How Do You Calculate the Thermal Energy to Melt a Copper Bar?

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The what?In summary, specific heat capacity is the amount of heat required to raise the temperature of a substance by 1 degree Kelvin per unit mass. Latent heat of fusion is the amount of heat required to change a substance from solid to liquid state. To calculate the thermal energy required to convert a 40 kg copper bar from a temperature of 280 K to molten metal at 1360 K, we need to consider both specific heat capacity and latent heat of fusion. The total thermal energy required is the sum of the energy needed to raise the temperature to 1360 K and the energy needed for the phase change.
  • #1
chameleon1
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Homework Statement



Define specific heat capacity and latent heat of fusion.
Copper melts at a temperature of 1356 K with a latent heat of fusion
of 210,000 J/kg (Joules per kilogram). Calculate the thermal energy required
to convert a copper bar with a mass of 40 kg at a temperature of 280 K into
molten metal at 1360 K if the average specific heat capacity of copper (liquid
and solid) is 385 J/kg/K (Joules per kilogram per Kelvin) in the temperature
range considered here

Homework Equations



energy SCH(specific heat capacity) * mass * change in temp

The Attempt at a Solution




i tried to use this formula but the numbers am gettin isn't write , if possible can someone guide me on how to do this or what am i doing wrong, am good with biology but am am also learning physics which is very difficult for me .
 
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  • #2
chameleon1 said:

Homework Statement



Define specific heat capacity and latent heat of fusion.
Copper melts at a temperature of 1356 K with a latent heat of fusion
of 210,000 J/kg (Joules per kilogram). Calculate the thermal energy required
to convert a copper bar with a mass of 40 kg at a temperature of 280 K into
molten metal at 1360 K if the average specific heat capacity of copper (liquid
and solid) is 385 J/kg/K (Joules per kilogram per Kelvin) in the temperature
range considered here

Homework Equations



energy SCH(specific heat capacity) * mass * change in temp

The Attempt at a Solution



i tried to use this formula but the numbers am gettin isn't write , if possible can someone guide me on how to do this or what am i doing wrong, am good with biology but am am also learning physics which is very difficult for me .

Welcome to PF.

Maybe if you show your calculation it will be easier to figure what may be going awry?
 
  • #3
Thanks for the welcome .

SHC heat change is 1360 k - 280 k = 1080K

1080 k * 40kg (mass) = 43200


43200 * 385 j/kg/k = 16632000

to calculate quantity of thermal energy, then multiply mass, the specific heat capacity (for liquid water this is 4.19J/g degrees Celsius), and the change in temperature. and this is what i used but the number is tooo high and loooks wrong , i don't know what i am doing wrong
 
  • #4
LowlyPion said:
Welcome to PF.

Maybe if you show your calculation it will be easier to figure what may be going awry?

Thanks for the welcome .

SHC heat change is 1360 k - 280 k = 1080K

1080 k * 40kg (mass) = 43200


43200 * 385 j/kg/k = 16632000

to calculate quantity of thermal energy, then multiply mass, the specific heat capacity (for liquid water this is 4.19J/g degrees Celsius), and the change in temperature. and this is what i used but the number is tooo high and loooks wrong , i don't know what i am doing wrong
 
  • #5
What about the 210,000 j/kg to melt it to liquid?
 
  • #6
LowlyPion said:
What about the 210,000 j/kg to melt it to liquid?

do i need to divide 210,000 by 40kg mass = 5250 OR??
 
  • #7
chameleon1 said:
do i need to divide 210,000 by 40kg mass = 5250 OR??

Didn't it say it was J/kg?

Is divide the right thing to do?
 
  • #8
LowlyPion said:
Didn't it say it was J/kg?

Is divide the right thing to do?

not sure, is there a formula which you can telll me , sorry i don't want to sound stupid, but maths and physics are not my thing but i have tooo to this its one of my modules please help me further and next time i would be able to this myself
 
  • #9
You figured the energy needed to raise the temp of the 40 kg.

But now you still need to consider the heat of fusion - the energy that will make it liquefy. If you need 210,000J to liquefy 1 kg of cooper already at 1360 K, then how much more to liquefy 40 kg?

Add the 2 together and you have your total thermal energy.
 
  • #10
LowlyPion said:
You figured the energy needed to raise the temp of the 40 kg.

But now you still need to consider the heat of fusion - the energy that will make it liquefy. If you need 210,000J to liquefy 1 kg of cooper already at 1360 K, then how much more to liquefy 40 kg?

Add the 2 together and you have your total thermal energy.


so 210,000 j/kg * 40kg = 8400000

but what do i need to add it too in order to get the thremal energy
 
  • #11
chameleon1 said:
so 210,000 j/kg * 40kg = 8400000

but what do i need to add it too in order to get the thremal energy

What about the energy to get if from 280K to 1360 K?
 
