5. A 28.4 g sample of aluminum is heated to 39.4 °C, and placed in a calorimeter containing 50.0 g of water. The

temperature of water increases from 21.00 °C to 23.00 °C. What is the specific heat capacity, C, of aluminum?

C Al = 0.8975 J/g. K

Explanation:

Q = mCΔT

∴ m Al = 28.4 g

∴ T Al = 39.4°C = 312.4 K

∴ m H2O = 50.0 g

∴ T1 H2O = 21°C = 294 K

∴ T2 H2O = 23°C = 296 K

∴ C H2O = 4,18 J/g. K

⇒ C Al = ?

in a calorimeter:

∴ Al give heat: Q Al < 0

∴ H2O revceives heat: Q H2O > 0

⇒ - Q Al = Q H2O

⇒ - (28.4 g) * (C Al) * (296 K - 312.4 K) = (50.0 g) * (4.18 J/g. K) * (296 K - 294 K)

⇒ - ( - 465.76 g. K) * (C Al) = 418 J

⇒ C Al = (418 J) / (465.76 g. K)

⇒ C Al = 0.8975 J/g. K = 897.5 J/Kg. K