A 129 g piece of metal is heated to 292 ∘C and dropped into 90.0 g of water at 26.0 ∘C. If the final temperature of the water and metal is 58.0 ∘C, what is the specific heat of the metal?

Answer 1

Heat will be transferred from the hot metal to the cooler water and heat will thus be gained by the water. At the end, both the metal and the water will be at the same temperature (26.0º).

The heat (q) will be equal to the mass (m) x the specific heat (C) times the change in temperature (∆T)

q = mC∆T

For the metal: q = (129 g)(C J/g/deg)(292 deg - 58.0 deg) = 30,186 C

For the water: q = (90.0 g)(4.184 J/g/deg)(58.0 deg - 26.0 deg) = 12,050 J

30186 C = 12050

C = 12050/30186

C = 0.399 J/g/deg --> specific heat of the metal