100 lines
1.7 KiB
Python
100 lines
1.7 KiB
Python
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# coding: utf-8
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from __future__ import print_function
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from lab import *
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import matplotlib.pyplot as plt
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import numpy as np
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##
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## Part 1
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## equivalent mass
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mt = 353.36
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mi = np.array([592.67, 576.99, 512.66, 479.75, 498.35])
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mf = np.array([682.53, 678.08, 634.87, 626.51, 626.81])
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T1 = np.array([20.2, 35.1, 21.1, 38.9, 22.5])
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T2 = np.array([83.1, 87.0, 89.9, 87.2, 95.0])
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Te = np.array([38.0, 51.0, 48.9, 61.9, 53.9])
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m1 = mi-mt
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m2 = mf-mi
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me = m2*(T2-Te)/(Te-T1) - m1
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me = me[2:].mean()
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print('m_e: ', me)
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##
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## Part 2
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## specific heat
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mt = 359.67
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mi = np.array([713.98, 717.14, 693.60])
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ms = np.array([130.02, 121.85, 38.95])
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m1 = mi-mt
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Ts = 100
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T1 = np.array([21.2, 23.2, 21.2])
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Te = np.array([23.6, 25.3, 23.0])
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cs = (Te-T1)*(m1+me)/((Ts-Te)*ms)
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print('''
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rame: {:.1f}
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ottone: {:.1f}
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alluminio: {:.1f}
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'''.format(*cs*4186))
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##
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## Part 3
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## Joule constant
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mt = 354.71
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m = 760.53 - mt
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V = 14
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I = 3
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T = sample(22.0, 23.4, 24.8, 26.4, 27.6, 29.0, 30.4)
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t = sample(0, 1, 2, 3, 4, 5, 6)*60
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# linear interpolation
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x = np.linspace(0,370)
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a,b = linear(t, T, 0.1)
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f = lambda x: a.n+b.n*x
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# plot T - t
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plt.xlabel('tempo (s)')
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plt.ylabel('temperatura (s)')
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plt.xlim(0,400)
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plt.scatter(t, T, color='#135964')
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plt.plot(t, f(t), '#317883')
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plt.show()
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# χ² test
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alpha = chi_squared_fit(t, T, f, 0.1)
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print('χ²: α={:.3f}, α>ε: {}'.format(alpha, alpha>epsilon))
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# find J from b
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J = I*V/(b*(m+me))
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print('J={}'.format(J))
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##
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## Part 4
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## latent heat
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mt = 355.12 # calorimeter tare
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mi = 386.61 # calorimeter + ice
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mf = 561.62 # calorimeter + ice + water
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m1 = mf-mi # water
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m2 = mi-mt # ice
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t1 = 91.1
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t2 = -17
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te = 66.8
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ca = 4.2045 # water specific heat at 91°C
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cg = 2.0461 # ice specific heat at -17°C
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l = ca*(m1+me)/m2*(t1-te) + cg*t2 - ca*te
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print(l)
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