Latent Heat and Heating Curve Questions in English

(B) During the phase transition from vapour to liquid,the substance releases energy in the form of latent heat of vaporization. Therefore,the system liberates heat to the surroundings while the temperature remains constant during the phase change process.

(D) The quantity of heat required to change the unit mass of a solid substance from the solid state to the liquid state at its melting point,while the temperature remains constant,is defined as the latent heat of fusion. Therefore,the correct option is $D$.

(A) The latent heat of vaporization $(L_v)$ is always greater than the latent heat of fusion $(L_f)$.
This is because,during the phase change from liquid to vapor,there is a significant increase in the volume of the substance,which requires a large amount of work to be done against external pressure.
Additionally,the intermolecular forces in a liquid must be completely overcome to transition into a gas,whereas in the solid-to-liquid transition,these forces are only partially weakened.
Therefore,more heat energy is required for vaporization compared to the fusion process.

(A) Steam at $100^{\circ}C$ contains an additional latent heat of vaporization,which is approximately $540 \ cal/g$,compared to water at the same temperature of $100^{\circ}C$.
When steam comes into contact with the skin,it releases this latent heat as it condenses into water at $100^{\circ}C$.
Therefore,the total energy released by steam is significantly higher than that released by water at the same temperature,making a burn from steam much more severe.

(A) Latent heat is a property of the material that depends on the mass of the substance undergoing a phase change.
It is defined as $Q = mL$,where $m$ is the mass and $L$ is the specific latent heat of fusion.
Since both the metallic ball and the spring are made of the same material,their specific latent heat $L$ is identical.
Given that their masses $m$ are also the same,the total latent heat required $Q$ must be the same for both.
The energy spent in stretching the spring is stored as elastic potential energy,which is an ordered form of energy.
Latent heat,however,relates to the energy required to overcome intermolecular forces during a phase transition,which is independent of the initial configuration or the internal elastic energy of the object.

(C) Energy supplied $E = 0.93 \, Wh = 0.93 \times 3600 \, J = 3348 \, J$.
Heat required to melt $10 \, g$ of ice at $0^{\circ}C$ is given by $Q = mL_f$,where $L_f = 80 \, cal/g = 80 \times 4.184 \, J/g = 334.72 \, J/g$.
$Q = 10 \, g \times 334.72 \, J/g = 3347.2 \, J$.
Since the energy supplied $(3348 \, J)$ is approximately equal to the heat required to melt the ice $(3347.2 \, J)$,the entire block of ice just melts.

(B) The correct answer is $B$.
At $100^{\circ}C,$ steam causes more severe burns than boiling water because steam contains additional latent heat of vaporization.
When steam comes in contact with the skin,it releases this latent heat $(L_v \approx 2260 \ J/g)$ in addition to the heat released by cooling to $100^{\circ}C$ water,resulting in a much higher total energy transfer to the skin.

(B) The formula for specific heat capacity is $c = \frac{Q}{m \cdot \Delta \theta}$.
During the phase change from water to steam (boiling),the temperature of the substance remains constant,meaning the change in temperature $\Delta \theta = 0$.
Substituting this into the formula,we get $c = \frac{Q}{m \cdot 0} = \infty$.
Therefore,the specific heat capacity of water during boiling is infinite.

(B) The power $P$ supplied is used to maintain the substance in the molten state by compensating for the heat lost to the surroundings.
When the power is turned off,the substance solidifies by releasing its latent heat of fusion $L$ over time $t$.
The total heat released during solidification is $Q = mL$.
The rate of heat loss is $\frac{Q}{t} = \frac{mL}{t}$.
Since the power $P$ was required to maintain the molten state,it must be equal to the rate of heat loss: $P = \frac{mL}{t}$.
Solving for $L$,we get $L = \frac{Pt}{m}$.

