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

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To calculate the energy required to melt 400 grams of lead from 25 oC to 328 oC, two steps are involved: heating the lead to its melting point and then melting it. The first step uses the specific heat formula, yielding 15,513.6 J to raise the temperature. The second step involves the latent heat of fusion, calculated as 10,000 J. Adding both energy amounts results in a total of 25,514 J, which appears reasonable when considering significant figures. The calculations confirm the correct approach to determining the heat required for melting lead.
physicskillsme
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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
 
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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.
 
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:
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:
Q = Q_1 + Q_2 = m c \Delta t + m L
 
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
 
Looks reasonable to me. (Be sure to round off your answer to a sensible number of significant figures.)
 
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.
 
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