matter Fusion And Solidification

Fusion or Melting: When heat is applied to a solid, initially its temperature rises. After reaching a particular temperature, the solid begins to melt. Until the whole substance converts to a liquid, the temperature does not change thought heat is continuously applied.

As soon as the solid completely melts, any further application of heat raises the temperature of the liquid again. The variation of temperature (due to application of heat at a constant rate) with time is shown in the graph.

Point A represents the initial temperature of the solid. On heating, it continues to be in solid state but its temperature increases. The line AB represents this stage.

Point B corresponds to a temperature θ where melting starts. The temperature remains at θ up to point C even though the heating continues. The part BC in the graph represents this stage. At this stage the specimen remains as a mixture of its solid and liquid forms θ is called the melting point of the solid.

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Point C represents the stage where the whole mass of the solid has changed to liquid, and on further heating, a rise in temperature is noticed. This is represented by the part CD of the graph.

Solidification or freezing: On absorbing heat, a solid changes into a liquid and on losing heat, a liquid changes into a solid. Freezing, therefore, can be considered as the reverse process of melting.

When heat is extracted from a liquid, its temperatilre decreases initially. On reaching a particular temperature, the liquid begins to freeze. Until the whole liquid freezes, its temperature remains constant despite extraction of heat.

Once the whole mass of liquid solidifies, further extraction of heat results in cooling of the solid substance. A graphical representation of the change in temperature of the liquid (due to extraction of heat at a constant rate) with respect to time is shown.

Point A denotes the initial temperature of the liquid. The line BC represents solidification at constant temperature θ, which is called the freezing point of the liquid and CD represents the cooling of the solid thus formed.

• Melting Point
• Freezing Point

Melting point: At a fixed pressure, the temperature at which a solid starts melting on absorbing heat, is called its melting point at that pressure.

Freezing point: At a fixed pressure, the temperature at which a liquid starts freezing into a solid by losing heat, is called its freezing point at that pressure.

• Melting point or freezing point for different substances are different. Melting point or freezing point of a substance depends on the superincumbent pressure. Melting point or freezing point of a substance at standard atmospheric pressure is called its normal melting point or normal freezing point.
• Melting point and freezing point are exactly equal and fixed only for a pure crystalline substance. A noncrystalline substance like glass, wax, fat, butter, pitctar etc., does not have any fixed freezing or melting point.
• This type of substance has a temperature range within which it starts melting. (For example, Butter melts within 28 °C to 33°C whereas freezes within 20°C to 23°C).
• On heating, non-crystalline substances, first attain a viscous stage which is neither purely a liquid state nor purely a solid state. So these substances, do not have any fixed melting or freezing point.

In pure substances and mixtures of substances do not have fixed melting points. Again, a few substances like magnesium oxide (MgO), calcium oxide (CaO) etc. do not melt at any temperature.

Normal freezing/melting points of some substances

Melting point of an alloy: Melting point of an alloy is generally less than that of its individual constituents. Solder, an alloy of lead and tin, has a melting point of 180°C, lower than the melting points of tin (232°C) and lead (327°C). Wood’s metal (an alloy of tin, lead, bismuth and cadmium) has a melting point of 60.5°C, whereas its constituents have melting points greater than 60.5°C.

An alloy of lead, tin and bismuth, called Rose’s metal, melts at 94.5°C, while none of its constituents has a melting point lower than 210°C. Lower melting point of all alloy finds its application in the construction of fuse wires an electrical circuit. Such alloys are also used in fire alarms and circuit breakers.

Freezing point Of a solution: Any pure liquid has a fixed freezing point. When a solute is dissolved in it, the freezing point of the solution becomes lower than that of the pure liquid. For example, while the normal freezing point of pure water is 0°C, saline sea water freezes at -2°C.

If the amount of salt in saline water increases, freezing point of the solution decreases. It has been observed that the lowest freezing point of saline water is -22 °C when salt and water are in the mass ratio of 1: 3. So, water freezes in lakes, tanks etc. in cold countries, but sea water does not freeze easily. For the same reason, freezing point of milk is a little less than that of pure water.

• In cold countries to prevent bursting of the radiator tube of a car due to the freezing of water, glycerine or glycol is mixed with water. As a result, the freezing point decreases and water does not freeze even at 0°C. Glycerine, glycol etc. are called antifreeze substances.
• When a solution is gradually cooled and the solution starts solidifying, the pure solvent gets separated in crystalline form, from the solution. Hence, the proportion of solute gradually increases in the rest of the solution and the solution becomes denser.
• At the same time the freezing point of the solution also decreases. If the cooling process is continued, then more solvent, in crystalline form, gets separated from the solution. Finally, at a certain temperature, the rest of the solution condenses as a whole into a solid. This particular temperature is called the eutectic temperature.

