Question

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?

Answer 1

The heat gained is given by;

$ΔH=m *s*(t2−t1)$

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,$ΔH=m*s*(t2−t1)$

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