Temperature & Heat: Equilibrium of 1 kg Aluminium & Water

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The discussion focuses on calculating the equilibrium temperature when a 0.1 kg piece of aluminum at 90 degrees Celsius is submerged in 1 kg of water at 10 degrees Celsius. The specific heat capacities are given as 920 J/(kg·°C) for aluminum and 4186 J/(kg·°C) for water. To find the equilibrium temperature, the heat lost by aluminum must equal the heat gained by water, leading to the equation mAl * cAl * ΔTAl = mwater * cwater * ΔTwater. This requires solving for two unknowns, ΔTAl and ΔTwater.

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superdave
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A .1 kg piece of aluminum at 90 degrees C is submerged in 1 KG of water at 10 degrees C. What is the temperature when they reach equalibrium?

Now I know, aluminum has a c of 920 J/(kg * degree C) and water has a c of 4186 J/(kg * degree C).

I'm not sure how to approach this.

Best I can think of is setting m*c*dT for the water equal to m*c*dT for the aluminum. But I have two unknown variables there. (The dT's) So that can't be right. Help is appreciated.
 
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What is the definition for your delta T? It is [tex]\Delta T = T_{end} - T_{start}[/tex]

Does that point you in the right direction? Also, think about the physical situation when the two have reached equillibrium...
 
superdave said:
A .1 kg piece of aluminum at 90 degrees C is submerged in 1 KG of water at 10 degrees C. What is the temperature when they reach equalibrium?

Now I know, aluminum has a c of 920 J/(kg * degree C) and water has a c of 4186 J/(kg * degree C).

I'm not sure how to approach this.

Best I can think of is setting m*c*dT for the water equal to m*c*dT for the aluminum. But I have two unknown variables there. (The dT's) So that can't be right. Help is appreciated.
This means you will need to work out two equations to determine the values of these two different delta T's.

AM
 

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