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- 1. OBJECTIVES Determine the heat capacity of the calorimeter . And to measure the calefaction of water at a temperature of 0 C in a calorimeter due to the action of the ambient temperature as a function of time. Determine the electrical conductivity of copper and aluminium by recording a current-voltage characteristic line. Test of the Wiedmann-Franz law.
- 2. APPARATUS Calorimeter vessel, 500 ml Immersion heater Heat conductivity rod, Cu Temperature meter digital Heat conductivity rod, Al Temperature probe Magnetic stirrer Surface temperature probe Heat conductive paste Stopwatch Gauze bag Supporting block Rheostat Glass beaker, short, 400 ml Universal measuring amplifier Digital multimeter Connecting cord, 500 mm, blue Connecting cord, 500 mm, red
- 3. THERMAL CONDUCTIVITY ESTIMATION
- 4. Methods of Heat Transfer 1. Convection Heat transfer by collective movement of molecules within fluids. Mass transfer takes place through diffusion 2. Radiation The transfer of energy to or from a body by means of the emission or absorption of electromagnetic radiation. It doesn't require a medium. 3. Conduction or diffusion
- 5. Thermal conductivity Thermal conductivity is the property of the material that indicates its ability to conduct heat. The thermal conductivity of this metal is, like electrical conductivity, determined largely by the free electrons. Suppose now that the metal has different temperatures at its ends. The electrons are moving slightly faster at the hot end and slower at the cool end. The faster electrons transmit energy to the
- 6. Thermal conductivity in metals Metals are particularly good conductors of heat because their particles are very closely packed so the vibrations are passed on very quickly. They also contain large numbers of "free electrons". As the metal is heated, the free electrons closest source are heated. to the heat This makes them move faster and they travel through the metal, colliding with both atoms and other electrons.
- 7. Heat capacity The heat capacity of a substance is the amount of heat required to change its temperature by one degree, and has units of energy per degree Q = C dt where Q = amount of heat supplied (J) C = heat capacity of the system or object (J/K) dt = temperature rise (K,) The SI unit for heat capacity is J/K (joule per kelvin).
- 8. Fourier’s law oF heat conduction If a temperature difference exists between different locations of a body, heat conduction occurs. The quantity of heat dQ transported with time dt is a function of the cross-sectional area A and the Temperature gradient dT/dx perpendicular to the surface. i.e., Can find λ.
- 9. As free electrons are mainly responsible for heat conduction in metals, They can be considered as a Fermi gas. They obey Fermi-Dirac distribution. So no of particles with momentum p, <np>=1/[z-1eβεp ]+1 z-1 =e- βμ =1/[eβ(εp-μ) ]+1 μ= εF (1-π2/12(KT/ εF )2 +………) So at T=0 and also at room temperature, μ=εF where εF = Fermi Energy So at T=0 and room temperature, <np>= 1/eβ(εp- εF)+1 So electronic energy can’t go beyond Fermi energy so the electrons which are sitting below Fermi energy are mainly responsible for conduction.
- 10. Experimental set-up for thermal conductivity.
- 11. Procedure for Thermal conductivity estimation 1.Measurement of heat capacity of lower calorimeter The heat capacity of the calorimeter is obtained from results of the mixing experiment and the following formula: cw = Specific heat capacity of water, mW = Mass of the water, Tw = Temperature of the hot water, T = Mixing temperature and T =
- 12. procedure • Weigh the calorimeter at room temperature. • Measure and record the room temperature and the temperature of the preheated water provided. • After filling the calorimeter with hot water, determine the mixing temperature in the calorimeter. • Reweigh the calorimeter to
- 13. 2)Determination of the influence of the surroundings The addition of heat from the surroundings is calculated from the temperature increase (Temperature (T) rise of the cold water in the calorimeter) ΔQ=(cwmw+C)(T-T0) Where T0 = Temperature at time t = 0 sec.
- 14. procedure Weigh the empty lower calorimeter. Add ice to the lower calorimeter till the temperature reaches to 0 C. Then remove the ice and take the temperature readings in 1 min interval till the temperature rises to 10 C. Reweigh the calorimeter to determine the mass of the water that it contains.