(C) In the heating curve,the horizontal regions represent phase changes where the temperature remains constant.
$1$. In the region between $H_1$ and $H_2$,the temperature is constant at $T_1$. This indicates a phase change from solid to liquid. The heat absorbed during this process is $(H_2 - H_1)$,which is the latent heat of fusion.
$2$. In the region between $H_3$ and $H_4$,the temperature is constant at $T_2$. This indicates a phase change from liquid to gas. The heat absorbed during this process is $(H_4 - H_3)$,which is the latent heat of vaporization.
Therefore,option $(C)$ is correct because $(H_2 - H_1)$ represents the latent heat of fusion.

(A) The process of heating ice from $-10^{\circ}C$ to steam at $100^{\circ}C$ involves several stages:
$1$. Heating ice from $-10^{\circ}C$ to $0^{\circ}C$: The temperature rises linearly as heat is supplied.
$2$. Melting ice at $0^{\circ}C$: The temperature remains constant at $0^{\circ}C$ while the phase changes from solid to liquid (latent heat of fusion).
$3$. Heating water from $0^{\circ}C$ to $100^{\circ}C$: The temperature rises linearly as heat is supplied.
$4$. Boiling water at $100^{\circ}C$: The temperature remains constant at $100^{\circ}C$ while the phase changes from liquid to gas (latent heat of vaporization).
Therefore, the heating curve must show two linear rising segments separated by a horizontal segment (melting) and followed by another horizontal segment (boiling). Option $A$ correctly represents this sequence.

(D) Let $P$ be the rate of heat loss (power) per unit time.
During the cooling phase from $A$ to $B$,the temperature drops from $90^{\circ}C$ to $80^{\circ}C$ in $t_1 = 2 \text{ min}$. The heat lost is given by $Q_1 = mC\Delta T = mC(90 - 80) = 10mC$.
Since $Q_1 = P \times t_1$,we have $10mC = P \times 2$,which implies $P = 5mC$.
During the solidification phase from $B$ to $D$,the temperature remains constant at $80^{\circ}C$ for $t_2 = 4 \text{ min}$. The heat lost is the latent heat,given by $Q_2 = mL$.
Since $Q_2 = P \times t_2$,we have $mL = P \times 4$.
Substituting the value of $P$ from the first equation into the second:
$mL = (5mC) \times 4$
$mL = 20mC$
$\frac{L}{C} = 20$.

(B) In a heating curve,the sloped portions represent a single phase where the temperature increases as heat is added,while the horizontal portions represent phase changes where the temperature remains constant.
$1$. The portion $AB$ represents the solid phase being heated.
$2$. The portion $BC$ represents the phase change from solid to liquid (melting) at a constant temperature of $60^{\circ}C$.
$3$. The portion $CD$ represents the liquid phase being heated from $60^{\circ}C$ to $210^{\circ}C$.
$4$. The portion $DE$ represents the phase change from liquid to gas (boiling) at a constant temperature of $210^{\circ}C$.
$5$. The portion $EF$ represents the gaseous phase being heated.
Therefore,the substance is in the liquid state during the portion $CD$.

(C) The horizontal parts of the heating curve represent phase changes where the temperature remains constant despite the continuous supply of heat. The length of these horizontal segments is proportional to the latent heat required for the phase change.
$1$. In the provided graph,there is one horizontal segment representing a phase change (e.g.,melting or vaporization). However,looking at the slope of the temperature-time graph,the slope is given by $\frac{dT}{dt} = \frac{P}{ms}$,where $P$ is the power (rate of heat supply),$m$ is the mass,and $s$ is the specific heat capacity.
$2$. Since $P$ and $m$ are constant,the specific heat capacity $s$ is inversely proportional to the slope $(s \propto \frac{1}{\text{slope}})$.
$3$. The graph shows a single sloping line. If we compare the latent heat segments (if multiple were present),we would compare their lengths. Given the options,we evaluate the latent heat of vaporization vs fusion. In standard heating curves for substances like water,the latent heat of vaporization is significantly larger than the latent heat of fusion. Thus,the horizontal segment corresponding to vaporization is longer than the one for fusion. Therefore,the conclusion that the latent heat of vaporization is greater than the latent heat of fusion is correct.

