# Melting and Latent Heat

1. Mar 30, 2006

### physicskillsme

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

2. Mar 30, 2006

### Staff: Mentor

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. Mar 30, 2006

### physicskillsme

OK so I have the step 1 part OK????

How do I set up part two? Latent heat (fusion) requires a seperate 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: Mar 30, 2006
4. Mar 30, 2006

### Staff: Mentor

Looks OK.

Heat required to melt a substance = mass X latent heat of fusion

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:
$$Q = Q_1 + Q_2 = m c \Delta t + m L$$

5. Mar 30, 2006

### physicskillsme

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

6. Mar 30, 2006

### Staff: Mentor

Looks reasonable to me. (Be sure to round off your answer to a sensible number of significant figures.)

7. Mar 30, 2006