Current Affairs JEE Main & Advanced

It is a method of determination of K of a metallic rod. (1) In this experiment a temperature difference \[({{\theta }_{1}}-{{\theta }_{2}})\] is maintained across a rod of length l and area of cross section A. If the thermal conductivity of the material of the rod is K, then the amount of heat transmitted by the rod from the hot end to the cold end in time t is given by, \[Q=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})\,t}{l}\]    ......(i) (2) In Searle's experiment, this heat reaching the other end is utilized to raise the temperature of certain amount of water flowing through pipes circulating around the other end of the rod. If temperature of the water at the inlet is \[{{\theta }_{3}}\] and at the outlet is \[{{\theta }_{4}}\], then the amount of heat absorbed by water is given by, \[Q=mc({{\theta }_{4}}-{{\theta }_{3}})\]                                                                                                            ......(ii) (3) Where, m is the mass of the water which has absorbed this heat and temperature is raised and c is the specific heat of the water Equating (i) and (ii), K can be determined i.e., \[K=\frac{mc({{\theta }_{4}}-{{\theta }_{3}})\,l}{A\,({{\theta }_{1}}-{{\theta }_{2}})\,t}\] (4) In numericals we may have the situation where the amount of heat travelling to the other end may be required to do some other work e.g.,  it may be required to melt the given amount of ice. In that case equation (i) will have to be equated to mL. i.e.  \[mL=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})\,t}{l}\]  

It is used to compare thermal conductivities of different materials. If \[{{l}_{1}},\,\,{{l}_{2}}\] and \[{{l}_{3}}\] are the lengths of wax melted on rods as shown in the figure, then the ratio of thermal conductivities is \[{{K}_{1}}:{{K}_{2}}:{{K}_{3}}=l_{1}^{2}:l_{2}^{2}:l_{3}^{2}\] \[\Rightarrow \] Thermal conductivity \[(K)\propto {{(\text{Melted length }l)}^{2}}\]    

