Calculate Total Energy Required to Melt 100g of Lead

[atOnz]

The specific heat capacity of solid lead is 0.159J/g°C, the molar enthalpy of fusion is 5.0kJ/mol, and the melting point of lead is 328°C. Calculate the total energy required to change 100g of lead at 25°C to molten lead at it's melting point.

I don't even know where to start, I know q=mc(delta)t and q=n(delta)H but I don't know which to use or when.

 

[Bystander]

(snip)I know q=mc(delta)t and q=n(delta)H but I don't know which to use or when.

At least make a stab at it --- what's q? m? c? deltaT? n? deltaH? Look at the definitions of those quantities.

 

[atOnz]

Fair enough, the least I can do is try it. Basically I will find the heat required to bring the lead to it's melting point using q = mc(delta)t.
So, q = (100g)(0.159/J/g°C)(328°C-25°C)...
q = 4.8x10^3 J.

After finding that out I need to find the energy required to actually melt the lead and change it's form from solid to a liquid. I would use q = n(delta)H.

n = the amount of moles per the 100g given. So 100g / 207.2g/mol = 0.483mol.

q = (0.483mol)(5.0kJ/mol) = 2.4kJ.
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So I have some numbers, and I assume that I add the two together. If I do infact have to do this, which unit (kJ or J) should I express it in?

 

[Bystander]

(snip)So I have some numbers, and I assume that I add the two together. If I do infact have to do this, which unit (kJ or J) should I express it in?

See? Not so hard was it? Units? Pick one, or use both --- delH, or q, = x.yzx kJ, or xyzx J. If you've got an obsessive-compulsive, anal-retentive teacher, use whatever you've been told --- "both" usually keeps such people happy.

 

[atOnz]

Haha alright. I'd normally follow your "obsessive-compulsive, anal-retentive teacher" advice but this is the first set of work I've passed into him, and he has yet to say what he preferrs.

It's not that I can't do the work, it's that I hate not knowing if I'm right or wrong or even in the same ball park.

Thank you for your help!

-Jay