## Heat capacity data for Problem 10.4 ## The experimental heat capacity data was in a per mole basis dividing by Avogadro's ## Number converts it to a per atom basis. Data is given for Al and Pb. ## Data are in the form (T,C). nA = 6.02e23 ## Avogadro's number Aldata =[(20,0.23/nA),(40,2.09/nA), (60,5.77/nA), (80,9.65/nA), (100,13.04/nA), (150,18.52/nA), (200, 21.58/nA), (250,23.25/nA), (300, 24.32/nA), (400,25.61/nA)] Pbdata = [(20,11.01/nA),(40,19.57/nA), (60,22.43/nA), (80,23.69/nA), (100,24.43/nA), (150,25.27/nA), (200, 25.87/nA), (250,26.36/nA), (300, 26.82/nA), (400,27.45/nA)] #Heat Capacity graphs gdisplay(xtitle='Temp',ytitle='Heat Capacity', x=0, y=600, width=600,height=300) ##Alg = gdots(color=color.magenta,pos=Aldata) ##Pbg = gdots(color=color.cyan, pos=Pbdata) CAlgraph = gcurve(color=color.magenta) CPbgraph = gcurve(color=color.cyan) ## To compare the experimental heat capacity data to what you have calculated ## theoretically, Cut and paste the computer code below into your program. ## Place it right before your while loop. Use the curves CAlgraph and CPbgraph ## (defined above) to plot your theoretical results. That is put these graphics ## commands into your while loop after you calculate the heat capacities: ## CAlgraph.plot( pos=(T1,CAl) ) # plots of heat capacity vs. temperature ## CPbgraph.plot( pos=(TPb,CPb) ) # Use whatever you named these variables ## HINTS for theoretical calculation: A good technique is to define additional variables, ## q1a = q1+1 and q1b = q1+2. ## Then use q1, q1a, and q1b to calculate three different values of the entropy. ## One can then calculate the two values of entropy difference needed to determine ## two temperatures T-lower and T-upper, from which the heat capacity can be ## calculated as C = dE/dT/Na, where Na = 35 atoms. Note that dE here corresponds ## to one quantum of energy, hbar*sqrt(4*ks/m). You will need to calculate dEAl ## and dEPb for aluminum and lead, and use these to calculate and plot the heat ## capacities for the two metals.