An 80.0-g piece of copper,initially at 285°C,is dropped into 250 g of water contained in a 300-g aluminum calorimeter;the water and calorimeter are initially at 12.0°C.What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 0 and 0.215 cal/g⋅°C,respectively.c_w = 1.00 cal/g⋅°C)

Question

Quizplus · Accepted Answer

First, we need to calculate the heat lost by the copper piece and the heat gained by the water and aluminum calorimeter.
qlost = heat lost by copper piece = mCΔT
qgained = heat gained by water and aluminum calorimeter = (m1C1 + m2C2)ΔT
where m1 and C1 are the mass and specific heat capacity of water, m2 and C2 are the mass and specific heat capacity of aluminum.

Using the above formula and taking the final temperature of the system to be T, we can set up an equation:
(80.0 g)(0.0920 cal/g°C)(285 - T) = (250 g)(1.00 cal/g°C)(T - 12.0°C) + (300 g)(0.215 cal/g°C)(T - 12.0°C)

Simplifying and solving for T, we get:
T = 27.8°C

Therefore, the final temperature of the system is 27.8°C. Answer choice B is the correct answer.

An 800-G Piece of Copper,initially at 285°C,is Dropped into 250 G