A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter containing water (the specific heat capacity of water is 4.18 J/g°C) initially at 21.1°C. The final temperature of water is 26.8°C. Calculate the mass of water in the calorimeter.

a. 1.7 g

b. 0.51 g

c. 150 g

d. 56 g

e. 61 g

m H2O = 56 g

Explanation:

Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT) Al = (mCΔT) H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g) * (mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol) (24.03 J/mol°C) (26.8 - 86.4) °C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / (C*ΔT) = 1327.64 J / (4.18 J/g.°C) (26.8 - 21.1) °C

⇒ mH2O = 55.722 g ≅ 56 g