(1) Series combination : Let n slabs each of cross-sectional area A, lengths \[{{l}_{1}},\,{{l}_{2}},\,{{l}_{3}},......{{l}_{n}}\] and conductivities \[{{K}_{1}},\,{{K}_{2}},\,{{K}_{3}}......{{K}_{n}}\] respectively be connected in the series (i) Heat current : Heat current is the same in all the conductors.i.e.,  \[\frac{Q}{t}={{H}_{1}}={{H}_{2}}={{H}_{3}}.........={{H}_{n}}\] \[\frac{{{K}_{1}}A({{\theta }_{1}}-{{\theta }_{2}})}{{{l}_{1}}}=\frac{{{K}_{2}}A({{\theta }_{2}}-{{\theta }_{3}})}{{{l}_{2}}}\]\[=\frac{{{K}_{n}}A({{\theta }_{n-1}}-{{\theta }_{n}})}{{{l}_{n}}}\] (ii) Equivalent thermal resistance :  \[R={{R}_{1}}+{{R}_{2}}+.....{{R}_{n}}\] (iii) Equivalent thermal conductivity : It can be calculated as follows From \[{{R}_{S}}={{R}_{1}}+{{R}_{2}}+{{R}_{3}}+...\] \[\frac{{{l}_{1}}+{{l}_{2}}+...{{l}_{n}}}{{{K}_{s}}}=\frac{{{l}_{1}}}{{{K}_{1}}A}+\frac{{{l}_{2}}}{{{K}_{2}}A}+....+\frac{{{l}_{n}}}{{{K}_{n}}A}\] \[\Rightarrow \] \[{{K}_{s}}=\frac{{{l}_{1}}+{{l}_{2}}+......\,{{l}_{n}}}{\frac{{{l}_{1}}}{{{K}_{1}}}+\frac{{{l}_{2}}}{{{K}_{2}}}+........\frac{{{l}_{n}}}{{{K}_{n}}}}\] (a) For n slabs of equal length \[{{K}_{s}}=\frac{n}{\frac{1}{{{K}_{1}}}+\frac{1}{{{K}_{2}}}+\frac{1}{{{K}_{3}}}+.....\frac{1}{{{K}_{n}}}}\]    (b) For two slabs of equal length, \[{{K}_{s}}=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}\] (iv) Temperature of interface of composite bar : Let the two bars are arranged in series as shown in the figure. Then heat current is same in the two conductors. i.e., \[\frac{Q}{t}=\frac{{{K}_{1}}A({{\theta }_{1}}-\theta )}{{{l}_{1}}}=\frac{{{K}_{2}}A(\theta -{{\theta }_{2}})}{{{l}_{2}}}\] By solving we get  \[\theta =\frac{\frac{{{K}_{1}}}{{{l}_{1}}}{{\theta }_{1}}+\frac{{{K}_{2}}}{{{l}_{2}}}{{\theta }_{2}}}{\frac{{{K}_{1}}}{{{l}_{1}}}+\frac{{{K}_{2}}}{{{l}_{2}}}}\] (a) If \[{{l}_{1}}={{l}_{2}}\] then \[\theta =\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\] (b) If \[{{K}_{1}}={{K}_{2}}\] and \[{{l}_{1}}={{l}_{2}}\] then \[\theta =\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] (2) Parallel Combination : Let n slabs each of length l, areas \[{{A}_{1}},{{A}_{2}},{{A}_{3}},.....{{A}_{n}}\] and thermal conductivities \[{{K}_{1}},{{K}_{2}},{{K}_{3}},.....{{K}_{n}}\] are connected in parallel then (i) Equivalent resistance : \[\frac{1}{{{R}_{s}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}+\frac{1}{{{R}_{3}}}+.....\frac{1}{{{R}_{n}}}\] For two slabs \[{{R}_{s}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] (ii) Temperature gradient : Same across each slab. (iii) Heat current : in each slab will be different. Net heat current will be the sum of heat currents through individual slabs. i.e., \[H={{H}_{1}}+{{H}_{2}}+{{H}_{3}}+....{{H}_{n}}\] \[\frac{K({{A}_{1}}+{{A}_{2}}+.....+{{A}_{n}})\,({{\theta }_{1}}-{{\theta }_{2}})}{l}\]          \[=\frac{{{K}_{1}}{{A}_{1}}({{\theta }_{1}}-{{\theta }_{2}})}{l}+\frac{{{K}_{2}}{{A}_{2}}({{\theta }_{1}}-{{\theta }_{2}})}{l}+...+\frac{{{K}_{n}}{{A}_{n}}\,({{\theta }_{1}}-{{\theta }_{2}})}{l}\] \[\Rightarrow \] \[K=\frac{{{K}_{1}}{{A}_{1}}+{{K}_{2}}{{A}_{2}}+{{K}_{3}}{{A}_{3}}+.....{{K}_{n}}{{A}_{n}}}{{{A}_{1}}+{{A}_{2}}+{{A}_{3}}+.....{{A}_{n}}}\] (a) For n slabs of equal area \[K=\frac{{{K}_{1}}+{{K}_{2}}+{{K}_{3}}+.....{{K}_{n}}}{n}\] (b) For two slabs of equal area \[K=\frac{{{K}_{1}}+{{K}_{2}}}{2}\].  

(1) Cooking utensils are provided with wooden handles, because wood is a poor conductor of heat. The hot utensils can be easily handled from the wooden handles and our hands are saved from burning. (2) We feel warmer in a fur coat. The air enclosed in the fur coat being bad conductor heat does not allow the body heat to flow outside. Hence we feel warmer in a fur coat. (3) Eskimos make double walled houses of the blocks of ice. Air enclosed in between the double walls prevents transmission of heat from the house to the cold surroundings. For exactly the same reason, two thin blankets are warmer than one blanket of their combined thickness. The layer of air enclosed in between the two blankets makes the difference. (4) Wire gauze is placed over the flame of Bunsen burner while heating the flask or a beaker so that the flame does not go beyond the gauze and hence there is no direct contact between the flame and the flask. The wire gauze being a good conductor of heat, absorb the heat of the flame and transmit it to the flask. Davy's safety lamp has been designed on this principle. The gases in the mines burn inside the gauze placed around the flame of the lamp. The temperature outside the gauze is not high, so the gases outside the gauze do not catch fire. (5) Birds often swell their feathers in winter. By doing so, they enclose more air between their bodies and the feathers. The air, being bad conductor of heat prevents the out flow of their body heat. Thus, birds feel warmer in winter by swelling their feathers.  

When one end of a metallic rod is heated, heat flows by conduction from the hot end to the cold end. (1) Variable state : In this state Temperature of every part of the rod increases. Heat received by each cross-section of the rod from hotter end used in three ways. (i) A part increases temperature of itself. (ii) Another part transferred to neighbouring cross-section. (iii) Remaining part radiates.

