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In gases the intermolecular forces are very weak and its molecule may fly apart in all directions. So the gas is characterized by the following properties. (i) It has no shape and size and can be obtained in a vessel of any shape or size. (ii) It expands indefinitely and uniformly to fill the available space. (iii) It exerts pressure on its surroundings. (iv) Intermolecular forces in a gas are minimum. (v) They can easily compressed and expand.

If to a given mass (m) of a solid, heat is supplied at constant rate P and a graph is plotted between temperature and time, the graph is as shown in figure and is called heating curve. From this curve it is clear that (1) In the region OA temperature of solid is changing with time so,\[Q=m{{c}_{S}}\Delta T\]\[\Rightarrow \]\[P\,\Delta t=m{{c}_{S}}\Delta T\][as \[Q=P\Delta t\]] But as \[(\Delta T/\Delta t)\] is the slope of temperature-time curve \[{{c}_{s}}\propto \frac{1}{\text{Slope of line }OA}\] i.e. specific heat (or thermal capacity) is inversely proportional to the slope of temperature-time curve. (2) In the region AB temperature is constant, so it represents change of state, i.e., melting of solid with melting point \[{{T}_{1}}\]. At A melting starts and at B all solid is converted into liquid. So between A and B substance is partly solid and partly liquid. If \[{{L}_{F}}\] is the latent heat of fusion. \[Q=m{{L}_{F}}\] or \[{{L}_{F}}=\frac{P({{t}_{2}}-{{t}_{1}})}{m}\]   [as \[Q=P({{t}_{2}}-{{t}_{1}})\]] or \[{{L}_{F}}\propto \] length of line AB i.e. Latent heat of fusion is proportional to the length of line of zero slope. [In this region specific heat \[\propto \]\[\frac{1}{\tan 0}=\infty \]] (3) In the region BC temperature of liquid increases so specific heat (or thermal capacity) of liquid will be inversely proportional to the slope of line BC i.e., \[{{c}_{L}}\propto \frac{1}{\text{Slope of line }B\text{C}}\] (4) In the region CD temperature is constant, so it represents the change of state, i.e., boiling with boiling point \[{{T}_{2}}\]. At C all substance is in liquid state while at D in vapour state and between C and D partly liquid and partly gas. The length of line CD is proportional to latent heat of vaporisation i.e., \[{{L}_{V}}\propto \] Length of line CD [In this region specific heat\[\propto \frac{1}{\tan 0}=\infty \]] (5) The line DE represents gaseous state of substance with its temperature increasing linearly with time. The reciprocal of slope of line will be proportional to specific heat or thermal capacity of substance in vapour state.  

Calorimetry means 'measuring heat'. When two bodies (one being solid and other liquid or both being liquid) at different temperatures are mixed, heat will be transferred from body at higher temperature to a body at lower temperature till both acquire same temperature. The body at higher temperature releases heat while body at lower temperature absorbs it, so that Heat lost = Heat gained i.e. principle of calorimetry represents the law of conservation of heat energy. (1) Temperature of mixture \[({{\theta }_{mix}})\] is always \[\ge \] lower temperature \[({{\theta }_{1}})\] and \[\le \] higher temperature \[({{\theta }_{H}})\], i.e., \[{{\theta }_{L}}\le {{\theta }_{mix}}\le {{\theta }_{H}}\]. It means the temperature of mixture can never be lesser than lower temperatures (as a body cannot be cooled below the temperature of cooling body) and greater than higher temperature (as a body cannot be heated above the temperature of heating body). Furthermore usually rise in temperature of one body is not equal to the fall in temperature of the other body though heat gained by one body is equal to the heat lost by the other. (2) Mixing of two substances when temperature changes only : It means no phase change. Suppose two substances having masses \[{{m}_{1}}\] and \[{{m}_{2}}\], gram specific heat \[{{c}_{1}}\] and \[{{c}_{2}}\], temperatures \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] \[({{\theta }_{1}}>{{\theta }_{2}})\] are mixed together such that temperature of mixture at equilibrium is \[{{\theta }_{mix}}\] Hence, Heat lost = Heat gained \[\Rightarrow \] \[{{m}_{1}}{{c}_{1}}({{\theta }_{1}}-{{\theta }_{mix}})={{m}_{2}}{{c}_{2}}({{\theta }_{mix}}-{{\theta }_{2}})\]\[\Rightarrow \]\[{{\theta }_{mix}}=\frac{{{m}_{1}}{{c}_{1}}{{\theta }_{1}}+{{m}_{2}}{{c}_{2}}{{\theta }_{2}}}{{{m}_{1}}{{c}_{1}}+{{m}_{2}}{{c}_{2}}}\] Temperature of mixture in different cases

