An aluminum cup contains 225 g of water and a 40-g copper stirrer, all at 27°C. A 470-g sample of silver at an initial temperature of 85°C is placed in the water. The stirrer is used to stir the mixture gently until it reaches its final equilibrium temperature of 32°C. Calculate the mass of the aluminum cup.

Mal = 0.232 kg = 232 g

Explanation:

mass of water (Mw) = 225 g = 0.225 kg

mass of copper stirrer (Mcu) = 40 g = 0.04 kg

initial temperature of water (Tw) = 27 degrees

mass of silver (Mag) = 470 g = 0.47 kg

initial temperature of silver (Ts) = 85 degrees

final temperature of the mixture (T) = 32 degrees

find the mass of the aluminum cup (Mal)

applying the conservation of energy

((Mal. cAl) + (Mw. cW) + (Mcu. cCu)) (ΔTw) = (Mag. cAg) (ΔTag)

we require the specific heat capacities of water (cW), aluminium (cAl), copper (cCu) and silver (cAg) which are as follows

water (cW) = 4186 J/kg. K

aluminium (cAl) = 900 J/kg. K

copper (cCu) = 387 J/kg. K

silver (cAg) = 234 J/kg. K

now we can put in our values to get the mass of the Aluminium cup (Mag)

((Mal. 900) + (0.225 x 4186) + (0.04 x 387)) (32-27) = (0.47 x 234) (85-32)

(900. Mal + 957.33) x 5 = 5828.94

900. Mal + 957.33 = 1165.79

900. Mal = 1165.79 - 957.33 = 208.5

Mal = 0.232 kg = 232 g