A 127 g piece of metal is heated to 282 ∘C and dropped into 90.0 g of water at 26.0 ∘C. If the final temperature of the water and metal is 57.5 °C, what is the specific heat of the metal?

Question

express the answer in J/(g•°C) using three significant figures

Answer 1

heat lost by hot metal = heat gained by cooler water

heat lost by metal = q = (m)(C)(∆T)

m = mass = 127 g

C = specific heat of metal = ?

∆T = change in temperature = 282º - 57.5º = 224.5º (this would be negative since heat is lost)

(127 g)(C)(224.5º) = heat lost by metal

heat gained by water = q = (m)(C)(∆T)

m = mass = 90.0 g

C = specific heat = 4.184 J/g/º

∆T = change in temperature = 57.5º - 26.0º = 31.5º

(90.0 g)(4.184 J/gº)(31.5º) = heat gained by water

Setting these two equation equal we have...

(127 g)(C)(224.5º) = (90.0 g)(4.184 J/gº)(31.5º)

28512 C = 11862

C = 0.416 J/gº = specific heat of the metal