Calculating the Heat Required to Melt Lead from 25 oC to 328 oC

In summary, you will need to increase the temperature to the melting point of lead and melt the lead to create the required energy.
  • #1
physicskillsme
24
0
Am I on the right track here?


How much energy is required to melt 400 grams of lead, if the initial temperature is 25 oC?

m=0.4kg deltaT 328-25= 303 SpecHeat = 128


Q=mCdeltaT

Q=.4*128*303

15513.6J
 
Physics news on Phys.org
  • #2
To melt that lead mass you'll need to do two things:
(1) Increase the temperature to the melting point of lead. (Hint: specific heat)
(2) Melt the lead. (Hint: latent heat of fusion)​
Find the energy for each step separately and add them.
 
  • #3
Doc Al said:
To melt that lead mass you'll need to do two things:
(1) Increase the temperature to the melting point of lead. (Hint: specific heat)
(2) Melt the lead. (Hint: latent heat of fusion)​
Find the energy for each step separately and add them.


OK so I have the step 1 part OK?

How do I set up part two? Latent heat (fusion) requires a separate equation presumably, I mean different to the Specific Heat in part 1?

So how does this look?

Heat Energy = m X L + m X c X q
 
Last edited:
  • #4
physicskillsme said:
OK so I have the step 1 part OK?
Looks OK.

How do I set up part two? Latent heat (fusion) requires a separate equation presumably, I mean different to the Specific Heat in part 1?
Heat required to melt a substance = mass X latent heat of fusion

So how does this look?

Heat Energy = m X L + m X c X q
I assume that is meant to represent the energy of both steps? If so, the first term is correct for step 2, but the second term has a q where it should have a delta t. I'll rewrite it:
[tex]Q = Q_1 + Q_2 = m c \Delta t + m L [/tex]
 
  • #5
m=0.4kg deltaT 328-25= 303 SpecHeat = 128

(1) Increase the temperature to the melting point of lead.


Q=mCdeltaT

Q=.4*128*303

Energy Required to heat: 15513.6J





(2) Melt the lead. (latent heat of fusion)

L=0.25x10^5 Taken for a list of properties where lead is listed as 0.25x10^5 for Melt ie:
Lt(J/kg)


Heat Energy = m * L = 10,000J

Add the energy for each step 10,000+15514=25514J

Answer: 25514J
 
  • #6
Looks reasonable to me. (Be sure to round off your answer to a sensible number of significant figures.)
 
  • #7
Doc Al said:
Looks reasonable to me. (Be sure to round off your answer to a sensible number of significant figures.)


OK thanks for your help.
 

What is melting?

Melting is the process by which a solid substance changes into a liquid when heated to its melting point.

What is the melting point?

The melting point is the temperature at which a solid substance changes into a liquid. It is a unique property of each substance and can be affected by factors such as pressure and impurities.

What is latent heat?

Latent heat is the amount of energy required for a substance to undergo a change in state without changing its temperature. In the case of melting, the latent heat of fusion is the energy needed to change a solid into a liquid.

Why does adding heat cause a substance to melt?

Adding heat increases the kinetic energy of the particles in a substance, causing them to vibrate and move more rapidly. When enough heat is added, the particles break free from their fixed positions in the solid and can move around freely, resulting in a liquid state.

Can a substance melt at a temperature below its melting point?

No, a substance will not melt at a temperature below its melting point. The melting point is the minimum temperature required for a substance to change from a solid to a liquid, so if the temperature is lower, the substance will remain a solid.

Similar threads

  • Introductory Physics Homework Help
Replies
3
Views
967
  • Introductory Physics Homework Help
Replies
13
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
Replies
3
Views
993
  • Introductory Physics Homework Help
Replies
2
Views
871
  • Introductory Physics Homework Help
Replies
11
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
Back
Top