• \[{{\theta }_{1}}>{{\theta }_{2}}>{{\theta }_{3}}>{{\theta }_{4}}>{{\theta }_{5}}\]
• \[\theta \to \text{Changing}\]

(2) Steady state : After sometime, a state is reached when the temperature of every cross-section of the rod becomes constant. In this state, no heat is absorbed by the rod. The heat that reaches any cross-section is transmitted to the next except that a small part of heat is lost to surrounding from the sides by convection & radiation. This state of the rod is called steady state. (3) Isothermal surface : Any surface (within a conductor) having its all points at the same temperature, is called isothermal surface. The direction of flow of heat through a conductor at any point is perpendicular to the isothermal surface passing through that point.   (4) Temperature gradient (T.G.) : The rate of change of temperature with distance between two isothermal surfaces is called temperature gradient. Hence (i) Temperature gradient \[=\frac{-\Delta \theta }{\Delta x}\] (ii) The negative sign show that temperature \[\theta \] decreases as the distance x increases in the direction of heat flow. (iii) For uniform temperature fall  \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{l}=\frac{\Delta \theta }{\Delta x}\] (iv) Unit : K/m or \[^{o}C/m\] (S.I.) and Dimensions \[[{{L}^{-1}}\theta ]\] (5) Law of thermal conductivity : Consider a rod of length l and area of cross-section A whose faces  are maintained at temperature \[{{\theta }_{1}}\] and \[{{\theta }_{2}}\] respectively. The curved surface of rod is kept insulated from surrounding to avoid leakage of heat (i) In steady state the amount of heat flowing from one face to the other face in time t is given by \[Q=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})\,t}{l}\] where K is coefficient of thermal conductivity of material of rod. (ii) Rate of flow of heat i.e. heat current \[\frac{Q}{t}=H\]\[=\frac{KA({{\theta }_{1}}-{{\theta }_{2}})\,}{l}\] (iii) In case of non-steady state or variable cross-section, a more general equation can be used to solve problems. \[\frac{dQ}{dt}=-\,KA\frac{d\theta }{dx}\] (6) More about K : It is the measure of the ability of a substance to conduct heat through it. (i) Units :  Cal/cm-sec \[^{o}C\]  (in C.G.S.), kcal/m-sec-K (in M.K.S.) and W/m- K (in S.I.). Dimension : \[[ML{{T}^{-3}}{{\theta }^{-1}}]\] (ii) The magnitude of K depends only on nature of the material. (iii) Substances in which heat flows quickly and easily are known as good conductor of heat. They possesses large thermal conductivity due to large number of free more...

  The process of transmission of heat energy in which the heat is transferred from one particle to other particle without dislocation of the particle from their equilibrium position is called conduction. (1) Heat flows from hot end to cold end. Particles of the medium simply oscillate but do not leave their place. (2) Medium is necessary for conduction (3) It is a slow process (4) The temperature of the medium increases through which heat flows (5) Conduction is a process which is possible in all states of matter. (6) When liquid and gases are heated from the top, they conduct heat from top to bottom. (7) In solids only conduction takes place (8) In non-metallic solids and fluids the conduction takes place only due to vibrations of molecules, therefore they are poor conductors. (9) In metallic solids free electrons carry the heat energy, therefore they are good conductor of heat.  

Heat energy transfers from a body at higher temperature to a body at lower temperature. The transfer of heat from one body to another may take place by one of the following modes. Conduction, Convection and Radiation                       

