Specific Latent Heat of Fusion

It's always tough when you have to take courses that are outside your major. But it's good that you're willing to put in the effort and ask for help when you need it. That's really important for any subject that you find challenging. Keep it up!Yeah, my major is computer science, but I'm taking physics, German, and math courses this semester. It's tough to balance everything, but it's important to get through it. In summary, the conversation was about a student trying to find the Specific Latent Heat of Fusion for an experiment. They provided the results for the experiment and equations they were using to find the answer. The student was having trouble rearranging the equations to find the final answer and
  • #1
xiao
8
0

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 [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
 
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  • #2
xiao said:
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
 
  • #3
"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
=======================

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 don't know wat i am doing
 
  • #5
learningphysics said:
0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (0.01258)(4200)(-11) = 4235.2563
0.01258L + (-581.196) = 4235.2563

learningphysics said:
subtract (0.01258)(4200)(-11) from both sides...
u mean subtract (-581.196) from 4235.2563
4235.2563 - (-581.196)
4816.4523
?
 
  • #6
xiao said:
=======================

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 don't 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.
 
  • #7
learningphysics said:
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)
 
  • #8
xiao said:
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.
 
  • #9
learningphysics said:
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 thanks a lot 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
xiao said:
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 [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.
 
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  • #11
learningphysics said:
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.

lol by bad wooooohooo i did it
 
Last edited by a moderator:
  • #12
xiao said:
lol by bad wooooohooo i did it

:) good job. seems like it was a tough experiment! I was terrible at chemistry.
 
  • #13
learningphysics said:
:) good job. seems like it was a tough experiment! I was terrible at chemistry.

the experiment was easy just melt ice in water in room temp until 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
xiao said:
the experiment was easy just melt ice in water in room temp until 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!
 
  • #15
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.
 

What is specific latent heat of fusion?

The specific latent heat of fusion is the amount of thermal energy required to change a substance from solid to liquid state without changing its temperature. It is a measure of the strength of the intermolecular forces holding the particles in a solid together.

How is specific latent heat of fusion different from specific heat capacity?

Specific latent heat of fusion is the amount of energy required to change the state of a substance, while specific heat capacity is the amount of energy required to change the temperature of a substance. Specific latent heat of fusion involves a change in state, while specific heat capacity involves a change in temperature.

What factors affect the value of specific latent heat of fusion?

The value of specific latent heat of fusion is affected by the type of substance, pressure, and temperature. Different substances have different intermolecular forces and therefore different values of specific latent heat of fusion. Changes in pressure and temperature can also affect the strength of intermolecular forces and thus the value of specific latent heat of fusion.

How do you calculate specific latent heat of fusion?

To calculate specific latent heat of fusion, you need to know the amount of heat energy required to change the state of a substance and the mass of the substance. The specific latent heat of fusion can then be calculated by dividing the amount of heat energy by the mass of the substance.

What are some real-life applications of specific latent heat of fusion?

Specific latent heat of fusion plays a crucial role in many real-life applications, such as refrigeration, cooking, and transportation of perishable goods. It is also used in industries that require precise temperature control, such as pharmaceuticals and chemical manufacturing. It is also important in understanding the Earth's climate and the melting of ice in polar regions.

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