(C) The graph shows a cooling curve where the temperature decreases with time.
There is a horizontal portion in the graph where the temperature remains constant even as time passes.
This constant temperature region indicates a phase change,specifically the release of latent heat during the transition from liquid to solid (freezing).
Therefore,the graph represents the change of state from liquid to solid.

(C) From the given heating curve,the heat absorbed during fusion is given by $H = P \times t$,where $P$ is the power (rate of heating) and $t$ is the time taken for the phase change (fusion).
For substance $A$:
The fusion process occurs at a constant temperature of $60^\circ C$. The time interval for this phase change is from $t = 2 \, min$ to $t = 6 \, min$.
Time taken for fusion,$t_A = 6 - 2 = 4 \, min = 4 \times 60 \, s = 240 \, s$.
Heat absorbed by $A$,$H_A = P \times t_A = 6 \, cal \, s^{-1} \times 240 \, s = 1440 \, cal$.
For substance $B$:
The fusion process occurs at a constant temperature of $20^\circ C$. The time interval for this phase change is from $t = 4 \, min$ to $t = 6.5 \, min$.
Time taken for fusion,$t_B = 6.5 - 4 = 2.5 \, min = 2.5 \times 60 \, s = 150 \, s$.
Heat absorbed by $B$,$H_B = P \times t_B = 6 \, cal \, s^{-1} \times 150 \, s = 900 \, cal$.
Ratio of heat absorbed:
$\frac{H_A}{H_B} = \frac{1440}{900} = \frac{144}{90} = \frac{8}{5}$.

(A) During the phase transition from liquid to gas,there is a significant increase in volume.
Therefore,to perform the phase transition at a constant temperature,more energy is required to overcome the intermolecular forces and expand the substance compared to the energy required for the phase transition from solid to liquid.

(C) Let $m$ be the mass of water. The heat required to raise the temperature from $0 \text{ °C}$ to $100 \text{ °C}$ is $Q_1 = m \cdot c \cdot \Delta T = m \cdot 1 \cdot (100 - 0) = 100m \text{ cal}$.
Given that this process takes $10 \text{ minutes}$,the rate of heat supply is $P = \frac{100m}{10} = 10m \text{ cal/min}$.
Now,the water is converted into steam at $100 \text{ °C}$ in $55 \text{ minutes}$. The heat required for this phase change is $Q_2 = m \cdot L$.
The heat supplied in $55 \text{ minutes}$ is $Q_2 = P \times 55 = 10m \times 55 = 550m \text{ cal}$.
Equating $Q_2 = mL$,we get $mL = 550m$,which implies $L = 550 \text{ cal/g}$.

(B) The process of converting boiling water into steam is a phase change process occurring at a constant temperature (isothermal process).
For any substance,the specific heat capacity $s$ is defined by the formula:
$Q = ms \Delta T$
Rearranging for $s$,we get:
$s = \frac{Q}{m \Delta T}$
In a phase change process,the temperature remains constant,so the change in temperature $\Delta T = 0$.
Substituting $\Delta T = 0$ into the formula:
$s = \frac{Q}{m \times 0} = \infty$
Therefore,the specific heat of water during the phase change from liquid to steam is infinite.

(C) The heat absorbed during the phase change (melting) is given by $H = mL$,where $m$ is the mass and $L$ is the latent heat of fusion.
Since the substances are heated at a constant rate $P = 6 \ cal \ s^{-1}$,the heat absorbed is $H = P \times \Delta t$,where $\Delta t$ is the time interval during which the substance is melting (the horizontal part of the graph).
For substance $A$,the melting occurs from $t = 2 \ s$ to $t = 6 \ s$,so $\Delta t_A = 6 - 2 = 4 \ s$.
For substance $B$,the melting occurs from $t = 4 \ s$ to $t = 6.5 \ s$,so $\Delta t_B = 6.5 - 4 = 2.5 \ s$.
Since the masses are equal,the ratio of heat absorbed is $H_A/H_B = (P \times \Delta t_A) / (P \times \Delta t_B) = \Delta t_A / \Delta t_B$.
$H_A/H_B = 4 / 2.5 = 40 / 25 = 8/5$.