Freezing point Of a solution Definition: The temperature at which a solution as a whole condenses into a solid, is called the eutectic temperature of that solution.

Following the method given above both salt and pure water can be prepared. As the temperature of saline water starts falling below 0°C, pure ice gets separated from the solution. It can be melted to obtain pure water.

In cold countries, salt is prepared from seawater. Sea water starts freezing at -2°C. At this temperature, some amount of pure water gets separated as ice. Hence, the amount of salt increases in the rest of sea water. This seawater is vaporised to obtain salt.

Change of State of Matter Notes for Class 11

Freezing mixture: Freezing mixture is a mixture of salt and ice. For instance, the temperature of a mixture consisting of 3 parts crushed ice and 1 part common salt is -22CC, and that of a mixture of 2 parts ice and 3 parts decreased to calcium chloride is almost -50°C.

Practical uses of freezing mixture: For preservation and transport of food materials like fish, and meat, that decay easily due to putrefaction, freezing mixture is extensively used. It is also used for making ice cream, kulfi etc. and in laboratories to create low temperatures.

Latent Heat of Fusion Definition: The temperature at which a solution as a whole condenses into a solid, is called the eutectic temperature of that solution.

Similarly, the amount of heat that needs to be extracted to change a liquid of unit mass into its solid state, at a constant temperature, is called the latent heat of solidification of that liquid.

The latent heat of fusion and the latent heat of solidification of a substance are equal. Latent heat is generally denoted by L.

Latent Heat of Fusion Explained

If H is the quantity of heat that is to be supplied to or extracted from a mass m of a substance to melt or solidify it then H = mL.

unit of heat unit of mass = \(\frac{\text { unit of heat }}{\text { unit of mass }}\)

Units of Latent Heat:

• CGS System: cal · g^-1
• SI: J · kg^-1

∴ \(1 \mathrm{cal} \cdot \mathrm{g}^{-1}=\frac{4.2 \mathrm{~J}}{10^{-3} \mathrm{~kg}}=4200 \mathrm{~J} \cdot \mathrm{kg}^{-1}\)

Latent heat of fusion of ice or latent heat of ice is 80 cal · g^-1: it means that to melt 1 g of ice at 0 °C into 1 g of water at 0 °C, 80 cal heat is to be supplied, or, to solidify 1 g of water 0°C into 1 g of ice at 0°C, 80 cal heat is to be extracted.

Latent heat of ice = 80 cal · g^-1 = 80 x 10³ cal · kg^-1

= 80 x 10³ x 4.2 J · kg^-1

= 3.36 x 10⁵ J · kg^-1

Therefore, in SI, the latent heat of ice is 3.36 x 10⁵ J · kg^-1.

Latent heat of fusion some substances

Examples of Phase Changes: Fusion and Solidification

Properties Of Bulk Matter – Change Of State Of Matter Fusion And Solidification Numerical Examples

Practice Questions on Change of State for Class 11

Example 1. A copper calorimeter of mass 300 g contains 270 g of water at 30° C. Now 20 g of ice at -10°C is added to it. What will be the final temperature? Specific heat capacity of copper = 0.1 cal · g^-1 · °C , specific heat capacity of ice = 0.5 cal · g^-1 · °C^-1, latent heat of fusion of ice = 80 cal · g^-1
Solution:

Given

A copper calorimeter of mass 300 g contains 270 g of water at 30° C. Now 20 g of ice at -10°C is added to it.

Heat lost by calorimeter and water in cooling from 30°C to 0°C = 300 x 0.1 x 30 + 270 x 30 = 9000 cal

Heat taken by ice to rise from -10°C to 0°C = 20 x 0.5 x 10 = 100 cal

Heat taken for melting of ice = 20 x 80 = 1600 cal.

Total heat required to raise 20 g of ice from -10°C to 0°C and to melt it = 100 + 1600 = 1700 cal.

As heat lost is greater than heat required, the whole of ice will melt and the extra (9000 – 1700) or 7300 cal of heat will increase the temperature of the calorimeter and the mixture. Let the final temperature = t°C.

∴ 300 x 0.1 x t+ (270 + 20) x 1 x t = 7300

or, \(320 t=7300 \text { or, } t=\frac{7300}{320}=22.81^{\circ} \mathrm{C}\)

∴ Final temperature of the mixture = 22.81 °C.

