An iron bar 0.50 m long and a copper bar 1.2 m long are joined ened to end. One of the iron bar is kept at 80 C while the far end of the copper bar is maintained at 0 C by a mixture of ice and water. The outer surfaces are of the bars arelagged so that there are no thermal energy losses. Both bars are circular cross - section, diameter 0.16 m .At thermal equilibruim the temperture at the juction of the metals is Tj.Calculate Tj and the rate of energy flow.Thermal conductivity of iron 75 Wm KThermal conductivity of copper 390 Wm K
Hi let the thermal conductivities be (i is for Iron and C is for copper): Ki 75 W/mK Kc 390 W/mK The temperatures: Ti 0 C Tc 80 C Tj ? (temperature of the junction AT EQUILIBRIUM) The lengths: Li 0.5 m Lc 1.2 m as Q K L (temperature difference) At equilibrium (heat gained by iron heat lost by copper) Qi -Qc (negative sign for loss in heat) Ki Li (Tj - Ti) -[Kc Lc (Tj - Tc)] 75 x 0.5 (Tj - 0) -390 x 1.2 (Tj - 80) 37.5 Tj 37440 - 468 Tj 468 Tj + 37.5 Tj 37440 505.5 Tj 37440 Tj 37440/505.5 74.0653 C the rate of energy flow: Qi Ki Li (Tj - Ti) 75 x 0.5 x (74.0653 - 0) 75 x 0.5 x 74.0653 2777.45 W Check Qc Kc Lc (Tj - Tc) 390 x 1.2 x (74.0653 - 80) 390 x 1.2 x (-5.9347) -2777.44 W The negative sign shows that heat is being lost by the copper reo. Hope to answer you well. Keep smiling. Bye.
Glass can be moulded Macor is a machinable ceramic Neither one affects magnetic fields. The insulation on the copper magnet wire will char if it gets to hot, though.
you utilize the Bernoulli Equations- that's particularly conservation of capability. The sum of rigidity (ability) capability and kinetic capability has to proceed to be consistent, so which you will say 0.5Dv^2+P consistent, the place P is the rigidity, D is the density and v is the fluid speed. (there is likewise a gravitational ability capability term, besides the undeniable fact that this is negligible consequently) you recognize the rigidity interior the bottle, the speed interior the bottle is 0, the rigidity outdoors is atmospheric rigidity, yet as that's the reference ingredient for the rigidity interior the bottle it disappears, so all would desire to verify is the speed outdoors the bottle. as quickly as you recognize this, you merely multiply the speed via the pass section (make specific your gadgets are an identical!) and you have the quantity pass value. P,interior0.5D(v,outdoors)^2 this would possibly not supply you with an exact answer, through fact the Bernoulli equations do no longer think with regard to the compressibility of the gas, yet you're entering into very final twelve months mechanical engineering stuff in case you initiate happening that highway!
