The specific heat of a certain type of metal is 0.128 J/(g⋅∘C). What is the final temperature if 305 J of heat is added to 92.9 g of this metal, initially at 20.0 ∘C?
?final=_____C
Given:
Q = 305 J
m = 92.9 g
C = 0.128 J/g.oC
Ti = 20 oC
use:
Q = m*C*(Tf-Ti)
305.0 = 92.9*0.128*(Tf-20.0)
Tf -20.0 = 25.6 oC
Tf = 45.6 oC
Answer: 45.6 oC
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