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1 Unit II Thermal Physics Introduction- Modes of Heat Transfer Normally there are three modes of transfer of heat from one place to another viz., conduction, convection and radiation. Conduction : Conduction refers to the heat transfer that occurs across the medium. Medium can be solid or a fluid. Convection: It is the process in which heat is transferred from hotter end to colder end by the actual movement of heated particles..radiation : In radiation, in the absence of intervening medium, there is net heat transfer between two surfaces at different temperatures in the form of electromagnetic waves. Specific Heat Capacity The specific heat capacity of a material is the amount of energy (in Joules) needed to increase the temperature of one kilogram of mass of the material by one elvin. RECTILINEAR FLOW OF HEAT THROUGH A ROD Consider a long rod AB of uniform cross section heated at one end A as shown in figure. Then there is flow of heat along the length of the bar and heat is also radiated from its surface. B is the cold end.

2 Consider the flow of heat between the sections P and Q at distance and +δ from the hot end. Ecess temperature above the surroundings at section P = 0 d Temperature gradient at section P = d d A d A d AS dt A E A d d S d Ep dt A Ecess temperature at section Q = Temperature gradient at Q = d d d d d d d d = d d

3 Heat flowing (entering) through P in one second Q = d A d () Heat flowing (leaving) through Q in one second Q = Q = Net heat gain by the element d d A d d d d A () d d in one second Q = Q- Q. (3) Q = = - = - d A d d A + d d A d d d A A d d d d A A d d..(4) Before the steady state is reached Before the steady state is reached, the amount of heat Q is used in two ways. A part of the heat is used in raising the temperature of the rod and the remaining heat is lost by radiation from the surface. Heat absorbed per second to raise the temperature of the rod

4 d = mass specific heat capacity dt = (A δ)ρ S (5) where A Area of the cross-section of the rod ρ Density of the rod S Specific heat capacity of the rod - Rate of rise in temperature Heat lost per second due to radiation = E p δ..(6) Where E Emissive power of the surface p - Perimeter of the bar δ Surface area of the element - Average ecess of temperature of the element over that of the surroundings Amount of heat (Q) = Amount of heat absorbed + Amount of heat lost Q = (A δ)ρ S + E p δ.(7) On Comparing the eqns (4) and (7) d A d = (A δ)ρ S + E p δ (8) Dividing both LHS and RHS of the above equation by A, we have, d A d A d AS dt A E A d d S d dt Ep A (9) The above equation is standard differential equation for the flow of heat through the rod. Special cases:-

5 Case : when heat lost by radiation is negligible. If the rod is completely covered by insulating materials, then there is no loss of heat due to radiation. Hence Epδ = 0 d S d d d dt h dt.(0) where, S h, thermal diffusivity of the rod. \ Case : After the steady state is reached. After the steady state is reached, there is no raise of temperature d Hence, = 0 dt equation (9) becomes d Ep d A Substituting, d d Ep A, we have.() d 0 d (This represent second order differential equation). The general solution of this equation is Ae Be..() Where A and B are two unknown constants which can be determined from the boundary conditions of the problem.

6 Suppose the bar is of infinite length, Ecess temperature above the surrounding of the rod of the hot end = 0 Ecess temperature above the surrounding of the rod at the cold end = 0 Boundary condition (i) When = 0, = 0 (ii) when =, = 0 0 = A+B 0 = Ae + Be - 0 = Ae As we know e cannot be zero, therefore A should be zero i.e., A=0 then, 0 = B Substituting A and B in equation (), we have = 0e -μ (3) The above equation represents the ecess temperature of a point at a distance from the hot end after the steady state is reached and it eponentially falls from hot end.

