Posted by **yolanda** on Friday, April 8, 2011 at 12:16am.

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 -
**drwls**, Friday, April 8, 2011 at 1:49am
Q = M*[C*(delta T) + H]

detlta T = 308 C

H = 5.5 kcal/kg

C = 0.028 kcal.kg*C

M = 250 kg

Do the calculation for the heat (Q) in kcal

- Physical Science -
**Jai**, Friday, April 8, 2011 at 1:59am
to get the heat released or absorbed,

Q = mc(T2-T1)

where

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)

where

m = mass

Lf = Latent Heat of Fusion (fusion means melting)

substituting:

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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