# Specific Latent Heat of Fusion

1. Oct 15, 2007

### xiao

1. The problem statement, all variables and given/known data
need to work out the Specific Latent Heat of Fusion
i did the expirment 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 (?)

2. Relevant equations
Mice l + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]

Ccal = Ccu = 390 Jkg-1K-1
C H2O = 4200 Jkg-1K-1
LH2O = 33x104 Jkg-1

3. 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 [Broken]

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

Last edited by a moderator: May 3, 2017
2. Oct 15, 2007

### xiao

Heres a lil correction i made

3. Oct 15, 2007

### learningphysics

"0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 4235.2563 J"

I'm going to leave out the units and rewrite:

0.01258L + (0.01258)(4200)(-11) = 4235.2563

subtract (0.01258)(4200)(-11) from both sides...
or equivalently add (0.01258)(4200)(11) to both sides...

4. Oct 15, 2007

### xiao

=======================

my attempt at the final answer

0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11°K = 4235.2563 J
0.01258kg x L (?) = 4235.2563 J + 581.196 J
L (?) = (4235.2563 J + 581.196 J) / 0.01258kg
l (?) = 382865.6041

which i think is wrong coz i dont know wat i am doing

5. Oct 15, 2007

### xiao

0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (-581.196) = 4235.2563

u mean subtract (-581.196) from 4235.2563
4235.2563 - (-581.196)
4816.4523
?????

6. Oct 15, 2007

### learningphysics

math wise it looks right to me. l = 382865.6041 J/kg.

I don't know if the answer is right... I'm rusty with my chemistry... but as far as the algebra for this part is concerned, the math is right.

7. Oct 15, 2007

### xiao

lol i know it means 382865.6041 Jouels per kilogram
382865.6041 Jouels is a HELL load of jouels to melt ice in water at room temprature
382865.6041 Jouels is like wat ur house used in 2 weeks (just guessing)

8. Oct 15, 2007

### learningphysics

yeah... it's basically the same as what you did...

0.01258L + (-581.196) = 4235.2563

0.01258L + (-581.196) - (581.196) = 4235.2563 - (-581.196)

0.01258L = 4816.4523

which gives the same answer you got.

9. Oct 15, 2007

### xiao

thanx mate even tho the answer is wrong atleast the math is write so i can put that in my Conclusion but thanx alot for the help

i still have to do the following experiment
combustion of fules
something on electricity
something on water turbines
something on solar energy
biodiesel

so i will be here much more to get help with my results :P

then after that i am freeeeeeeeeeeeeeeeeeeeeeeeeeee from anykind of science

10. Oct 15, 2007

### learningphysics

I don't know. This webpage:

http://www.physchem.co.za/Heat/Latent.htm [Broken]

gives the latent heat of ice as 334 kJ.kg-1 = 334,000J/kg, which is not that far off from what you have.

Last edited by a moderator: May 3, 2017
11. Oct 15, 2007

### xiao

lol by bad wooooohooo i did it

Last edited by a moderator: May 3, 2017
12. Oct 15, 2007

### learningphysics

:) good job. seems like it was a tough experiment! I was terrible at chemistry.

13. Oct 15, 2007

### xiao

the expirment was easy just melt ice in water in room temp untill u reach a certian temp

but i find physics hard even the basics like this coz this is ment to be basic and i am not even a science student i am doing my BA in international relations and politics

14. Oct 15, 2007

### learningphysics

You'll usually be able to get lots of help here. good luck!

15. Oct 15, 2007

### learningphysics

shouldn't the left side of this equation be:

12.58g x L (?) +12.58g x 4200 Jkg-1K-1 x (21°K) ?

because the ice is initially at 0°C... and this water that is from the melted ice goes to a final temperature of 21°C from 0°C.