- #1

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i used the equation [tex]Q = mc \Delta T[/tex]

-(199g)(4.186J/g°C)(T - 80.9°C) + (178g)(0.13J/g°C)(T - 24.5°C) = 0

T = T final = 82.5°C

can someone please tell mw what i did wrong?

can someone also confirm if the specific heat for lead is 0.13J/g°C?