Chemisty

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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?

  • Chemisty -

    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.

  • Chemisty -

    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?

  • Chemisty -

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

  • Chemisty -

    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?

  • Chemisty -

    okay is this correct

    I took 250.0 * 5.5 = 1375

    1375+2156=3531kcal

  • Chemisty -

    That looks ok to me.

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