- #1
gibson101
- 47
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I solved the problem and have the correct answer just not sure about something.
Question:What will be the equilibrium temperature when a 274 g block of copper at 317°C is placed in a 137 g aluminum calorimeter cup containing 829 g of water at 13.0°C?
So heat lost by copper = heat gained by aluminum and water
So mcΔT= mcΔT + mcΔT
(274g)(.093 cal/g*C)(317-Tfinal)=(137g)(.22 cal/g*C)(Tfinal-12.6)+(829g)(1 cal/g*C)(Tfinal-12.6)
Solving this equation gives me 21.75 degrees C.
But why am i subtracting the final temperature from the initial temperature for copper? Whereas for the aluminum and water I am subtracting the initial from the final temp? I though ΔT was final temp - initial temp?
Question:What will be the equilibrium temperature when a 274 g block of copper at 317°C is placed in a 137 g aluminum calorimeter cup containing 829 g of water at 13.0°C?
So heat lost by copper = heat gained by aluminum and water
So mcΔT= mcΔT + mcΔT
(274g)(.093 cal/g*C)(317-Tfinal)=(137g)(.22 cal/g*C)(Tfinal-12.6)+(829g)(1 cal/g*C)(Tfinal-12.6)
Solving this equation gives me 21.75 degrees C.
But why am i subtracting the final temperature from the initial temperature for copper? Whereas for the aluminum and water I am subtracting the initial from the final temp? I though ΔT was final temp - initial temp?