Condition	Temperature of mixture
If bodies are of same material i.e. \[{{c}_{1}}={{c}_{2}}\]	\[{{\theta }_{mix}}=\frac{{{m}_{1}}{{\theta }_{1}}+{{m}_{2}}{{\theta }_{2}}}{{{m}_{1}}+{{m}_{2}}}\]
If bodies are of same mass \[{{m}_{1}}={{m}_{2}}\]	\[{{\theta }_{mix}}=\frac{{{\theta }_{1}}{{c}_{1}}+{{\theta }_{2}}{{c}_{2}}}{{{c}_{1}}+{{c}_{2}}}\]
If \[{{m}_{1}}={{m}_{2}}\] and \[{{c}_{1}}={{c}_{2}}\]	\[{{\theta }_{mix}}=\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\]

(3) Mixing of two substances when temperature and phase both changes or only phase changes: A very common example for this category is ice-water mixing. Suppose water at temperature \[{{\theta }_{W}}^{o}C\] is mixed with ice at \[{{0}_{i}}^{o}C,\] first ice will melt and then it's temperature rises to attain thermal equilibrium. Hence;  Heat given = Heat taken \[\Rightarrow \] \[{{m}_{W}}{{C}_{W}}({{\theta }_{W}}-{{\theta }_{mix}})={{m}_{i}}{{L}_{i}}+{{m}_{i}}{{C}_{W}}({{\theta }_{mix}}-0{}^\circ )\] \[\Rightarrow \] \[{{\theta }_{mix}}=\frac{{{m}_{W}}{{\theta }_{W}}-\frac{{{m}_{i}}{{L}_{i}}}{{{C}_{W}}}}{{{m}_{W}}+{{m}_{i}}}\] (i) If \[{{m}_{W}}={{m}_{i}}\] then \[{{\theta }_{mix}}=\frac{{{\theta }_{W}}-\frac{{{L}_{i}}}{{{C}_{W}}}}{2}\] (ii) By using this formulae if \[{{\theta }_{mix}}

  Whenever heat is converted into mechanical work or mechanical work is converted into heat, then the ratio of work done to heat produced always remains constant. i.e. \[W\propto Q\] or \[\frac{W}{Q}=J\] This is Joule's law and J is called mechanical equivalent of heat. (1) From W = JQ if Q = 1 then J = W. Hence the amount of work done necessary to produce unit amount of heat is defined as the mechanical equivalent of heat. (2) J is neither a constant, nor a physical quantity rather it is a conversion factor which used to convert Joule or erg into calorie or kilo calories vice-versa. (3) Value of  \[J=4.2\,\frac{Joule}{cal}=4.2\times {{10}^{7}}\frac{erg}{cal}\] \[=4.2\times {{10}^{3}}\frac{Joule}{kcal}\]. (4) When water in a stream falls from height h, then its potential energy is converted into heat and temperature of water rises slightly. From     \[W=JQ\] \[\Rightarrow \]  mgh = J (mc \[\Delta \theta \])           [where m = Mass of water, c = Specific heat of water, \[\Delta \theta =\] temperature rise] \[\Rightarrow \] Rise in temperature \[\Delta \theta =\frac{gh}{Jc}{}^\circ C\] (5) The kinetic energy of a bullet fired from a gun gets converted into heat on striking the target. By this heat the temperature of bullet increases by\[\Delta \theta \]. From     W = JQ    \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=J(\,m\,s\,\Delta \theta )\] [where m = Mass of the bullet, v = Velocity of the bullet,     c = Specific heat of the bullet] \[\Rightarrow \] Rise in temperature \[\Delta t=\frac{{{v}^{2}}}{2Jc}{}^\circ C\] If the temperature of bullet rises upto the melting point of the bullet and bullet melts then. From     \[W=J({{Q}_{Temperature\text{ }change}}+{{Q}_{Phase\text{ }change}})\] \[\Rightarrow \] \[\frac{1}{2}m{{v}^{2}}=J(mc\,\Delta \theta +mL)\];    L = Latent heat of bullet \[\Rightarrow \]  Rise in temperature   \[\Delta \theta =\left[ \frac{\left( \frac{{{v}^{2}}}{2J}-L \right)}{c} \right]\,{}^\circ C\] (6) If m kg ice-block falls down through some height (h) and melts partially (m' kg) then its potential energy gets converted into heat of melting. From  W = JQ  \[\Rightarrow \] \[mgh=J\,m'L\] \[\Rightarrow \] \[h=\frac{m'}{m}\left( \frac{JL}{g} \right)\] If ice-block melts completely then \[m'=m\Rightarrow h=\frac{JL}{g}meter\]  

