# Specific Latent Heat of Fusion

by xiao
Tags: fusion, heat, latent, specific
 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
P: 8
 Quote by xiao 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 = 4235.2563 J
Heres a lil correction i made
 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...
 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
P: 8
 Quote by learningphysics 0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (-581.196) = 4235.2563

 Quote by learningphysics subtract (0.01258)(4200)(-11) from both sides...
u mean subtract (-581.196) from 4235.2563
4235.2563 - (-581.196)
4816.4523
?????
HW Helper
P: 4,124
 Quote by 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
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.
P: 8
 Quote by 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.
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)
HW Helper
P: 4,124
 Quote by 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 ?????
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.
P: 8
 Quote by 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.
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
HW Helper
P: 4,124
 Quote by 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)
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.
P: 8
 Quote by learningphysics 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
HW Helper
P: 4,124
 Quote by xiao lol by bad wooooohooo i did it
:) good job. seems like it was a tough experiment! I was terrible at chemistry.
P: 8
 Quote by learningphysics :) 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
HW Helper
P: 4,124
 Quote by 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
You'll usually be able to get lots of help here. good luck!
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.

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