(A) Fire is extinguished primarily by the vaporization of water,which absorbs a large amount of latent heat from the burning body,thereby lowering its temperature below the ignition point.
Additionally,the water vapor produced forms a blanket around the burning material,cutting off the supply of oxygen,which is necessary for combustion.
Hot water is more effective than cold water because it is already closer to its boiling point,allowing it to vaporize much more rapidly upon contact with the fire,thus creating a larger volume of steam to displace oxygen and absorb heat more quickly.

(B) The heat required to melt the ice is given by $Q = mL$,where $m = 60 \, g$ and $L = 80 \, cal/g$.
$Q = 60 \times 80 = 4800 \, cal$.
To convert this energy into Joules,we use the mechanical equivalent of heat $J = 4.2 \, J/cal$.
$Q_{joules} = 4800 \times 4.2 = 20160 \, J$.
The time taken is $t = 1 \, minute = 60 \, seconds$.
The power $P$ is given by $P = \frac{Q_{joules}}{t} = \frac{20160}{60} = 336 \, W$.

(D) The graph shows the temperature versus heat input for a substance.
$OA$: The temperature increases while the substance is in the solid state.
$AB$: The temperature remains constant,indicating a phase change from solid to liquid (melting).
$BC$: The temperature increases while the substance is in the liquid state.
$CD$: The temperature remains constant,indicating a phase change from liquid to gas (boiling).
$DE$: The temperature increases while the substance is in the gaseous phase. Since the rate of heat supply is constant,the slope of the temperature-heat graph is given by $\frac{dT}{dQ} = \frac{1}{ms}$,where $m$ is the mass and $s$ is the specific heat capacity. Thus,the slope is inversely proportional to the heat capacity $(ms)$. However,in the context of such problems,the slope represents the reciprocal of the heat capacity of the substance in that phase. Given the options,$D$ is the most appropriate description as the slope is $\frac{1}{ms}$.

(A) Let the block of ice fall from a height $h$.
When the block of ice falls into the lake,its gravitational potential energy is converted into heat energy upon impact.
This heat energy is responsible for melting a portion of the ice.
The potential energy lost by the block is $PE = mgh$.
The heat required to melt a mass $m/5$ of ice is $Q = (m/5)L$.
Equating the potential energy to the heat energy,we get:
$mgh = \frac{m}{5}L$
Canceling $m$ from both sides,we have:
$gh = \frac{L}{5}$
Solving for $h$,we get:
$h = \frac{L}{5g}$

(C) From the graph,the substance undergoes a phase change (vaporization) at a constant temperature of $125^{\circ} C$. This phase change occurs from $t = 20 \, min$ to $t = 30 \, min$.
The duration of the vaporization process is $\Delta t = 30 - 20 = 10 \, min$.
The rate of heat absorption is $P = 42 \, kJ \, min^{-1}$.
The total heat absorbed during vaporization is $Q = P \times \Delta t = 42 \times 10 = 420 \, kJ$.
The mass of the substance is $m = 5 \, kg$.
The latent heat of vaporization $L$ is given by $Q = m \times L$.
Therefore,$L = \frac{Q}{m} = \frac{420}{5} = 84 \, kJ \, kg^{-1}$.

(B) When an ice cube at $-20^oC$ is placed in a room at $20^oC$,it absorbs heat from the surroundings.
First,the temperature of the ice increases from $-20^oC$ to $0^oC$. This is represented by a rising curve.
At $0^oC$,the ice starts melting into water. During the phase change (melting),the temperature remains constant at $0^oC$ even as heat is absorbed. This is represented by a horizontal line (plateau) on the temperature-time graph.
After all the ice has melted,the temperature of the water starts increasing from $0^oC$ towards the room temperature of $20^oC$. This is represented by another rising curve.
Therefore,the correct graph shows an initial rise,a horizontal plateau at $0^oC$,and a subsequent rise.

