Register to reply

Specific Latent Heat of Fusion

by xiao
Tags: fusion, heat, latent, specific
Share this thread:
xiao
#1
Oct15-07, 12:48 PM
P: 8
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]

we already know the following
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



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
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
xiao
#2
Oct15-07, 01:15 PM
P: 8
Quote Quote by xiao View Post
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 -11K = -[116.87g x 390 Jkg-1K-1 x -11K + 80.82g x 4200 Jkg-1K-1 x -11K]

Convert grams into kilo grams
0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11K = -[0.11687kg x 390 Jkg-1K-1 x -11K +0.08082kg x 4200 Jkg-1K-1 x -11K]


0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11K = 4235.2563 J
Heres a lil correction i made
learningphysics
#3
Oct15-07, 01:20 PM
HW Helper
P: 4,124
"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...

xiao
#4
Oct15-07, 01:22 PM
P: 8
Specific Latent Heat of Fusion

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

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
xiao
#5
Oct15-07, 01:26 PM
P: 8
Quote Quote by learningphysics View Post
0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (-581.196) = 4235.2563

Quote Quote by learningphysics View Post
subtract (0.01258)(4200)(-11) from both sides...
u mean subtract (-581.196) from 4235.2563
4235.2563 - (-581.196)
4816.4523
?????
learningphysics
#6
Oct15-07, 01:30 PM
HW Helper
P: 4,124
Quote Quote by xiao View Post
=======================

my attempt at the final answer

0.01258kg x L (?) + 0.01258kg x 4200 Jkg-1K-1 x -11K = 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
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.
xiao
#7
Oct15-07, 01:32 PM
P: 8
Quote Quote by learningphysics View Post
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.
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)
learningphysics
#8
Oct15-07, 01:32 PM
HW Helper
P: 4,124
Quote Quote by xiao View Post
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
?????
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.
xiao
#9
Oct15-07, 01:36 PM
P: 8
Quote Quote by learningphysics View Post
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.
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
learningphysics
#10
Oct15-07, 01:36 PM
HW Helper
P: 4,124
Quote Quote by xiao View Post
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)
I don't know. This webpage:

http://www.physchem.co.za/Heat/Latent.htm

gives the latent heat of ice as 334 kJ.kg-1 = 334,000J/kg, which is not that far off from what you have.
xiao
#11
Oct15-07, 01:37 PM
P: 8
Quote Quote by learningphysics View Post
I don't know. This webpage:

http://www.physchem.co.za/Heat/Latent.htm

gives the latent heat of ice as 334 kJ.kg-1 = 334,000J/kg, which is not that far off from what you have.
lol by bad wooooohooo i did it
learningphysics
#12
Oct15-07, 01:39 PM
HW Helper
P: 4,124
Quote Quote by xiao View Post
lol by bad wooooohooo i did it
:) good job. seems like it was a tough experiment! I was terrible at chemistry.
xiao
#13
Oct15-07, 01:41 PM
P: 8
Quote Quote by learningphysics View Post
:) good job. seems like it was a tough experiment! I was terrible at chemistry.
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
learningphysics
#14
Oct15-07, 02:15 PM
HW Helper
P: 4,124
Quote Quote by xiao View Post
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
You'll usually be able to get lots of help here. good luck!
learningphysics
#15
Oct15-07, 02:59 PM
HW Helper
P: 4,124
Mice L(?) + Mice CH2O ∆T = -[Mcal Ccal ∆T + MH2O CH2O ∆T]

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]
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.


Register to reply

Related Discussions
Specific Heat and Phase Change/Latent Heat Problems PLEASE HELP Introductory Physics Homework 5
Specific heat and latent heat of fusion Introductory Physics Homework 2
Specific heat capacity and latent heat of fusion Introductory Physics Homework 4
Specific & Latent Heat Introductory Physics Homework 1