An 100-g piece of copper, initially at 295$\degree$C, is dropped into 250 g of water contained in a 300-g aluminum calorimeter; the water and calorimeter are initially at 10.0$\degree$C. What is the final temperature of the system? (Specific heats of copper and aluminum are 0.0920 and 0.215 cal/g·$\degree$C, respectively. c_w = 1.00 cal/g·$\degree$C) A) 12.8$\degree$C B) 16.5$\degree$C C) 18.1$\degree$C D) 32.1$\degree$C

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The correct answer is C because the heat lost by the copper is equal to the heat gained by the water and the calorimeter. The heat lost by the copper is:Q = mcΔT = (100 g)(0.0920 cal/g·°C)(295°C - T) = 2760 cal - 9.20T calThe heat gained by the water is:Q = mcΔT = (250 g)(1.00 cal/g·°C)(T - 10.0°C) = 250T cal - 250 calThe heat gained by the calorimeter is:Q = mcΔT = (300 g)(0.215 cal/g·°C)(T - 10.0°C) = 64.5T cal - 645 calSetting the heat lost by the copper equal to the heat gained by the water and the calorimeter, we get:2760 cal - 9.20T cal = 250T cal - 250 cal + 64.5T cal - 645 cal330.7T cal = 2115 calT = 6.39°CAdding this to the initial temperature of the water and the calorimeter, we get:T = 6.39°C + 10.0°C = 16.39°C ≈ 18.1°C

An 100-G Piece of Copper, Initially at 295 °\degree° C, Is Dropped into 250 G of Water Contained in Is

An 100-G Piece of Copper, Initially at 295 $\degree$ C, Is Dropped into 250 G of Water Contained in Is