(C) The boiling point of oxygen at $1\ atm$ is approximately $90\ K$.
When liquid oxygen is heated from $50\ K$ to $300\ K$ at constant pressure, it first heats up as a liquid until it reaches its boiling point $(90\ K)$.
During this phase, the temperature increases linearly with time because the rate of heating is constant ($Q = mc\Delta T$, so $\frac{dT}{dt} = \frac{P}{mc} = \text{constant}$).
Once it reaches the boiling point, the temperature remains constant while the liquid undergoes a phase change (boiling) to become gas. This is represented by a horizontal line on the temperature-time graph.
After the phase change is complete, the oxygen is in the gaseous state and its temperature continues to rise as it is heated further to $300\ K$.
Graph $C$ shows a linear increase in temperature, followed by a constant temperature region during the phase change, and then a linear increase again as a gas. Therefore, graph $C$ is the correct representation.

(D) When a substance is heated at a constant rate,its temperature increases until it reaches its boiling point. During the phase change (boiling),the temperature remains constant as the heat supplied is used to overcome the intermolecular forces (latent heat of vaporization). After the substance has completely turned into gas,the temperature increases again. Therefore,the temperature-time graph shows a linear increase,followed by a horizontal plateau during the phase change,and then another linear increase. This behavior is correctly represented by graph $D$.

(B) In a heating curve,the temperature $\theta$ remains constant during a phase change because the heat supplied is used as latent heat to change the state of the substance rather than increasing its kinetic energy.
$1$. The segment $AB$ represents heating of the solid.
$2$. The segment $BC$ is a horizontal line where the temperature is constant,representing the melting process where the solid converts into liquid.
$3$. The segment $CD$ represents heating of the liquid.
$4$. The segment $DE$ is another horizontal line representing the boiling process where the liquid converts into vapour.
$5$. The segment $EF$ represents heating of the vapour.
Therefore,the portion $BC$ represents the conversion of solid into liquid.

(B) The power $P$ supplied to the substance to keep it in the molten state at its melting point is equal to the rate at which heat is lost to the surroundings to maintain the phase change equilibrium.
When the power is turned off,the energy released by the substance as it solidifies is equal to the energy it would have absorbed to melt,which is given by $Q = P \times t$.
The heat required for the phase change (solidification) is given by $Q = mL$,where $L$ is the latent heat of fusion.
Equating the two expressions for energy: $mL = P \times t$.
Therefore,the latent heat of fusion is $L = \frac{Pt}{m}$.

(C) The rate of heat supply is constant for both substances,given as $P = 6 \, cal \, s^{-1}$.
For a substance,the heat absorbed during the phase change (fusion) is given by $H = P \times \Delta t$,where $\Delta t$ is the time interval during which the temperature remains constant (the horizontal part of the graph).
For substance $A$,the temperature remains constant from $t = 2 \, s$ to $t = 6 \, s$. Thus,the time interval for fusion is $\Delta t_A = 6 - 2 = 4 \, s$.
The heat absorbed by $A$ is $H_A = P \times \Delta t_A = 6 \times 4 = 24 \, cal$.
For substance $B$,the temperature remains constant from $t = 4 \, s$ to $t = 6.5 \, s$ (as seen from the graph,the horizontal section ends between $6$ and $7$,specifically at $6.5$). Thus,the time interval for fusion is $\Delta t_B = 6.5 - 4 = 2.5 \, s$.
The heat absorbed by $B$ is $H_B = P \times \Delta t_B = 6 \times 2.5 = 15 \, cal$.
The ratio of heat absorbed is $\frac{H_A}{H_B} = \frac{P \times \Delta t_A}{P \times \Delta t_B} = \frac{4}{2.5} = \frac{40}{25} = \frac{8}{5}$.

