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Specific Latent Heat of Fusion 
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#1
Oct1507, 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 Jkg1K1 C H2O = 4200 Jkg1K1 LH2O = 33x104 Jkg1 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 Jkg1K1 x 11°K = [116.87g x 390 Jkg1K1 x 11°K + 80.82g x 4200 Jkg1K1 x 11°K] Convert grams into kilo grams 0.01258kg x L (?) + 0.01258kg x 4200 Jkg1K1 x 11°K = 0.11687kg x 390 Jkg1K1 x 11°K +0.08082kg x 4200 Jkg1K1 x 11°K 0.01258kg x L (?) + 0.01258kg x 4200 Jkg1K1 x 11°K = 501.3723 + 3733.884 0.01258kg x L (?) + 0.01258kg x 4200 Jkg1K1 x 11°K = 4235.2563 J how do i switch over the 0.01258kg + 0.01258kg x 4200 Jkg1K1 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 


#2
Oct1507, 01:15 PM

P: 8




#3
Oct1507, 01:20 PM

HW Helper
P: 4,124

"0.01258kg x L (?) + 0.01258kg x 4200 Jkg1K1 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
Oct1507, 01:22 PM

P: 8

Specific Latent Heat of Fusion
=======================
my attempt at the final answer 0.01258kg x L (?) + 0.01258kg x 4200 Jkg1K1 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
Oct1507, 01:26 PM

P: 8

0.01258L + (581.196) = 4235.2563 4235.2563  (581.196) 4816.4523 ????? 


#6
Oct1507, 01:30 PM

HW Helper
P: 4,124

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
Oct1507, 01:32 PM

P: 8

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
Oct1507, 01:32 PM

HW Helper
P: 4,124

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
Oct1507, 01:36 PM

P: 8

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
Oct1507, 01:36 PM

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P: 4,124

http://www.physchem.co.za/Heat/Latent.htm gives the latent heat of ice as 334 kJ.kg1 = 334,000J/kg, which is not that far off from what you have. 


#11
Oct1507, 01:37 PM

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#12
Oct1507, 01:39 PM

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#13
Oct1507, 01:41 PM

P: 8

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
Oct1507, 02:15 PM

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#15
Oct1507, 02:59 PM

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P: 4,124

12.58g x L (?) +12.58g x 4200 Jkg1K1 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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