Entropy is a measure of disorder of molecular motion of a system. Greater is the disorder, greater is the entropy. The change in entropy i.e. \[dS=\frac{\text{Heat absorbed by system}}{\text{Absolute temperature}}\] or \[dS=\frac{dQ}{T}\] The relation is called the mathematical form of Second Law of Thermodynamics. (1) For solids and liquids (i) When heat given to a substance changes its state at constant temperature, then change in entropy \[dS=\frac{dQ}{T}=\pm \frac{mL}{T}\] where positive sign refers to heat absorption and negative sign to heat evolution. (ii) When heat given to a substance raises its temperature from \[{{T}_{1}}\] to \[{{T}_{2}},\] then change in entropy \[dS=\int{\frac{dQ}{T}}=\int_{\,{{T}_{1}}}^{\,{{T}_{2}}}{mc}\frac{dT}{T}=mc{{\log }_{e}}\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)\] \[\Rightarrow \] \[\Delta S=2.303\,mc{{\log }_{10}}\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)\]. (2) For a perfect gas : Perfect gas equation for n moles is \[PV=nRT\] \[\Delta S=\int{\frac{dQ}{T}}=\int{\frac{\mu {{C}_{V}}dT+P\,dV}{T}}\]                [As \[dQ=dU+dW\]] \[\Rightarrow \] \[\Delta S=\int{\frac{\mu {{C}_{V}}dT+\frac{\mu RT}{V}dV}{T}}\] \[=\mu {{C}_{V}}\int_{\,{{T}_{1}}}^{\,{{T}_{2}}}{\frac{dT}{T}}+\mu R\int_{\,{{V}_{1}}}^{\,{{V}_{2}}}{\frac{dV}{V}}\]                                          [As \[PV=\mu RT\]] \[\therefore \]\[\Delta S=\mu {{C}_{V}}{{\log }_{e}}\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)+\mu R{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] In terms of T and P,  \[\Delta S=\mu {{C}_{P}}{{\log }_{e}}\left( \frac{{{T}_{2}}}{{{T}_{1}}} \right)-\mu R{{\log }_{e}}\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)\] and in terms of P and V \[\Delta S=\mu {{C}_{V}}{{\log }_{e}}\left( \frac{{{P}_{2}}}{{{P}_{1}}} \right)+\mu {{C}_{P}}{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\]

