A30.0 g sample of a metal was heated in a hot water bath to 80°c. it was then quickly transferred to a coffee-cup calorimeter. the calorimeter contained 100.0 g of water at a temperature of 20°c. the final temperature of the contents of the calorimeter was 25°c. what is the specific heat capacity of the metal?

when the metal  is lost heat and the calorimeter of water is gained the heat 

and when the heat lost = the heat gained so,

(M*C*ΔT)m =  (M*C*ΔT)w

when Mm= mass of the metal = 30 g 

Δ Tm = (80-25) = 55 °C

and Mw = mass of water = 100 g  

Cw is the specific heat of water = 4.181 J/g.°C

ΔTw = (25-20) = 5 °C

so by substitution:

∴ 30* Cm*55 = 100 * 4.181 * 5 

∴Cm (specific heat of metal) = (100*4.181*5)/(30*55) 

∴C of metal = 1.267 J/g.°C

Idon't know hope someone you