If the final temperature of the water and metal is 58.4 ∘C , what is the specific heat of the metal? Express your answer in J/(g⋅∘C) using three significant figures.

Question

A 127 g piece of metal is heated to 286  ∘C and dropped into 90.0 g of water at 26.0 ∘C.

Answer 1

heat lost by hot metal MUST equal heat gained by cooler water

heat lost by hot metal = q = mC∆T = (127 g)(C)(286º - 58.4º) where Tf is final temperature

heat gained by water = q = mC∆T = (90.0 g)(4.184 J/gº)(58.4º- 26º) C = specific heat = 4.184 J/gº

Setting these two equal we have...

(127 g)(C)(286º - 58.4º) = (90.0 g)(4.184 J/gº)(58.4º- 26º)

36,322C - 7417C = 12,201

24905 C = 12,201

C = 0.490 J/gº

EDIT: the calculations should look like this, after correcting the math error:

28,905C = 12, 201 and C = 0.422 J/gº.