(1) Carnot designed a theoretical engine which is free from all the defects of a practical engine. This engine cannot be realised in actual practice, however, this can be taken as a standard against which the performance of an actual engine can be judged. It consists of the following parts (i) A cylinder with perfectly non-conducting walls and a perfectly conducting base containing a perfect gas as working substance and fitted with a non-conducting frictionless piston (ii) A source of infinite thermal capacity maintained at constant higher temperature \[{{T}_{1}}\]. (iii) A sink of infinite thermal capacity maintained at constant lower temperature \[{{T}_{2}}\]. (iv) A perfectly non-conducting stand for the cylinder. (2) Carnot cycle : As the engine works, the working substance of the engine undergoes a cycle known as Carnot cycle. The Carnot cycle consists of the following four strokes (i) First stroke (Isothermal expansion) (curve AB) : The cylinder containing ideal gas as working substance allowed to expand slowly at this constant temperature \[{{T}_{1}}\]. Work done = Heat absorbed by the system \[{{W}_{1}}={{Q}_{1}}=\int_{\,{{V}_{1}}}^{\,{{V}_{2}}}{P\,dV}=R{{T}_{1}}{{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)=\]Area ABGE  (ii) Second stroke (Adiabatic expansion) (curve BC) : The cylinder is then placed on the non conducting stand and the gas is allowed to expand adiabatically till the temperature falls from \[{{T}_{1}}\] to \[{{T}_{2}}\]. \[{{W}_{2}}=\int_{\,{{V}_{2}}}^{\,{{V}_{3}}}{P\,dV}=\frac{R}{(\gamma -1)}[{{T}_{1}}-{{T}_{2}}]=\]Area BCHG (iii) Third stroke (Isothermal compression) (curve CD) : The cylinder is placed on the sink and the gas is compressed at constant temperature \[{{T}_{2}}\]. Work done = Heat released by the system \[{{W}_{3}}={{Q}_{2}}=-\int_{\,{{V}_{3}}}^{\,{{V}_{4}}}{\,P\,dV}=-R{{T}_{2}}{{\log }_{e}}\frac{{{V}_{4}}}{{{V}_{3}}}\] \[=R{{T}_{2}}{{\log }_{e}}\frac{{{V}_{3}}}{{{V}_{4}}}=\text{Area }CDFH\,\] (iv) Fourth stroke (adiabatic compression) (curve DA) : Finally the cylinder is again placed on non-conducting stand and the compression is continued so that gas returns to its initial stage. \[{{W}_{4}}=-\int_{\,{{V}_{4}}}^{\,{{V}_{1}}}{P\,dV}=-\frac{R}{\gamma -1}({{T}_{2}}-{{T}_{1}})\]\[=\frac{R}{\gamma -1}({{T}_{1}}-{{T}_{2}})=\text{Area }ADFE\] (3) Efficiency of Carnot cycle : The efficiency of engine is defined as the ratio of work done to the heat supplied i.e. \[\eta =\frac{\text{Work done}}{\text{Heat input}}=\frac{W}{{{Q}_{1}}}\] Net work done during the complete cycle \[W={{W}_{1}}+{{W}_{2}}+(-{{W}_{3}})+(-{{W}_{4}})\]\[={{W}_{1}}-{{W}_{3}}=\text{Area }ABCD\]                                          [As \[{{W}_{2}}={{W}_{4}}\]] \[\therefore \] \[\eta =\frac{W}{{{Q}_{1}}}=\frac{{{W}_{1}}-{{W}_{3}}}{{{W}_{1}}}=\frac{{{Q}_{1}}-{{Q}_{2}}}{{{Q}_{1}}}=1-\frac{{{W}_{3}}}{{{W}_{1}}}=1-\frac{{{Q}_{2}}}{{{Q}_{1}}}\]   or  \[\eta =1-\frac{R{{T}_{2}}{{\log }_{e}}({{V}_{3}}/{{V}_{4}})}{R{{T}_{1}}{{\log }_{e}}({{V}_{2}}/{{V}_{1}})}\] Since points B and C lie on same adiabatic curve  \[\therefore \] \[{{T}_{1}}V_{2}^{\gamma -1}={{T}_{2}}V_{3}^{\gamma -1}\] or \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{V}_{3}}}{{{V}_{2}}} \right)}^{\gamma -1}}\]                           ...(i) Also point D and A lie on the same adiabatic curve  \[\therefore \] \[{{T}_{1}}V_{1}^{\gamma -1}={{T}_{2}}V_{4}^{\gamma -1}\] or \[\frac{{{T}_{1}}}{{{T}_{2}}}={{\left( \frac{{{V}_{4}}}{{{V}_{1}}} \right)}^{\gamma -1}}\]                           ...(ii) From (i) and (ii), \[\frac{{{V}_{3}}}{{{V}_{2}}}=\frac{{{V}_{4}}}{{{V}_{1}}}\] or \[\frac{{{V}_{3}}}{{{V}_{4}}}=\frac{{{V}_{2}}}{{{V}_{1}}}\]\[\Rightarrow \]\[{{\log }_{e}}\left( \frac{{{V}_{3}}}{{{V}_{4}}} \right)={{\log }_{e}}\left( \frac{{{V}_{2}}}{{{V}_{1}}} \right)\] So efficiency of Carnot engine \[\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}\] (i) Efficiency of a heat engine depends only on temperatures of source and sink and is independent of all other factors. (ii) All reversible heat engines working between same temperatures are equally efficient and no heat engine can be more efficient than Carnot engine (as it is ideal). (iii) As on Kelvin scale, temperature can never be negative (as 0 K is more...

First law of thermodynamics merely explains the equivalence of work and heat. It does not explain why heat flows from bodies at higher temperatures to those at lower temperatures. It cannot tell us why the converse is possible. It cannot explain why the efficiency of a heat engine is always less than unity. It is also unable to explain why cool water on stirring gets hotter whereas there is no such effect on stirring warm water in a beaker. Second law of thermodynamics provides answers to these questions. Statement of this law is as follows (1) Clausius statement : It is impossible for a self acting machine to transfer heat from a colder body to a hotter one without the aid of an external agency. From Clausius statement it is clear that heat cannot flow from a body at low temperature to one at higher temperature unless work is done by an external agency. This statement is in fair agreement with our experiences in different branches of physics. For example, electrical current cannot flow from a conductor at lower electrostatic potential to that at higher potential unless an external work is done. Similarly, a body at a lower gravitational potential level cannot move up to higher level without work done by an external agency. (2) Kelvin's statement : It is impossible for a body or system to perform continuous work by cooling it to a temperature lower than the temperature of the coldest one of its surroundings. A Carnot engine cannot work if the source and sink are at the same temperature because work done by the engine will result into cooling the source and heating the surroundings more and more. (3) Kelvin-Planck's statement : It is impossible to design an engine that extracts heat and fully utilises into work without producing any other effect. From this statement it is clear that any amount of heat can never be converted completely into work. It is essential for an engine to return some amount of heat to the sink. An engine essentially requires a source as well as sink. The efficiency of an engine is always less than unity because heat cannot be fully converted into work.