lab 5 enthalpy of ice
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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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Question 5 of 18
Submit
What is the quantity of heat (in kJ) associated with cooling 150.3 g of
water from 25.60°C to ice at -10.70°C?
kJ
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