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8 HEAT CONDUCTION THROUGH A COMPOUND MEDIA (SERIES AND PARALLEL) Consider a composite slab of two different materials, A & B of thermal conductivity & respectively. Let the thickness of these two layers A & B be d and d respectively Let the temperature of the end faces be θ & θ and temperature at the contact surface be θ, which is unknown. Heat will flow from A to B through the surface of contact only if θ > θ. After steady state is reached heat flowing per second (Q) through every layer is same. A is the area of cross section of both layers Amount of heat flowing per sec through A Q = A( ) () Amount of heat flowing per sec through B Q= A( ) () The amount of heat flowing through the materials A and B is equal in steady conditions Hence () and () are equal

9 ) ( A = ) ( A.(3) Rearranging the (3), we have A(-) = A(-) - = - + = + + = ( + ) =..(4) This is the epression for interface temperature of two composite slabs in series. Substituting from equation (4) in equation (), we get Q = A = A = A = A = ) ( A = ) ( A

10 Q = A( )..(5) Q is the amount of heat flowing through the compound wall of two materials. This method can also be etended to comp[osite slab with more than two slabs. Generally, the amount of heat conducted per sec for any number of slabs is given by, Q = A ( ) BODIES IN PARALLEL Let us consider a compound wall of two different materials A and B of thermal conductivities and and of thickness d and d respectively. These two material layers are arranged in parallel.

11 The temperatures θ is maintained at one faces of the material A and B and opposite faces of the material A and B are at temperature θ. A & A be the areas of cross-section of the materials. Amount of heat flowing through the first material (A) in one second. Q = A ( ).() Amount of heat flowing through the second material (B) in one second. Q = A ( ).() The total heat flowing through these materials per second is equal to the sum of Q and Q Q = Q+Q.(3) Substituting equations () and () in (3), we get, Q= A ( ) + Amount of heat flowing per second A A ( ) A Q = In general, the net amount of heat flowing per second parallel to the composite slabs is given by Q ) ( A

12 RADIAL FLOW OF HEAT In this method heat flows from the inner side towards the other side along the radius of the cylindrical shell. This method is useful in determining the thermal conductivity of bad conductors taken in the powder form. CYLINDRICAL SHELL METHOD (or) RUBBER TUBE METHOD Consider a cylindrical tube of length l, innerradius r andouter radius r. The tube carries steam or some hot liquid. After the steady state is reached, the temperature on the innersurface is θ and on the outer surface is θ in such a way θ >θ. Heat is conducted radially across the wall of the tube. Consider an element of thickness dr and length l at a distance r from the ais. Working: Steam is allowed to pass through the ais of the cylindrical shell. The heat flows from the inner surface to the other surface radially.after the steady state is reached, the temperature at the inner surface is noted as and on the outer surface is noted as.

13 Calculation: The cylinder may be considered to consists of a large number of coaial cylinders of increasing radii. Consider such an elemental cylindrical shell of thr thickness dr at a distance r from the ais. Let the temperatures of inner and outer surfaces of the elemental shell be and +d. Then, d The Amount ofheat conducted per second Q A dr Here Area of cross section A = πrl d Q rl dr Rearranging the above equation we have dr l d r Q () The Thermal conductivity of the whole cylinder can be got by, integrating equation () within the limits r to r and to, log r r e dr r r r l Q Rearranging we get, = l Q d r Q.log e r l

14 =.306 log r Q 0 r l W m - - By knowing the values in RHS, the thermal conductivity of the given material can be found. DETERMINATION OF THERMAL CONDUCTIVITY OF RUBBER It is based on the principle of radial flow of heat through a cylindrical shell. Description: It consists of a calorimeter, stirrer with a thermometer. The setup is kept inside the woodenbo. The space between the calorimeter and the bo is filled with insulating materials such as cotton, wool, etc. to avoid radiation loss, as shown in fig. Working: The empty calorimeter is weighed, let it be (w). It is filled with two third of water and is again weighed, let it be (w) A known length of rubber tube is immersed inside the water contained in the calorimeter. Steam is passed through one end of the rubber tube and let out through the other end of the tube. The heat flows from the inner layer of the rubber tube to the outer layer and is radiated. The radiated heat is gained by the water in the calorimeter. The time taken for the steam flow to raise the temperature of the water about 0C is noted, let it be t seconds.

