- #1
xiao
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Homework Statement
need to work out the Specific Latent Heat of Fusion
i did the experiment and here are the results
Results:
Mass of calorimeter
Mcal 116.87g
Mass of H2O
MH2O = [Mcal + H2O] - Mcal 80.82g (197.69g)
Mass of Ice
Mice = [Mcal + Mice + H2O] - MH2O 12.58g (210.27g)
Room Temperature (Tr) 26°c
6°c above (Tr) = Ti 32°c
Aprox. 6°c below (Tr) = Tf 21°c
Change in Temperature ∆T -11°K
Specific Latent Head of Ice L (?)
Homework Equations
Mice l + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]
we already know the following
Ccal = Ccu = 390 Jkg-1K-1
C H2O = 4200 Jkg-1K-1
LH2O = 33x104 Jkg-1
The Attempt at a Solution
so there where i have gotten so far with
Mice L(?) + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]
I need to find out what L(?) is
12.58g x L (?) +12.58g x 4200 Jkg-1K-1 x -11°K = -[116.87g x 390 Jkg-1K-1 x -11°K + 80.82g x 4200 Jkg-1K-1 x -11°K]
Convert grams into kilo grams
0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 0.11687kg x 390 Jkg-1K-1 x -11°K +0.08082kg x 4200 Jkg-1K-1 x -11°K
0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = -501.3723 + -3733.884
0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = -4235.2563 J
how do i switch over the 0.01258kg + 0.01258kg x 4200 Jkg-1K-1 x -11°K to the other side to have just L (?) on one side
I need help to find out L(?)
how do i rearange to find L(?)
Thais is wat i mean
http://img87.imageshack.us/img87/2803/71713937qj0.jpg
PS i know that i am not very good at science, i am a International Relations and Politics student but they r making me do this
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