a piece 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