(1) Evaporation : Vaporisation occurring from the free surface of a liquid is called evaporation. Evaporation is the escape of molecules from the surface of a liquid. This process takes place at all temperatures and increases with the increase of temperature. Evaporation leads to cooling because the faster molecules escape and, therefore, the average kinetic energy of the molecules of the liquid (and hence the temperature) decreases. (2) Melting (or fusion)/freezing (or solidification) : The phase change of solid to liquid is called melting or fusion. The reverse phenomenon is called freezing or solidification. When pressure is applied on ice, it melts. As soon as the pressure is removed, it freezes again. This phenomenon is called regelation. (3) Vaporisation/liquefication (condensation) : The phase change from liquid to vapour is called vaporisation. The reverse transition is called liquefication or condensation. (4) Sublimation : Sublimation is the conversion of a solid directly into vapours. Sublimation takes place when boiling point is less than the melting point. A block of ice sublimates into vapours on the surface of moon because of very very low pressure on its surface. Heat required to change unit mass of solid directly into vapours at a given temperature is called heat of sublimation at that temperature. (5) Hoar frost : Direct conversion of vapours into solid is called hoar frost. This process is just reverse of the process of sublimation, e.g., formation of snow by freezing of clouds. (6) Vapour pressure : When the space above a liquid is closed, it soon becomes saturated with vapour and a dynamic equilibrium is established. The pressure exerted by this vapour is called Saturated Vapour Pressure (S.V.P.) whose value depends only on the temperature ? it is independent of any external pressure. If the volume of the space is reduced, some vapour liquefies, but the pressure is unchanged. A saturated vapour does not obey the gas law whereas the unsaturated vapour obeys them fairly well. However, a vapour differs from a gas in that the former can be liquefied by pressure alone, whereas the latter cannot be liquefied unless it is first cooled. (7) Boiling : As the temperature of a liquid is increased, the rate of evaporation also increases. A stage is reached when bubbles of vapour start forming in the body of the liquid which rise to the surface and escape. A liquid boils at a temperature at which the S.V.P. is equal to the external pressure. It is a fast process. The boiling point changes on mixing impurities. (8) Dew point : It is that temperature at which the mass of water vapour present in a given volume of air is just sufficient to saturate it, i.e. the temperature at which the actual vapour pressure becomes equal to the saturated vapuor pressure. (9) Humidity : Atmospheric air always contains some more...

(1) Thermal capacity : It is defined as the amount of heat required to raise the temperature of the whole body (mass m) through \[{{0}^{o}}C\] or 1K. Thermal capacity \[=mc=\mu C=\frac{Q}{\Delta \theta }\] The value of thermal capacity of a body depends upon the nature of the body and its mass. Dimension : \[[M{{L}^{2}}{{T}^{-2}}{{\theta }^{-1}}]\], Unit : cal/\[^{o}C\] (practical)  Joule/k (S.I.) (2) Water Equivalent : Water equivalent of a body is defined as the mass of water which would absorb or evolve the same amount of heat as is done by the body in rising or falling through the same range of temperature. It is represented by W. If m = Mass of the body, c = Specific heat of body, \[\Delta \theta =\] Rise in temperature. Then heat given to body \[\Delta Q=mc\Delta \theta \]                 ... (i) If same amount of heat is given to W gm of water and its temperature also rises by  \[\Delta \theta \]. Then heat given to water \[Q=W\times 1\times \Delta \theta \] ...(ii)     [As \[{{c}_{\text{water}}}=1\]] From equation (i) and (ii)  \[\Delta Q=mc\Delta \theta =W\times 1\times \Delta \theta \] \[\Rightarrow \] Water equivalent (W) = mc gm (i) Unit : Kg (S.I.)               Dimension : \[[M{{L}^{0}}{{T}^{0}}]\] (ii) Unit of thermal capacity is J/kg while unit of water equivalent is kg. (iii) Thermal capacity of the body and its water equivalent are numerically equal. (iv) If thermal capacity of a body is expressed in terms of mass of water it is called water-equivalent of the body.