- 15. Sample reading Calefaction of cold water, Time(min) Temperature(0C) ΔQsurr(J) 1 1 0 2 1.2 294.3 3 1.3 515.2 Slope of the graph gives us dQsurr/dt
- 16. 3. Determination of the heat flow through metal rod thermal conductivity Equation for the measurement of the thermal conductivity of Aluminium/copper rod is given by
- 17. Procedure Weigh the empty, lower calorimeter. Insert the insulated end of the metal rod into the upper calorimeter vessel. To improve the heat transfer, cover the end of the metal rod with heat-conduction paste. When doing so, care must be taken to ensure that the non-insulated end of the rod remains completely immersed in the cold water during the experiment. Using an immersion heater, bring the water in
- 18. Keep the water in the lower calorimeter at 0 C with the help of ice. The measurement can be begun when a constant temperature gradient has become established between the upper and lower surface probes. At the onset of measurement, remove the ice from the lower calorimeter. Measure and record the change in the differential temperature and the temperature of the water in the lower calorimeter for a period of 5 minutes.
- 19. Sample Calculation Time(s) Temp of lower calorimeter(0C) ΔT(0C) ΔQ(J) 0 3 30.57 0 20 3.2 30.48 190.6 40 3.3 30.32 270.8 ΔQ=(cwmw+C)(T-T0) Slope of the graph gives, dQtotal/dt. So (dQ/dt)rod=(dQ/dt)tot-(dQ/dt)surr Since at steady state, dT/dx=ΔT/ Δx We can find λ using Fourier's conduction formula.
- 20. Electrical Conductivity Estimation
- 21. ohm’s law Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. Here the constant for proportionality is R, is the Resistance. V=I R
- 22. Conductivity Resistance can also be defined as, R=ρL / A Where, ρ = Resistivity L= Length A = Cross sectional area The inverse of resistivity is called Conductivity. Electrical conductivity σ = 1/ρ,
- 23. Experimental Setup for Electrical conductivity measurement
- 24. Measurement of the electrical conductivity. This is the Circuit diagram for the measurement of Electrical Conductivity.
- 25. Procedure Perform the experimental set-up according to the circuit diagram. Adjust amplifier and voltage in variable transformer described as in the manual. Set the rheostat to its maximum value and slowly decrease the value during the experiment.
- 26. Sample reading For copper, Voltage(mV) Current(mA) 2.3 0.40 3.6 0.45 8.9 0.50 From the graph, Slope will give us R So Electrical conductivity, σ= l / AR;
- 27. Relation between electrical and thermal conductivity???
- 28. Wiedemann-Franz law At a given temperature, the thermal and electrical conductivities of metals are proportional, but raising the temperature increases the thermal conductivity while decreasing the electrical conductivity. This behaviour is quantified in the WiedemannFranz Law. λ / σ = LT or L= λ / σT where,
- 29. Results Heat Capacity of Calorimeter = 85.23 J/K Thermal conductivity of Aluminium =315.56 Wm-1K-1 Literature value = 205 Wm-1K-1 Thermal conductivity of Copper =952.94 Wm-1K-1 Literature value = 386 Wm-1K-1 Electrical conductivity of Aluminium =3.651*107 Ω-1m-1 Literature value = 3.75*107 Ω-1m-1 Electrical conductivity of Copper =5.764*107 Ω-1m-1 Literature value = 5.882*107 Ω-1m-1
- 30. PRECAUTIONS & Discussion Care should be taken to ensure the immersion heater used is not exposed to air for long. Always maintain minimum water level in the upper calorimeter while maintaining at 100 C. Handle temperature probes with care. Bending front portion too much damage
- 31. Obtained values for electrical conductivity are very close to the literature values. But for thermal conductivity it is significantly different. This is because, It is very difficult to obtain a constant temperature gradient since we have to keep both the calorimeter in fixed temperatures. Even though after coming to steady state the system will get disturbed due
- 32. Presented by, Suhas K Ramesh Rollno-10094 IISER BHOPAL
- 33. Links •http://www.engineeringtoolbox.com/heat-capacity-d_338.html •http://www.chm.davidson.edu/vce/calorimetry/heatcapacity.html •http://www.ndted.org/EducationResources/CommunityCollege/Materials/Physical_Chemi cal/ThermalConductivity.htm •http://www.eos.ubc.ca/ubcgif/iag/foundations/properties/resistivity.htm

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