Chemisty
posted by Jessica on .
LEAD IS A SOFT, DENSE METAL WITH A SPECIFIC HEAT OF 0.028KCAL/KGC, A MELTING POINT OF 328.0C, AND A HEAT FUSION OF5.5KCAL/KG. HOW MUCH HEAT MUST BE PROVIDED TO MELT A 250.0KG SAMPLE OF LEAD WITH A TEMPERATURE OF 20.0C?

q1 = heat to move temperature from 20 C to 328.
q1 = mass Pb x specific heat lead x delta T.
q2 = heat to melt Pb at 328.
q2 = mass Pb x heat fusion.
Total heat = q1 + q2.
Watch the units. Post your work if you get stuck. 
Tell me if this is right
For the first step this is what I am doing
m=250.0kg
c=0.028kcal
deltaT 32820?? Is this correct
this gives me 2156
Am I on the right track for the first step? 
Yes, but your number doesn't have units. Most profs will count off for that.

I know.. I am just making sure I am on the right track.. I am going to try and finish it now.. do you mind if I send you my answer and you tell me how I did?

okay is this correct
I took 250.0 * 5.5 = 1375
1375+2156=3531kcal 
That looks ok to me.