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?

Answer 1

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.

Answer 2

Tell me if this is right

For the first step this is what I am doing

m=250.0kg
c=0.028kcal
deltaT- 328-20?? Is this correct

this gives me 2156

Am I on the right track for the first step?

Answer 3

Yes, but your number doesn't have units. Most profs will count off for that.

Answer 4

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?

Answer 5

okay is this correct

I took 250.0 * 5.5 = 1375

1375+2156=3531kcal

Answer 6

That looks ok to me.