(1) Phase : We use the term phase to describe a specific state of matter, such as solid, liquid or gas. A transition from one phase to another is called a phase change. (i) For any given pressure a phase change takes place at a definite temperature, usually accompanied by absorption or emission of heat and a change of volume and density. (ii) In phase change ice at \[{{0}^{o}}C\] melts into water at \[{{0}^{o}}C\]. Water at \[{{100}^{o}}C\] boils to form steam at \[{{100}^{o}}C\]. (iii) In solids, the forces between the molecules are large and the molecules are almost fixed in their positions inside the solid. In a liquid, the forces between the molecules are weaker and the molecules may move freely inside the volume of the liquid. However, they are not able to come out of the surface. In vapours or gases, the intermolecular forces are almost negligible and the molecules may move freely anywhere in the container. When a solid melts, its molecules move apart against the strong molecular attraction. This needs energy which must be supplied from outside. Thus, the internal energy of a given body is larger in liquid phase than in solid phase. Similarly, the internal energy of a given body in vapour phase is larger than that in liquid phase. (iv) In case of change of state if the molecules come closer, energy is released and if the molecules move apart, energy is absorbed. (2) Latent heat : The amount of heat required to change the state of the mass m of the substance is written as : Q = mL, where L is the latent heat. Latent heat is also called as Heat of Transformation. It's unit is cal/gm or J/kg and Dimension: \[[{{L}^{2}}{{T}^{-2}}]\] (i) Latent heat of fusion : The latent heat of fusion is the heat energy required to change 1 kg of the material in its solid state at its melting point to 1 kg of the material in its liquid state. It is also the amount of heat energy released when at melting point 1 kg of liquid changes to 1 kg of solid. For water at its normal freezing temperature or melting point \[({{0}^{o}}C),\] the latent heat of fusion (or latent heat of ice) is \[{{L}_{F}}={{L}_{\text{ice}}}\approx 80\,cal/gm\approx 60\,kJ/mol\approx 336\,kilo\,joule/kg\] (ii) Latent heat of vaporisation : The latent heat of vaporisation is the heat energy required to change 1 kg of the material in its liquid state at its boiling point to 1 kg of the material in its gaseous state. It is also the amount of heat energy released when 1 kg of vapour changes into 1 kg of liquid. For water at its normal boiling point or condensation temperature \[({{100}^{o}}C),\] the latent heat of vaporisation (latent heat of steam) is \[{{L}_{V}}={{L}_{\text{steam}}}\approx 540\,cal/gm\approx 40.8\,kJ/mol\approx 2260\,kilo\,joule/kg\] (iii) Latent heat of vaporisation is more than the latent heat of fusion. This is because when a substance gets more...