(C) When water at $100\,^{\circ}C$ turns into steam at $100\,^{\circ}C$,it absorbs an additional amount of energy known as the latent heat of vaporization. This energy is stored in the steam without any change in temperature. Therefore,when steam comes into contact with skin,it releases this extra latent heat in addition to the heat released by cooling the water,causing more severe burns compared to water at the same temperature.

(A) The latent heat $(L)$ required to melt a substance is given by the formula $Q = mL$,where $m$ is the mass and $L$ is the specific latent heat of fusion.
Since both the metallic ball and the spring are made of the same material,their specific latent heat of fusion $(L)$ is identical.
Given that both objects have the same mass $(m)$,the total heat required $(Q = mL)$ to melt them must be the same.
Therefore,the latent heat required is the same for both.

(C) When the container is placed in a large evacuated chamber,the pressure above the water surface drops significantly.
This causes rapid evaporation of the water molecules.
Since evaporation is an endothermic process,the latent heat required for vaporization is taken from the remaining water.
As the remaining water loses energy,its temperature drops below $0\,^{\circ}C$,causing it to freeze.
Thus,a portion of the water vaporizes,and the remaining portion freezes.

(B) Let $P$ be the power of the heater (heat supplied per unit time).
For heating water from $0\,^oC$ to $100\,^oC$:
$Q_1 = P \times t_1 = m \cdot c \cdot \Delta T$
$P \times 10 = m \times 1 \times (100 - 0) = 100m \quad ... (i)$
For converting water at $100\,^oC$ to steam at $100\,^oC$:
$Q_2 = P \times t_2 = m \cdot L$
$P \times 55 = m \cdot L \quad ... (ii)$
Dividing equation $(ii)$ by $(i)$:
$\frac{P \times 55}{P \times 10} = \frac{m \cdot L}{100m}$
$5.5 = \frac{L}{100}$
$L = 5.5 \times 100 = 550\, cal/g$.

(B) The energy supplied is $Q_{\text{supplied}} = 0.93 \, W \cdot h = 0.93 \times 3600 \, J = 3348 \, J.$
The energy required to melt $10 \, g$ of ice at $0 \, ^\circ C$ is given by $Q_{\text{req}} = m \cdot L_f,$
where $L_f = 80 \, cal/g = 335.2 \, J/g$ (using $1 \, cal = 4.19 \, J$).
Thus,$Q_{\text{req}} = 10 \, g \times 335.2 \, J/g = 3352 \, J.$
Comparing the values,$Q_{\text{supplied}} \approx Q_{\text{req}} \approx 3350 \, J.$
Since the supplied energy is almost exactly equal to the latent heat required for phase change,the entire block of ice just melts.

(A) The process of heating ice from $-10\,^oC$ to steam at $100\,^oC$ involves several stages:
$1$. Heating ice from $-10\,^oC$ to $0\,^oC$: The temperature increases linearly with heat supplied.
$2$. Melting ice at $0\,^oC$: The temperature remains constant at $0\,^oC$ as phase change occurs.
$3$. Heating water from $0\,^oC$ to $100\,^oC$: The temperature increases linearly with heat supplied.
$4$. Boiling water at $100\,^oC$: The temperature remains constant at $100\,^oC$ as phase change occurs.
Thus,the temperature-heat graph should show two regions of increasing temperature separated by two horizontal plateaus. Graph $A$ correctly depicts this behavior.

(A) When water at $100\,^{\circ}C$ turns into steam at $100\,^{\circ}C$,it absorbs an additional amount of heat known as the latent heat of vaporization $(L_v = 2260\,kJ/kg)$.
Because steam contains this extra latent heat energy compared to liquid water at the same temperature,it releases significantly more energy upon contact with the skin.
Therefore,a burn caused by steam at $100\,^{\circ}C$ is more dangerous than a burn caused by water at the same temperature.

(C) The specific heat capacity $s$ is defined as $s = \frac{Q}{m \Delta T}$,where $Q$ is the heat supplied,$m$ is the mass,and $\Delta T$ is the change in temperature.
During a phase transition (like boiling water changing into steam),the temperature of the substance remains constant,meaning $\Delta T = 0$.
Since the denominator becomes zero,the specific heat capacity $s$ becomes infinite $(s = \infty)$.

