Answer to Question #98049 in General Chemistry for fatemeh

Question #98049
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 68.6 g of this metal, initially at 20.0 ∘C?
1
Expert's answer
2019-11-05T07:39:54-0500

The heat gained is given by;

"\u0394H=m *s*(t2\u2212t1)"

Where ΔH=305J

Mass of metal, m=68.6g

Specific heat of metal, s=0.128Jg∘C

t1→initial temperature=20∘C

t2→final temperature=?

Therefore,"\u0394H=m*s*(t2\u2212t1)"

305J= 68.6g×0.128Jg∘C×(t2−20)

305=8.7808t2-175.616

t2=480.616/8.7808

t2=54.734876093

Hence t2=54.73∘C


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