A piece of Al(s) at 98 degree C is placed in 62g of H_2O at 25 degree C. The Alu

Question

A piece of Al(s) at 98 degree C is placed in 62g of H_2O at 25 degree C. The Aluminum cools and after equilibrium the final temperature of Aluminum and water is 32 degree C. What was the mass of the aluminum? C_s(H_2O)= 4.184J/(g degree C) C_s(Al)= 0.90 J/(g. degree C) Consider a glass of 239 mL of water at 26 degree C. Calculate the mass of ice at -15 degree C that must be added to cool the water to 10 degree C after thermal equilibrium is achieved. To find the mass of water use the density of water = 1.0 g/mL.

Explanation / Answer

The energy amount lost from aluminium is equal to the energy gained by water.

qlost = qgain

Substitute the values,

(X) (98 C - 32 C)(0.90 J /g C) = (62 g) (32 C - 25 C) (4.184 J/ g C)

X = 30.57 g

2. The energy amount lost from water is equal to the energy gained by ice.

qlost = qgain

Substitute the values,

(239 g) (26 C - 10 C) = (X) (10 C + 15 C)

X = 152.96 g

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