I was hoping that you could check to make sure I’m using the correct units. For some reason I am having a hard time understanding when to use 4186 for specific heat of water or 4.186. For the latent heat of fusion, I have it in kg•cal degrees C. In my book it has it in J/kg. Since the book says it is 3.33 X 10^5, how do I get my answer to show in those terms? Thank you!

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Chapter2: The Kinetic Theory Of Gases
Section: Chapter Questions
Problem 48P: Nuclear fusion, the energy solute of Sun, hydrogen bombs, and fusion reactors, occurs much more...
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I was hoping that you could check to make sure I’m using the correct units. For some reason I am having a hard time understanding when to use 4186 for specific heat of water or 4.186. For the latent heat of fusion, I have it in kg•cal degrees C. In my book it has it in J/kg. Since the book says it is 3.33 X 10^5, how do I get my answer to show in those terms? Thank you!
(MCAT)
H20
O°c Hzo
Part Il: Heat of Fusion of lce
Starting with equation (5), derive an expression for the heat of fusion of ice using
equations (1) and (2). Express the result in terms of the mass and temperature of
the calorimeter water (mwater, Th), mass of ice (m), and final equilibrium temperature
(Teg). Clearly show the steps you took to get your result.
Heat that
= - Q
H2O = -10389J
ICE
goes into ia
(meAT).
-ice
mL
H20
ice
(HL\(4.1816ca) (2416-41I.6)= (6025) L¢
MCAT =
mL
H10
10.389 =(.025) (Lf)
MHeo Heo( Teq-Tn)=mice CLf)
.025
Lf=415 kg.calc
H2O H2D
Mice
Mass of dry Styrofoam cup:
24.79
=,025 kg
mcup =
Data Table for Heat of Fusion of Ice Experiment (enter units for L; in table)
mwater
mtotal
mice
Th
(°C)
Teg
(°C)
Lf
(kg)
(kg)
(kg)
41.6
415 Kg.cal'c
30g trial
146
.1712
.025
24.6
Jour tesult
38:7
2.6
357 kg.cal'c
60g trial
.134
.159
.06
(.134) (41126)(2.6-58.7) =
.06
1196
157
(560.924)(-36.1) - 20249.35.
-06
• 219
-337489
Transcribed Image Text:(MCAT) H20 O°c Hzo Part Il: Heat of Fusion of lce Starting with equation (5), derive an expression for the heat of fusion of ice using equations (1) and (2). Express the result in terms of the mass and temperature of the calorimeter water (mwater, Th), mass of ice (m), and final equilibrium temperature (Teg). Clearly show the steps you took to get your result. Heat that = - Q H2O = -10389J ICE goes into ia (meAT). -ice mL H20 ice (HL\(4.1816ca) (2416-41I.6)= (6025) L¢ MCAT = mL H10 10.389 =(.025) (Lf) MHeo Heo( Teq-Tn)=mice CLf) .025 Lf=415 kg.calc H2O H2D Mice Mass of dry Styrofoam cup: 24.79 =,025 kg mcup = Data Table for Heat of Fusion of Ice Experiment (enter units for L; in table) mwater mtotal mice Th (°C) Teg (°C) Lf (kg) (kg) (kg) 41.6 415 Kg.cal'c 30g trial 146 .1712 .025 24.6 Jour tesult 38:7 2.6 357 kg.cal'c 60g trial .134 .159 .06 (.134) (41126)(2.6-58.7) = .06 1196 157 (560.924)(-36.1) - 20249.35. -06 • 219 -337489
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