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March 31, 2015

March 31, 2015

Posted by **J.M.** on Saturday, September 23, 2006 at 3:02pm.

I will be happy to critique your work or thinking. See the other posts in this series for hints.

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 ammount 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?

IT is correct. It does not matter which unit 2.4kJ is 2400 joules. The common practice in scientific notation would be to write it as 2.4kJ

Thanks for the help bobpursley :)

ok i just wanted to know the name of the course/class using graphics for the change over solids to gases using the (m.c.delta T)formula and what grade it's taught :P thanks....

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