Trial 1 Trial 2 Mass of ice M.. (grams) 79 79 Mass of warm water 120 119 Mer (grams) Initial temperature of water Tinitial ("C) 41 42 Final temperature of mixture Tfinal ("C) 4 Calculated heat of fusion of ice 72.4 60.8 amerinental (cal/g) Standard value of heat of fusion of ice Lutmdard (cal/g) 79.7 79.7 Percentage error 9.22% 23.76%

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter8: Thermochemistry
Section: Chapter Questions
Problem 12QAP: The heat of neutralization, Hneut, can be defined as the amount of heat released (or absorbed), q,...
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1:13 0
* ll 66%i
Cyra Garces
Active Now
CALORIMETRY
HEAT OF FUSION
Trial 1
Trial 2
Mass of ice M (grams)
79
79
Mass of warm water
Mter (grams)
120
119
Initial temperature of water
Tinitial ("C)
41
42
Final temperature of mixture
Tfinal ("C)
-4
Calculated heat of fusion of ice
Leperinemtal (cal/g)
72.4
60.8
Standard value of heat of
fusion of ice Lemdard
(cal/g)
79.7
79.7
Percentage error
9.22%
23.76%
Calculations
According to the principle of the conservation of energy, the quantity of heat absorbed by the ice as
it melts (Quained melring) and then heats up to the final equilibrium temperature (Qpain by melted
ice) must equal the quantity of heat released by the warm water (Qlast by warm water) as it cools down
to the final equilibrium temperature.
Mathematically:
le
guined metingl iast by warm water
pained meltingl + o
cat
(M(L) + (M)(1)T - OC) = (Mt(1)(Tie- T
fina
- 0C) - (Mt(1
final
water
initial
More
Edit
II
-
Transcribed Image Text:1:13 0 * ll 66%i Cyra Garces Active Now CALORIMETRY HEAT OF FUSION Trial 1 Trial 2 Mass of ice M (grams) 79 79 Mass of warm water Mter (grams) 120 119 Initial temperature of water Tinitial ("C) 41 42 Final temperature of mixture Tfinal ("C) -4 Calculated heat of fusion of ice Leperinemtal (cal/g) 72.4 60.8 Standard value of heat of fusion of ice Lemdard (cal/g) 79.7 79.7 Percentage error 9.22% 23.76% Calculations According to the principle of the conservation of energy, the quantity of heat absorbed by the ice as it melts (Quained melring) and then heats up to the final equilibrium temperature (Qpain by melted ice) must equal the quantity of heat released by the warm water (Qlast by warm water) as it cools down to the final equilibrium temperature. Mathematically: le guined metingl iast by warm water pained meltingl + o cat (M(L) + (M)(1)T - OC) = (Mt(1)(Tie- T fina - 0C) - (Mt(1 final water initial More Edit II -
3. Make the step by step
computation for trial 2,
Transcribed Image Text:3. Make the step by step computation for trial 2,
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