Question

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PHYS 202 L-Latent Heat of Fusion of Ice Aim: The aim of this laboratory exercise is to of this laboratory exercise is to investigate the effects of the transfer of heat on temperature and phase and to determine an empirical value for the heat of fusion of ice, L. NAME: DATE: Table of data TRIAL 1 200 ml TRIAL 2 TRIAL 3 200mL 2ooml O. a hal 0.24 25° o ang Volume of tap water Vw, ml Mass of tap water m., kg Initial Temperature of water Ti.C Final Temperature of water and melted ice Tr. C Total volume of melted ice and water VT, mL Volume of ice added V. (VT-V) ml Mass of ice added mi, kg L. kcal/kg 30° 15° 240mi 4 Ome 04 ng 250ML 23om 30ML 03hg 05hg ANALYSIS Use your data a equation (2) to determine the value of the latent heat of fusion. Show your work for all three trials with the appropriate units. TRIAL 1: L m .c.(T-T;)- .T, LF ( 924 (80= 15)x9200) 414200 (IS) -200001/1g TRIAL 2 [2= 0.2-26-10)89200)-442001103 5920;000.-72000 = 370,000 JIJ 1831 035 ( 28 5) x4200)-(9200x6) = 319200-25200 =21400011 10.05 TRIAL 3 24 Find the experimental value of the latent heat of fusion - the average for the three trials. 070 6780001 21800 9 87000329X 3 L- 9 CW=4201/17 A Sourcing

Answer 1

Trial 1 (mwcw(T; - Ty) - moty)_0.2(4186)(30 – 15) - 0.04 * 4186 * 15 mi 0.04 L. = 2.51 * 105 kg Trial 2 (m.c(T; -T.) - moT) 0.2(4186) (25 – 10) - 0.03 * 4186 * 10 mi 0.03 Lg = 3.77 * 105 kg Trial 3: (mwcw(T; – T;) – mcwT;)_0.2(4186)(25 – 6) – 0.05 * 4186 * 6 mi 0.05 L; = 2.93 * 105, Part b: The average is: Li + L2 + L3_ L; = 3 = 3.07 * 105 - The percentage of error is: we 3.34 – 3.07 * * 100 = 8.1 % Using the given data: 0.4(4186) (30 - T) -0.1 * 4186 *T 3.34 x 105 = 0.1 T = 8.04°C %E= 3.34