Question

A 126 piece of metal is heated to 285 and dropped into 83.0 of water at 26. If the final temperature of the water and metal is 58.3, what is the specific heat of the metal ?

Answer 1

Q = c*m*(t₂-t₁)
Q(metal) = Q(H₂O)
Q(H₂O) = c(H₂O)*m(H₂O)*(t₂(H₂O) – t₁(H₂O)) = 4.186 * 83 * (58.3 – 26)= 11.2 kJ
Q(metal) = c(metal)*m(metal)*(t₁(metal) – t₂(metal)) = c(metal) * 126 * (285 – 58.3) = 28.56 kJ = 11.2 kJ
c(metal) = 0.392 J/(g*K)

Answer:
c(metal) = 0.392 J/(g*K)