(C) In a heating curve,the horizontal sections represent phase changes where temperature remains constant. Thus,$AB$ and $CD$ represent phase changes.
$AB$ corresponds to the melting process (solid to liquid),and $CD$ corresponds to the boiling process (liquid to vapour).
The heat supplied during a phase change is given by $Q = mL$,where $m$ is the mass and $L$ is the latent heat.
Since the heat is supplied at a constant rate,the length of the horizontal segment is proportional to the heat supplied $(Q)$.
From the graph,the length of $CD$ is twice the length of $AB$ $(CD = 2AB)$,which implies that the latent heat of vaporization $(L_v)$ is twice the latent heat of fusion $(L_f)$.
Therefore,the statement that 'latent heat of fusion is twice the latent heat of vaporization' is false.

(D) During the melting process,the temperature of the substance remains constant,but the heat supplied (latent heat) is used to overcome the intermolecular forces of attraction between the particles of the solid.
This process increases the potential energy of the molecules,which in turn increases the total internal energy of the system.
Therefore,the Assertion is incorrect.
Latent heat is defined as the amount of heat energy required to change the state of a unit mass of a substance without changing its temperature,which makes the Reason correct.
Thus,the correct option is $D$.

(C) Perspiration involves the exchange of heat from the body to the surroundings.
Water absorbs latent heat from the body to undergo phase change from liquid to vapour.
This process removes thermal energy from the skin,causing the body to cool down.
Emissivity is a property of the surface material. $A$ thin layer of water on the skin does not enhance its emissivity; in fact,it does not significantly change the emissivity of the skin in the context of cooling.
Therefore,the Assertion is correct,but the Reason is incorrect.

(N/A) Matter typically exists in three distinct states: $Solid$,$Liquid$,and $Gas$.
$A$ transition from one state to another is referred to as a change of state.
Common examples include the transition from $Solid$ to $Liquid$ (melting) and $Liquid$ to $Gas$ (vaporization),as well as their reverse processes ($Freezing$ and $Condensation$).
These changes occur due to the exchange of heat energy between the substance and its surroundings,which alters the internal energy of the molecules without changing the temperature during the phase transition.

(N/A) To understand the change in the state of matter,consider the heating curve of water.
$1$. Take some ice cubes in a beaker and note the initial temperature,which is $0^{\circ} C$.
$2$. Place the beaker on a constant heat source and start heating it slowly.
$3$. Continuously stir the mixture of ice and water to ensure uniform temperature distribution.
$4$. Note the temperature of the mixture at regular intervals (e.g.,every minute) and plot a graph of temperature versus time.
$5$. You will observe that the temperature remains constant at $0^{\circ} C$ as long as there is ice present in the beaker,even though heat is being continuously supplied.
$6$. This constant temperature phase occurs because the heat supplied is being utilized as latent heat of fusion to overcome the intermolecular forces of attraction,changing the state from solid (ice) to liquid (water).
$7$. Once all the ice has melted,the temperature of the water will begin to rise until it reaches the boiling point $(100^{\circ} C)$,where another phase change (liquid to gas) occurs at a constant temperature.

(N/A) The change of state from solid to liquid is called melting,and the change of state from liquid to solid is called freezing (or fusion).
It is observed that the temperature remains constant until the entire amount of the solid substance melts.
Both the solid and liquid states of the substance coexist in thermal equilibrium during the change of states from solid to liquid.
The temperature at which the solid and the liquid states of the substance are in thermal equilibrium with each other is called its melting point.
It is a characteristic property of the substance and also depends on pressure.
The melting point of a substance at standard atmospheric pressure is called its normal melting point.
An activity to explain the process of melting of ice:
Take a slab of ice. Take a metallic wire and fix two weights,say $5 \ kg$ each,at its ends. Put the wire over the slab as shown in the figure. You will observe that the wire passes through the ice slab. This happens because,just below the wire,the ice melts at a lower temperature due to an increase in pressure.
When the wire has passed,the water above the wire freezes again. Thus,the wire passes through the slab,and the slab does not split. This phenomenon of refreezing is called regelation.
Skating is possible on snow due to the formation of water below the skates. Water is formed due to the increase of pressure and it acts as a lubricant.

