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Physical Science

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lead has specific heat of .028kcal/kgC, melting point of 328 degrees C and a heat fusion of 5.5 kcal/kg. how much heat must be provided to melt a 250kg sample of lead with a temperature of 20 degrees Celsius?

  • Physical Science -

    Q = M*[C*(delta T) + H]

    detlta T = 308 C
    H = 5.5 kcal/kg
    C = 0.028*C
    M = 250 kg

    Do the calculation for the heat (Q) in kcal

  • Physical Science -

    to get the heat released or absorbed,
    Q = mc(T2-T1)
    m = mass of substance
    c = specific heat capacity
    T2 = final temperature
    T1 = initial temperature
    **note: if Q is (-), heat is released and if (+), heat is absorbed

    now we can only apply this to substances that did not change its phase, but in the problem, we see that phase change occurs. from solid lead->liquid lead.
    thus we need another data called Latent Heat of Fusion to calculate for the heat required to change its phase:
    H = m(Lf)

    m = mass
    Lf = Latent Heat of Fusion (fusion means melting)

    Q = mc(T2-T1)
    Q = 250*(0.028)*(328-20)
    Q = 2156 cal
    *note that is is only for the temp change,, it's phase is still solid.
    now to change its phase,
    H = m*Lf
    H = 250*5.5
    H = 1375 cal

    thus the total heat (Q,total) needed is
    Q,total = Q + H = 2156 + 1375
    Q,total = 3531 calories

    hope this helps~ :)

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