  • #12
LowlyPion said:
What about the energy to get if from 280K to 1360 K?

so 1350k - 280k = 1070k

so 1070k needed to get it to 1360k

then add 1360k to 8400000 but most likely not OR?
 
  • #14
LowlyPion said:
Have you just completely forgotten about what you calculated here already?
https://www.physicsforums.com/showpost.php?p=2129871&postcount=3

am lost with all this numbers now , tell me how to do this as i have tried already, once you tell me it will stick into my head , what's the answer to this ridddle :)
 
  • #15
chameleon1 said:
am lost with all this numbers now , tell me how to do this as i have tried already, once you tell me it will stick into my head , what's the answer to this ridddle :)

You figured out in post 3 how much energy it took to raise the copper to the melting point of 1360K.

Then you figured how much to melt the block, once it got to 1360K

Total energy required then is the sum of those 2 energy requirements.
 
  • #16
LowlyPion said:
You figured out in post 3 how much energy it took to raise the copper to the melting point of 1360K.

Then you figured how much to melt the block, once it got to 1360K

Total energy required then is the sum of those 2 energy requirements.

and what was The i had figured how much to melt the block, once it got to 1360K
 
  • #17
chameleon1 said:
and what was The i had figured how much to melt the block, once it got to 1360K

Are you asking about what you found in post 10?

The results of post 3 and post 10 added together should be your answer.
 
  • #18
LowlyPion said:
Are you asking about what you found in post 10?

The results of post 3 and post 10 added together should be your answer.

16632000 + 8400000 = 33432000 j/kg

Thats the answer and that's the right sysmbol i put at the end?
 
  • #19
chameleon1 said:
16632000 + 8400000 = 33432000 j/kg

Thats the answer and that's the right sysmbol i put at the end?

No. They asked for energy.

That would just be in Joules.
 
  • #20
LowlyPion said:
No. They asked for energy.

That would just be in Joules.


16632000 + 8400000 = 33432000Q

Q = m x T x Cp

Q = change in thermal energy
m= mass of substance
T = change in temperature (Tf – Ti)
Cp = specific heat of substance

i got this from a website

so is it 33432000 Q OR what's the symbol for the thermal energy
 
  • #21
chameleon1 said:
what's the symbol for the thermal energy

J for joules.

Btw:
16632000 + 8400000 = 33432000Q

Check your math again.
 
  • #22
LowlyPion said:
J for joules.

Btw:


Check your math again.

1080 k * 40kg (mass) = 43200 from post 3

so 210,000 j/kg * 40kg = 8400000 from post 10

= 43200 + 8400000 = 8443200 is that right if it is what is the symbol i should put at the end
 
  • #23
chameleon1 said:
1080 k * 40kg (mass) = 43200 from post 3

Not quite.

43200 is merely kg*K. You've forgotten the 385 J/(K*kg)

You found the right answer to that part of it - raising the temperature of 40 kg to the melting point in 3. Why resistant to accepting it now?
 
  • #24
This problem is fairly straightforward if you pay attention to the units.

They give you the specific heat needed to raise the temp of 1 kg of copper by 1 degree K. That's 385 J/(K*kg)

Well you have 40 kg and you have 1080 degrees to raise it.

That's simply 385*40*1080. That gets the number of J's needed to get it to the melting point.

Even though it is at the melting point it is still solid. That's where the heat of fusion comes in. 210,000 J/kg is needed to get it to melt. And of course that is in addition to what you already found. That is given as J/kg so to figure J then you need to multiply 210,000J/kg*40kg to yield the Joules required.
 
  • #25
LowlyPion said:
This problem is fairly straightforward if you pay attention to the units.

They give you the specific heat needed to raise the temp of 1 kg of copper by 1 degree K. That's 385 J/(K*kg)

Well you have 40 kg and you have 1080 degrees to raise it.

That's simply 385*40*1080. That gets the number of J's needed to get it to the melting point.

Even though it is at the melting point it is still solid. That's where the heat of fusion comes in. 210,000 J/kg is needed to get it to melt. And of course that is in addition to what you already found. That is given as J/kg so to figure J then you need to multiply 210,000J/kg*40kg to yield the Joules required.

385*40*1080 = 16632000

210,000J/kg*40kg = 8400000


16632000 + 8400000 = 25032000j/kg this is thermal energy needed

thats right now
 
  • #26
chameleon1 said:
385*40*1080 = 16632000

210,000J/kg*40kg = 8400000

16632000 + 8400000 = 25032000j/kg this is thermal energy needed

thats right now

At last. The Emerald City.
 
  • #27
hahaahaha 26 posts to get it right yesssssssssss
 

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