Data Initial Mass (kg) = U.0 9225 Evaporated Mass (kg) = 0. 00 705 kg Final Mass (kg) = 0 . 08520 Elapsed Time (s) = 300.56 Calculations Background Rate (kg/sec) = Evaporated Mass Elapsed Time = 2. 35 x10 KO/s Total Collection Time (s) 20 x 10-3 kg Background Time 851. 06 Time for each phase (s) = 283. 61 5 Part II: Data Current: I(A) = 0 . 408 AI= 10 0041 5 Voltage: V (V) = 11. 83 AV - 0. 118 4 Heating time: t (s) = 283.21 , At = 10. 3 s Mass of evaporated LN2: M (kg) = 0. 0 644 kq AM =0. 24200 Time: 190.7 s Mass (g) vs Time (s) Mass: 78.432 Mass: 70 815 Linear Fit for. Latest | Mass m = mt+b m (Slope): -0.02731 +/- 2.436E-05 g/s b (Y-Intercept): 83.64 +/-0.004194 g Correlation: -0.9999 150- RMSE: 0.03635 g Linear Fit for. Latest | Mass m = mt+b m (Slope): -0.02271 +/-1.146E-05 g/s b (Y-Intercept): 75. 15 +/-0.008275 g Correlation: -1.000 RMSE: 0.01633 g Mass (9) 100 4 50- 600 800 200 400 (190.7, 189.22) Time (s)Analysis: (1) What does the slope of a line on the mass vs time graph represent? (2) Do the slopes of Phase I and Phase III agree, based on the uncertainties in the linear fit of the two lines? Show your work. (3) Do the slopes of Phase I and Phase III agree with the rate of Background loss calculated in Part I? Show your work. (4) Using equation (C), calculate the Latent Heat of Vaporization L. Show your work. (5) Using Rule 4 for combining errors, calculate the error AL . Show your work. (6) Express L,+ AL, to the appropriate number of significant places. (7) Calculate the % difference between your measured value and the accepted value for Ly. Do the two values agree or disagree? Explain. (8) What possible experimental errors could account for the observed difference between your measured value and the accepted value for Ly