# Change of phase

Method to determine it by method of mixture:

The latent heat of fusion of ice can be determined by method of mixture. To determine a calorimeter with it stirrer is taken and is weighted first. Now, the calorimeter is fitted with water about $\frac{2}{3}$of its volume and again weighted. Now, the calorimeter is place inside the wooden box. The initial temperature of calorimeter and water noted with the help of thermometer${{\rm{t}}_1}{\rm{\: }}$and now piece of ice taken and water in it is soaked with the help of blotting paper in the calorimeter. The final temperature of mixture is noted. Determination of Latent heat of fusion of ice according to method of mixture

Mass of calorimeter + stirrer = ${\mu _1}$

Calorimeter + stirrer +water = μ2

Mass of water = ${\mu _2} - {\rm{\: }}{\mu _1} = {\mu _{\rm{i}}}{\rm{\: \: \: \: }}$(say)

Mass of calorimeter + stirrer + water + Ice = μ3

Mass of ice = ${\mu _3} - {\mu _2} = {\mu _{\rm{i}}}{\rm{\: }}\left( {{\rm{say}}} \right)$

Initial temperature of calorimeter of water = θ1o C

Initial temperature of ice = ${0^{\rm{o}}}{\rm{C}}$

Final temperature of mixture $= {\theta ^{\rm{o}}}{\rm{C}}$

Specific heat capacity of calorimeter = Sc

Let,

Latent heat of fusion of ice = Lf

According to principle of calorimetry,

Heat gained by ice = heat lost of calorimeter and water.

Or, ${{\rm{H}}_{\rm{i}}}{\rm{*}}{{\rm{L}}_{\rm{f}}} + {\rm{\: }}{\mu _{\rm{i}}}{{\rm{s}}_{\rm{w}}}\left( {\theta - 0} \right) = {\mu _1}{{\rm{S}}_{\rm{c}}}\left( {{\theta _1} - \theta } \right) + {\mu _{\rm{w}}}{\rm{\: }}{{\rm{S}}_{\rm{w}}}\left( {{\theta _1} - \theta } \right)$

Or, ${{\rm{L}}_{\rm{f}}} = \frac{{{\mu _1}{{\rm{S}}_{\rm{c}}} + {\mu _{\rm{w}}}{\rm{\: }}{{\rm{S}}_{{\rm{w}}\left( {{\theta _1} - \theta } \right) - {{\rm{S}}_{\rm{w}}}\theta }}}}{{{\mu _{\rm{i}}}}}$

This is required expression for latent heat of fusion of ice by method of mixture.

The method to determine a latent heat of vaporization;

The specific latent heat of stem or specific latent heat of vaporization of water is defined as the amount of heat required to change unit mass of water at steam point to steam at the same temperature. When we boil water to evaporate it then what we observe. We observe that when we boil water to evaporate it at 100 °C, then all the given heat is absorbed by the liquid. This absorbed heat is called latent heat of vaporization and taken by vapour molecules. measurement latent heat of vaporization of water

Let, us consider a clean and dry calorimeter and fill it about one third with water.

${{\rm{m}}_1}$ is the mass of calorimeter with stirrer

S is the mass specific heat of calorimeter with stirrer

${{\rm{m}}_2}$ is the mass of water

When steam changes from 100oC to water at 100oC i.e. suppose θ3oC to water at θ2oC

Then amount of heat lost stem = mL

According to principle of calorimetry,

Heat gained by steam = heat gained by water then amount of heat lost stem = mL

I.e. ${\rm{\: mL}} + {\rm{m}}\left( {{\theta _3} - {\theta _2}} \right){{\rm{S}}_{\rm{w}}} = \left( {{{\rm{m}}_1}{\rm{S}} + {{\rm{m}}_2}{\rm{\: }}{{\rm{S}}_{\rm{w}}}} \right)\left( {{\theta _2} - {\theta _1}} \right)$

Or, Or, ${\rm{L}} = \frac{{\left( {{{\rm{m}}_1}{\rm{S}} + {{\rm{m}}_2}{\rm{\: }}{{\rm{S}}_{\rm{w}}}} \right)\left( {{\theta _2} - {\theta _1}} \right)}}{{\rm{m}}} - {\rm{\: }}(\left( {{\theta _3} - {\theta _2}} \right){{\rm{S}}_{\rm{w}}}$

This is required expression for latent heat of vaporization.

Super cooling:

Super cooling, also known as under cooling, is the process of lowering the temperature of a liquid or a gas below its freezing point without it becoming a solid. When a liquid is cooled gently, it found experimentally that the liquid may cool even several degrees below its freezing point without solidification. Such a liquid at a temperature below its freezing point is called super cooling.

Freezing point depression is when a solution can be cooled below the freezing point of the corresponding pure liquid due to the presence of the solute. cooling curve

Regelation;

The phenomenon of melting of ice due to compression and freezing after removal of applied pressure is called regelation. Many sources state that regelation can be demonstrated by looping a fine wire around a block of ice, with a heavy weight attached to it. The pressure exerted on the ice slowly melts it locally, permitting the wire to pass through the entire block. The wire's track will refill as soon as pressure is relieved, so the ice block will remain solid even after wire passes completely through. This experiment is possible for ice at −10 °C or cooler, and while essentially valid, the details of the process by which the wire passes through the ice are complex. The phenomenon works best with high thermal conductivity materials such as copper, since latent heat of fusion from the top side needs to be transferred to the lower side to supply latent heat of melting. a loaded thin wire cutting through an ice block without melting

Latent heat of vaporization of a material greater that than of latent heat of fusion:

Latent heat of vaporization of a material greater that than of latent heat of fusion because the volume of the substance increases as the liquid change from liquid state vapour state as compared to the change from solid to liquid. Hence against the surrounding pressure large amount work is to be done.

Steam at 100°C burns more than water at 100°C;

One gram of stem at 100oC loses 540 calories of heat on condensing in to water at the same temperature while 1 g of water at 100oC loses 1 cal of heat on cooling to 99oC. since heat lost on condensing the steam is greater than that lost by cooling the water, so one can get a more sver burn from steam at 100oC than from water at 100oC.

The temperature remains constant during the change of state of substance:

The temperature remains constant during the change of state of substance because heat has been used up in changing the state. This energy is used up in separating the molecules farther apart and as work must be done against the surrounding pressure to change the state of substance.

An ice cube at 0°C more effective in cooling a drink than the same mass of water at 0°C:

An ice cube at 0°C more effective in cooling a drink than the same mass of water at 0°C In case of because the impurities like sugar in milk are present which lower the freezing point below 0°C which is effective in cooling.

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