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)H20O°c HzoPart Il: Heat of Fusion of lceStarting with equation (5), derive an expression for the heat of fusion of ice usingequations (1) and (2). Express the result in terms of the mass and temperature ofthe 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= - QH2O = -10389JICEgoes into ia(meAT).-icemLH20ice(HL\(4.1816ca) (2416-41I.6)= (6025) L¢MCAT =mLH1010.389 =(.025) (Lf)MHeo Heo( Teq-Tn)=mice CLf).025Lf=415 kg.calcH2O H2DMiceMass of dry Styrofoam cup:24.79=,025 kgmcup =Data Table for Heat of Fusion of Ice Experiment (enter units for L; in table)mwatermtotalmiceTh(°C)Teg(°C)Lf(kg)(kg)(kg)41.6415 Kg.cal'c30g trial146.1712.02524.6Jour tesult38:72.6357 kg.cal'c60g trial.134.159.06(.134) (41126)(2.6-58.7) =.061196157(560.924)(-36.1) - 20249.35.-06• 219-337489

Specific heat is the amount of heat required for a unit mass to increase the temperature by one…

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College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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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

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