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Example 2. 90 g of water at 40°C is kept in a calorimeter of water equivalent lOg. What is the effect of immersion 100 g of ice at -10°C in it? The specific heat capacity of ice is = 0.5 cal · g^-1 °C^-1, latent heat of fusion of ice = 80 cal · g^-1.
Solution:

Given

90 g of water at 40°C is kept in a calorimeter of water equivalent lOg.

Maximum heat that can be lost by the calorimeter and the contained water in cooling up to 0°C from 40°C = (10 + 90) x 40 = 4000 cal.

Heat taken by 100 g of ice to rise from -10°C to 0°C = 100 x 0.5 x 10 = 500 cal.
0°C

∴ Total heat required = 500 + 8000 = 8500 cal

Since, heat required is more than the maximum heat available, some ice will not melt. Let us assume that mg of ice melts.

∴ \(500+80 m=4000 \quad \text { or, } 80 m=3500\)

or, m = \(\frac{3500}{80}=43.75\)

Hence, only 43.75 g ice will melt. Mass of ice that remains unchanged = 100-43.75 = 56.25 g. The final temperature of the mixture will be 0°C.

Therefore, the calorimeter will finally contain 56.25 g of ice and (90 + 43.75) g or 133.75 g of water.

Example 3. Densities of ice and water at 0°C are 0.916 g- cm-3 and 1 g · cm^-3 respectively. A metal piece of mass 10 g at 100°C is dropped in a mixture of ice and water. Some ice melts and the volume of the mixture decreases by 0.1cm³ without any change in temperature. If latent heat of fusion of ice is 80 cal · g^-1, find the specific heat of the metal.
Solution:

Given

Densities of ice and water at 0°C are 0.916 g- cm-3 and 1 g · cm^-3 respectively. A metal piece of mass 10 g at 100°C is dropped in a mixture of ice and water. Some ice melts and the volume of the mixture decreases by 0.1cm³ without any change in temperature. If latent heat of fusion of ice is 80 cal · g^-1

Volume of 1 g ice at 0°C = 1/0.916 = 1.092 cm³

Volume of 1g of water at 0°C = 1/1 = 1 cm³

Hence, contraction in volume on melting of 1 g of ice = 1.092- 1 = 0.092 cm³

∴ Contraction of 0.1 cm³ in volume is due lo melting of \(\frac{0.1}{0.092} \mathrm{~g}\) or 1.087 8 of ice.

Heat needed to melt 1.087 g of ice = 1.087 x 80 cal

Let specific heat of the metal be s.

⇔ 10 x s x 100 = 1.087 x 80

or, \(s=\frac{1.087 \times 80}{1000}=0.087 \mathrm{cal} \cdot \mathrm{g}^{-1} \cdot{ }^{\circ} \mathrm{C}^{-1} \text {. }\)

Example 4. A calorimeter of mass 50 g contains 200 g of water j at 20°C. 20 g ice at 0°C is mixed with it. Final temperature of the mixture becomes 11 °C. What is the latent heat of fusion of ice? The specific heat of the material of the calorimeter is 0.095 cal · g^-1 · °C^-1.
Solution:

Given

A calorimeter of mass 50 g contains 200 g of water j at 20°C. 20 g ice at 0°C is mixed with it.

Let latent heat of fusion of ice = L cal • g^-1.

Heat taken to change 20 g of ice at 0°C to water at 0°C = 20 L cal

Heat taken by 20 g of water to rise to 11°C from 0°C = 20x 1 x 11 = 220 cal.

Heat lost by the calorimeter and its contents in cooling from 20°C to 11°C

= 50 x 0.095 x (20 – 11)+ 200 x 1 x (20-11)

= 1842.75 cal

Heat gain = heat lost

∴ 220 + 20L = 1842.75 or, L = \(\frac{1622.75}{20}=\) = 81.137

Therefore, the latent heat of fusion of ice is 81.137 cal · g^-1.

Example 5. When a solid of mass 50 g is heated at a constant rate of 5 cal · s^-1, there is a rise in temperature of 11°C in the first minute. For the next 13 min of heating, the temperature remains constant. Rate of rise of temperature after this is 6°C ·min^-1. Find

1. specific heat in the solid state,
2. specific heat in the liquid state and
3. latent heat of fusion of the solid.

Solution:

Given

When a solid of mass 50 g is heated at a constant rate of 5 cal · s^-1, there is a rise in temperature of 11°C in the first minute. For the next 13 min of heating, the temperature remains constant. Rate of rise of temperature after this is 6°C ·min^-1.

