LatticeVibrationsinSolidsEPSolnME501F2011

# LatticeVibrationsinSolidsEPSolnME501F2011 - 1 ME 501...

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1 ME 501 Lattice Vibrations in Solids Example Problem Fall 2011 A 1.0-kg slab of aluminum at 300 K is quenched by placing it in a bath of liquid nitrogen at 77 K. What is the heat loss for the aluminum slab? (Answer: Q = -155 kJ) 0 0.5 1 1.5 2 2.5 3 3.5 50 100 150 200 250 300 350 400 C V /R Temperature (K) Specific Heat for Aluminum D = 398 K Cv/R = -1.0276 + 0.034647 T - 1.0831x10 -4 T 2 +1.1549x10 -7 T 3  3 34 exp 1 DD B D D VB D D xh k T T x CN k D x x      For aluminum, 398 D K . Using the tabulated values of   D Dx , the heat capacity is calculated as a function of temperature from 39.8 K to 398 K. The curve fit to the calculated data gives us 42 73 23 3 3 4 1.0276 0.034647 1.0831 10 1.1549 10 exp 1 V D D BD C x T T T Nk x ab Tc T d T    Integrating from initial temperature 1 T to final temperature 2 T we obtain

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2    2 2 1 1 23 4 22 33 44 2 1 21 5 8 2 1 234 224 1.0276 0.017324 3.6103 10 2.8873 10 T T V T BB T C Ub T c T d T dT aT Nk bcd aT T T T T T TT    
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LatticeVibrationsinSolidsEPSolnME501F2011 - 1 ME 501...

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