15 Observation and calculation: Let w Weight of calorimeter w Weight of calorimeter and water w w Weight of the water alone Initial temperature of the water Final temperature of the water - Rise in temperature of the water S Temperature of the steam l Length of the rubber tube (immersed) r Inner radius of the rubber tube r Outer radius of the rubber tube s Specific heat capacity of the calorimeter s Specific heat capacity of the water Average temperature of the rubber tube. 3 3 = We know from the theory of cylindrical shell method the amount of heat conducted by the rubber tube per second is given by

16 l Q = S 3 r loge r The amount of heat gained by calorimeter per second = The amount of heat gained by.() w s t w s w () water per second = (3) t The amount of heat gained by the water and calorimeter per second is obtained by () +(3) Under steady state Q = Q = ( w w ) s( ) ws ( ) t ( ) w s w w s...(4) t The amount of heat conducted by The amount of heat gained by the rubber tube per second = the water and the calorimeter per escond Hence, equation () = Equation (4) l S 3 r loge r Substituting 3 = = ( ) ws w w s t ( )log r e w s w w s = r Wm - - ( ) lt s By substituting the values in RHS, the thermal conductivity of the rubber can be determined.

17 Methods to determine thermal conductivity The thermal conductivity of a material is determined by various methods. Searle s method for good conductors like metallic rods. Forbe s method - for determining the absolute conductivity of metals 3. Lee s disc method for bad conductors 4. Radial flow method for bad conductors LEE S DISC METHOD FOR DETERMINATION OF THERMAL CONDUCTIVITY OF BAD CONDUCTOR The thermal conductivity of bad conductor like ebonite or card board is determined by this method. Description: The given bad conductor (B) is shaped with the diameter as that of the circular slab (or) disc D. The bad conductor is placed inbetween the steam chamber (S) and the disc (D), provided the bad conductor, steam chamber and the slab should be of same diameter. Holes are provided in the steam chamber (S) and the disc (D) in which

18 thermometer are inserted to measure the temperatures. The total arrangement is hanged over the stand as shown in fig. Working: Steam is passed through the steam chamber till the steady state is reached. Let the temperature of the steam chamber (hot end) and the disc (cold end) be and respectively. Observation and Calculation: Let be the thickness of the bad conductor (B), m is the mass of the slab, s be the specific heat capacity of the slab. r is the radius of the slab and h be the height of the slab, then Under steady state Amount of heat conducted by the Bad conductor per second = A ( ).() Area of the cross section is = πr () Amount of heat conducted per second = The amount of heat lost by slab per second r ( ) (3) = m s Rate of cooling = msr c..(4)

19 The amount of heat conducted by the Bad conductyor (B) per second = Amount of heat lost by the slab (D) Per second Hence, we can write equation (3) = equation (4) r ( ) = msr c = msr c.(5) r ( ) To find the rate of cooling Rc Rc in equation (3) represents the rate of cooling of the disc along with the steam chamber. To find the rate of cooling for the disc alone, the bad conductor is removed and the steam chamber is directly placed over the disc and heated. When the temperature of the slab attains 5C higher than, the steam chamber is removed. The slab is allowed to cool, simulataneously a stop watch is switched ON. A graph is plotted taking time along ais and temperature along y ais, the d rate of cooling for the disc alone (i.e) is found from the graph as shown in fig. dt The rate of cooling is directly proportional to the surface area eposed. Case(i) Steam chamber and bad conductor are placed over slab, in which radiation takes place from the bottom surface of area (πr ) of the slab and the sides of the of area (πrh). Rc = πr + πrh Rc = πr(r+h) (6)

20 Case(ii) The heat is radiated by the slab alone, (i.e) from the bottom of area(πr ), top surface of the slab of area (πr ) and also through the sides of the slab of area πrh. d dt = πr + πr + πrh d dt = πr + πrh From (6) and (7) d dt = πr(r+h)..(7) r r c r h) dr rh) dt Rc = ( rh) d ( rh) dt (8) Substituting (8) in (5) we have d ms ( rh) dt = Wm - - r ( )( rh) Hence, thermal conductivity of the given bad conductor can be determined from the above relation.

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