(1) In case of gases, heat energy supplied to a gas is spent not only in raising the temperature of the gas but also in expansion of gas against atmospheric pressure. (2) Hence specific heat of a gas, which is the amount of heat energy required to raise the temperature of one gram of gas through a unit degree shall not have a single or unique value. (3) If the gas is compressed suddenly and no heat is supplied from outside i.e. \[\Delta Q=0,\] but the temperature of the gas raises on the account of compression. \[\Rightarrow \] \[c=\frac{Q}{m(\Delta \theta )}=\frac{0}{m\Delta \theta }=0\] (4) If the gas is heated and allowed to expand at such a rate that rise in temperature due to heat supplied is exactly equal to fall in temperature due to expansion of the gas. i.e. \[\Delta \theta =0\] \[\Rightarrow \]\[c=\frac{Q}{m(\Delta \theta )}=\frac{Q}{0}=\infty \] (5) If rate of expansion of the gas were slow, the fall in temperature of the gas due to expansion would be smaller than the rise in temperature of the gas due to heat supplied. Therefore, there will be some net rise in temperature of the gas i.e. \[\Delta T\] will be positive. \[\Rightarrow \]\[c=\frac{Q}{m(\Delta \theta )}=\] Positive  (6) If the gas were to expand very fast, fall of temperature of gas due to expansion would be greater than rise in temperature due to heat supplied. Therefore, there will be some net fall in temperature of the gas i.e. \[\Delta \theta \] will be negative. \[\Rightarrow \] \[c=\frac{Q}{m(-\Delta \theta )}=\]Negative  Hence the specific heat of gas can have any positive value ranging from zero to infinity. Further it can even be negative. The exact value depends upon the mode of heating the gas. Out of many values of specific heat of a gas, two are of special significance, namely \[{{C}_{P}}\] and \[{{C}_{V}},\] in the chapter ?Kinetic theory of gases? we will discussed this topic in detail. Specific heat of steam : \[{{c}_{\text{steam}}}=0.47\,cal/gm\times {}^\circ C\]

(1) Among all known solids and liquids specific heat of water is maximum i.e. water takes more time to heat and more time to cool w.r.t. other solids and liquids. (2) It is observed that by increasing temperature, initially specific heat of water goes on decreasing, becomes minimum at \[37{}^\circ C\] and then it start increasing. Specific heat of water is - \[\frac{1\,cal}{gm\times {}^\circ C}=1000\frac{cal}{kg\times {}^\circ C}=4200\frac{J}{kg\times {}^\circ C}\] (This value is obtained between the temperature \[14.5{}^\circ C\] to \[15.5{}^\circ C\]) (3) The variation of specific heat with temperature for water is shown in the figure. Usually this temperature dependence of specific heat is neglected. (4) As specific heat of water is very large; by absorbing or releasing large amount of heat its temperature changes by small amount. This is why, it is used in hot water bottles or as coolant in radiators.  

When a solid is heated through a small range of temperature, its volume remains more or less constant. Therefore specific heat of a solid may be called its specific heat at constant volume \[{{C}_{V}}\]. (1) From the graph it is clear that at \[T=0,\,\,{{C}_{V}}\] tends to zero (2) With rise in temperature, \[{{C}_{V}}\] increases and at a particular temperature (called Debey's temperature) it becomes constant = 3R = 6 cal/mole \[\times \] kelvin = 25 J/mole \[\times \] kelvin (3) For most of the solids, Debye temperature is close to room temperature. (4) Dulong and Petit law : Average molar specific heat of all metals at room temperature is constant, being nearly equal to 3R = 6 cal. \[mol{{e}^{-1}}\,{{K}^{-1}}\] = 25 J \[mol{{e}^{-1}}\,{{K}^{-1}}\], where R is gas constant for one mole of the gas. This statement is known as Dulong and Petit law. (5) Debey's law : It was observed that at very low temperature molar specific heat \[\propto {{T}^{3}}\] exception are Sn, Pb and Pt) (6) Specific heat of ice : In C.G.S. \[{{c}_{\text{ice}}}=0.5\,\frac{cal}{gm\times {}^\circ C}\] In S.I. \[{{c}_{ice}}==500\,\frac{cal}{kg\times {}^\circ C}=2100\,\frac{Joule}{kg\times {}^\circ C}\].   Specific heat of some solids at room temperature and atmospheric pressure

Substance	Specific heat \[\mathbf{(J-k}{{\mathbf{g}}^{\mathbf{-1}}}\,{{\mathbf{K}}^{\mathbf{-1}}}\mathbf{)}\]	Molar specific heat \[\mathbf{(J-g}\,\,\mathbf{mol}{{\mathbf{e}}^{\mathbf{-1}}}\,{{\mathbf{K}}^{\mathbf{-1}}}\mathbf{)}\]
Aluminium	900.0	24.4
Copper	386.4	24.5
Silver	236.1	25.5
Lead	127.7	26.5
Tungsten	134.4	24.9