(N/A) On heating,when ice is converted into water and we continue heating,the temperature begins to rise. The temperature continues to rise until it reaches nearly $100^{\circ} C$,at which point it becomes steady.
The heat supplied is now utilized to change water from the liquid state to the vapour or gaseous state.
The change of state from liquid to vapour (or gas) is called vaporization.
It is observed that the temperature remains constant until the entire amount of the liquid is converted into vapour.
Both the liquid and vapour states of the substance coexist in thermal equilibrium during the change of state from liquid to vapour.
The temperature at which the liquid and the vapour states of the substance coexist is called its boiling point.
The process of boiling of water:
$1$. Take a round-bottom flask,more than half-filled with water.
$2$. Place it over a burner and fix a thermometer and a steam outlet through the cork of the flask.
$3$. As water gets heated in the flask,note first that the air,which was dissolved in the water,will come out as small bubbles.

(N/A) certain amount of heat energy is transferred between a substance and its surroundings when it undergoes a change of state. The amount of heat per unit mass transferred during the change of state of the substance is called the latent heat of the substance for the process.
For example,if heat is added to a given quantity of ice at $-10^{\circ} C$,the temperature of the ice increases until it reaches its melting point $\left(0^{\circ} C\right)$. At this temperature,the addition of more heat does not increase the temperature but causes the ice to melt,or changes its state. Once the entire ice melts,adding more heat will cause the temperature of the water to rise. $A$ similar situation occurs during a liquid-gas change of state at the boiling point. Adding more heat to boiling water causes vaporisation,without an increase in temperature.
The heat required during a change of state depends upon the heat of transformation and the mass of the substance undergoing a change of state.
Thus,if mass $m$ of a substance undergoes a change from one state to the other,then the quantity of heat required is given by:
$Q = m L$ or $L = \frac{Q}{m}$
where $L$ is known as latent heat and is a characteristic of the substance.
Its $SI$ unit is $J \ kg^{-1}$.
The value of $L$ also depends on the pressure. Its value is usually quoted at standard atmospheric pressure.
The latent heat for a solid-liquid state change is called the latent heat of fusion $\left(L_{f}\right)$,and that for a liquid-gas state change is called the latent heat of vaporisation $\left(L_{v}\right)$.
These are often referred to as the heat of fusion $\left(L_{f}\right)$ and the heat of vaporisation $\left(L_{v}\right)$.
$A$ plot of temperature versus heat energy for a quantity of water is shown in the figure. When heat is added (or removed) during a change of state,the temperature remains constant.

(B) The process of changing the state of matter from one form (solid,liquid,or gas) to another is known as a phase transition or a change of state. This process typically occurs at a constant temperature when heat is added or removed,and the energy involved is known as latent heat.

(N/A) Fusion is the process of phase transition in which a substance changes from a solid state to a liquid state by the absorption of heat at a constant temperature.
The melting point is the specific temperature at which a solid substance changes into its liquid state at a given pressure.
The value of the melting point depends on the following factors:
$1$. Nature of the substance: Different materials have different intermolecular forces,leading to different melting points.
$2$. External pressure: For most substances,an increase in pressure increases the melting point. However,for substances that contract upon melting (like ice),an increase in pressure decreases the melting point.

(N/A) Vaporization is the phase transition of a substance from the liquid state to the gaseous or vapor state. It occurs when molecules in a liquid gain enough kinetic energy to overcome the intermolecular forces holding them together,allowing them to escape into the surrounding space as gas.
The normal boiling point of a substance is defined as the temperature at which its vapor pressure equals the standard atmospheric pressure,which is $1.013 \times 10^5 \ Pa$ or $1 \ atm$.