1. Let specific heat in the solid state = s₁

As per data, 50 x s₁ x 11 = 5 x 60

or, s₁ = \(\frac{300}{550}\) = 0.545 cal · g^-1 · °C^-1.

2. The second phase of heating, is a case of fusion of the solid since the temperature remains constant.

Let latent heat of fusion = L

∴ 50 x L = 13 x 60 x 5

∴ L = 78 cal · g^-1.

3. In the third phase of heating, the temperature of the liquid state of the specimen increases.

Let s₂ = specific heat of the substance in its liquid state.

∴ 50 x s₂ x 6 = 5 x 60 or, s₂ = 1 cal · g^-1 · °C^-1.

Change of Volume During Melting and Solidification: The change from the liquid to the solid state and vice versa involves a change in volume, even though the temperature remains constant.

• For most materials, the volume decreases when a liquid changes to its solid state, and so the density increases. Conversely a change from solid to liquid state results in an increase in volume and therefore a decrease in density.
• But there are a few exceptions to this phenomenon—ice, cast iron, brass, antimony, bismuth etc., decrease in volume on melting and increase in volume on solidification. So, these objects have lower densities in their solid states.

Advantages and disadvantages of change in volume with the change of state: it is seen that the volume of water increases when it solidifies. This expansion may create some problems. In cold countries, the water inside the pipes may freeze into ice during winters.

• The expansion in volume exerts a huge force on the pipes causing them to burst. Again, hot water pipes burst more often than cold water pipes do. The amount of air dissolved in cold water is more than in hot water.
• So when cold water freezes the dissolved air comes out of the ice. If it is possible for air to move out of the pipe the ice can occupies the space of that air. But hot water has less dissolved air. So hot water pipes are more prone to burst compared to cold water pipes.
• In the same way, water deposits in the hill crevices expand on solidification and produce cracks on the stone surface.
• The expansion of volume on solidification has some advantages too. It facilitates the survival of fish and other marine creatures even in extreme cold weather.

Metals such as cast iron and brass expand on solidification. This property makes it suitable for casting in foundries. As molten iron solidifies inside the cast, it expands and fills up the volume evenly. The object conforms to the shape of the cast perfectly without creating any defects.

Effect of Pressure on Melting Point

1. For substances whose volume decreases on melting: The melting point of a substance whose volume decreases on melting, decreases with increase in pressure i.e. the substance melts at a lower temperature.

• Such substances are ice, brass, cast iron, antimony, bismuth etc. Increase in pressure helps in the decrease in volume. This makes melting easier and therefore the melting point decreases.
• It is found that the melting point of ice decreases by 0.0073°C when the pressure on it is increased by one standard atmosphere.

Regelation: Regelation is defined as melting of ice under application of pressure and resolidification of the molten ice on withdrawal of the pressure. When two pieces of ice at 0°C are pressed together, the melting point at the surfaces in contact falls below 0°C .

• But the temperature of ice remains 0°C; so ice melts at the junction. As the necessary latent heat is drawn from the ice itself, the temperature of the contact surface and of the molten ice falls below 0°C. On release of pressure, the melting point rises to 0°C again.
• Water at the surface of contact, being at a temperature lower than 0°C, freezes and the two pieces stick together to form a single piece of ice. This process of resolidification of water into ice is known as regelation.

Factors Affecting Phase Changes in Matter

2. For substances whose volume increases on melting: Substances like wax, copper, naphthalene etc. increase in volume on melting. An increase in pressure on them increases the melting point i.e. they melt at a higher temperature.

As increased pressure opposes expansion in volume, melting becomes difficult and thus melting point increases. One standard atmosphere rise in pressure raises the melting point of wax by 0.04°C.

Laws of Fusion: The facts related to melting can be summarised as the following laws. These laws are known as laws of fusion.

1. Any solid, at a fixed pressure, melts at a fixed tempera¬ture. The temperature remains constant until the whole solid specimen melts. This temperature is called the melting point of the solid at that pressure. Melting point of a solid is the same as tire freezing point of its liquid state. Different substances have different melting points.
2. For solids which decrease in volume on melting, the melting point decreases on increase in pressure. Increase in pressure, however, increases the melting point of those solids which expand on melting.
3. Unit mass of a solid absorbs a fixed quantity of heat during melting at a constant temperature. This quantity of heat is called its latent heat of fusion. Unit mass of the same material in liquid state releases the same amount of heat while freezing, The latent heat of fusion or latent heat of solidification is fixed for every substance but is different for different substances.
4. Freezing point of a solution is always less than the freez¬ing point of the pure solvent.
5. The melting point of an alloy is always less than the individual melting points of all of its constituent metals.