lab 5 enthalpy of ice

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Seattle Central College *

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163

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Chemistry

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Jan 9, 2024

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pdf

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Uploaded by DrAtomBison13

Enthalpy of Fusion of Ice Purpose/Question To determine the enthalpy of fusion of ice by measuring the temperature. Prelab questions 1. a. V2 = V1(1.087) = 1.087 cm^3 Δ V = V2 - V1 = 1.086 - 1.000 = .087 cm^3 b. P = 101.3 kPa = 1 atm V = 0.87 cm^3 = 8.7 * 10^-5 L PV = (1 atm)(8.7 *10^-5)(101.3 J) = .08813 this value is insignificant 2. Q = m • Lf -> 7.18 = 56.668 * Lf -> .1267 kJ/g = 126.7 J/g Hypothesis If the enthalpy of fusion of ice is calculated correctly, then my expectations of the rate of heat loss of the ice should align with that. Materials/Equipment 150 mL barker 2 8 oz polystyrene foam cups and covers Digital balance Digital thermometer Tongs Cup Ice cubes Paper towels Cold and hot tap water Procedure First, I put 2 small ice cubes into a cup with room temp water and let it melt. Then I mixed the melted cold water with hot water in a beaker until it was around 45 degrees Celsius. Then I stacked my two polystyrene foam cups together, putting the cover on top. I put this on the balance and recorded the mass, which was 4.212 grams. Then I put the calorimeter on the balance and tared it, and then I put around 90 mL of my 45 degree Celsius water into the calorimeter, and I recorded the mass of the water, which was 5.486 grams. Then I put the thermometer into a hole in my cup cover, and recorded the temperature, which was 48.31 degrees Celsius. Then I took my ice cubes and dried them with paper towels, and after removing the cover and thermometer from my calorimeter (my cup), I put the ice into it. Then I recorded the mass of this (calorimeter, water, and ice), which was 8.41 grams. Then I put the cover back on the cup and the thermometer back inside, and swirled the cup around, taking care to hear the ice hitting the sides of the cup and watching the thermometer to take note of the lowest temperature, which was 49 degrees Celsius. Results/Observations/Data/Calculations Mass (g) Volume (cm^3) 4.212 107.02
5.486 109.27 8.41 114.63 Temp (i) (Celsius) Temp (f) (Celsius) 48.31 .48 49.28 .48 49 .38 Q1 = mCw T Q = 4.212 (4.18)(-47.83) Q1 = -842.103 Q2 = mCw T Q2 = 8.41 (4.18)(48.62) Q2 = 1709.178 Water loss (J) Ice gain (J) -842.103 1704.178 -842.103 + Qf + 1709.178 = 0 Qf = 867.075 J Qf = mLf -> 867.075 = 4.212 Lf -> 205.858 = Lf H = 5.938 kJ/mol wasn’t necessary, because the only values needed are the heat values of formation of reactants and products Accepted value: 6.01 kJ/mol Hfusion = 6.01(1000/1)(1/18) = 333.9 J/gram (5.938 - 6.01)/6.01 * 100% = -.072/6.01 * 100% = 1.198% Conclusion Yes, because the value reached aligns with the rate of melting of the ice. Post lab questions Post lab questions also answered in the results section of this report.
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