what is the specific heat of this metal object in J/g °C?

Question

A metal object with a mass of 21 g is heated to 100 °C and then transferred to a calorimeter containing 57 g of water at 22 °C. If the water and metal object reach a final temperature of 27 °C, what is the specific heat of this metal object in J/g °C?

Answer 1

heat lost by the hot object must equal the heat gained by the cooler object.

heat lost by metal = heat gained by water

heat = q = mC∆T where m is mass; C is specific heat and ∆T is the change in temperature

For the metal we have...

q = (21 g)(C)(100º - 27º)

For the water we have...

q = (57 g)(4.184 J/gº)(27º - 22º)

Setting these two equal we have..

(21 g)(C)(100º - 27º) = (57 g)(4.184 J/gº)(27º - 22º)

(21 g)(73º)(C) = (238.5 J/º)(5º)

C = 0.778 J/gº = specific heat of the metal