A5.0 piece of metal is heated to 100 c, then placed in a beaker containing 20.0 g of water at 10 c. the temperature of the water rises to 15 c. assuming that the heat lost by the metal = heat gained by the water. calculate the specific heat of the metal.

This metal has a specific heat of 0.9845J/ g °C

Explanation:

Step 1: Given data

q = m*ΔT *Cp

⇒with m = mass of the substance

⇒with ΔT = change in temp = final temperature T2 - initial temperature T1

⇒with Cp = specific heat (Cpwater = 4.184J/g °C) (Cpmetam = TO BE DETERMINED)

Step 2:  Calculate specific heat

For this situation : we get for q = m*ΔT *Cp

q(lost, metal) = q(gained, water)

- mass of metal(ΔT)(Cpmetal) = mass of water (ΔT) (Cpwater)

-5 * (15-100)(Cpmetal) = 20* (15-10) * (4.184J/g °C =

-5 * (-85)(Cpmetal) = 418.4

Cpmetal = 418.4 / (-5*-85) = 0.9845 J/g °C

This metal has a specific heat of 0.9845J/ g °C