If 10.0g of Pb at 25.0oC absorbs 1.00kJ of heat, what is it's final temperature? SH Pb = 0.128 J/goC
Given:
Q = 1000 J
m = 10 g
C = 0.128 J/g.oC
Ti = 25 oC
use:
Q = m*C*(Tf-Ti)
1000.0 = 10.0*0.128*(Tf-25.0)
Tf -25.0 = 781.25 oC
Tf = 806 oC
Answer: 806 oC
If 10.0g of Pb at 25.0oC absorbs 1.00kJ of heat, what is it's final temperature? SH Pb = 0.128 J/goC
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