Question

An aluminum cup contains 225 g of water and a 40 g copper stirrer, all at 27°C. A 410 g sample of silver at an initial temperature of 88°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.

Answer 1

Answer:

130.22 g

Explanation:

Parameters given:

Mass of water Mw = 225 g

Mass of stirrer Ms = 40 g

Mass of silver M(S) = 410 g

By applying the law of conservation of energy:

(McCc + MsCs + MwCw)ΔTw = M(S)C(S)ΔT(S)

where Mc = Mass of cup

Cc = Specific heat capacity of aluminium cup = 900 J/gC

Cs = Specific heat capacity of copper stirrer = 387 J/gC

Cw = Specific heat capacity of water = 4186 J/gC

ΔTw = change in temperature of water = 32 - 27 = 5 °C

C(S) = Specific heat capacity of silver = 234 J/gC

ΔT(S) = change in temperature of silver = 88 - 32 = 56 °C

Therefore:

[(Mc * 900) + (40 * 387) + (225 * 4186)] * 5 = 410 * 234 * 56

(900Mc + 957330) * 5 = 5276700

900Mc + 957330 = 5276700 / 5 = 1074528

900Mc = 1074528 - 957330

900Mc = 117198

Mc = 117198/ 900

Mc = 130.22 g

The mass of the cup is 130.22 g.