Apiece of metal is heated to 100℃ and placed into 75.0g of water in a coffee-cup calorimeter. initially, the temperature of the water in the calorimeter was 23.1℃. the final temperature of he system was 30.6℃. what is the mass of the piece of metal? the specific heat of the water is 4.184 j/g℃. the specific heat of the metal is 0.900j/g℃.

The mass of this piece of metal is 37.68g

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 = 0.9J/ g °C)

Step2 :

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)

-m (30.6 -100) * 0.9 = 75g *( 30.6 - 23.1) * 4.184J / g °C

-m * (-69.4) * 0.9 = 75g * (7.5) * 4.184J / g °C

-m * (-69.4) *0.9 = 2353.5

-m = 2353.5 / (-69.4 *0.9) = -37.68

m = 37.68g

The mass of this piece of metal is 37.68g

the answer is a! (2.9 x 10^-18 j) hope i could someone. : )