# Determine the final temperature when 10.0 g of aluminum at 130.0 celcius mixes with 200.0 g of water at 25.0 Celsius.

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In order to answer this question we first need to look up the specific heat of aluminum and liquid water:

`c_(pw)=4.1813 J/(gK)`

`c_(pa)=0.897 J/(gK)`

Next we can use the knowledge that the heat lost by aluminium will equal the heat gained by water, and that the final temperature will be the same for both the water and the aluminium once equilibrium has been achieved. The equation for change in heat is:

`DeltaQ=mc_pDeltaT`

Therefore:

`(200)(4.1813)(T_f-25)=(10)(0.897)(130-T_f)`

`836.26T_f-20906.5=1166.1-8.97T_f`

`836.26T_f+8.97T_f=1166.1+20906.5`

`845.23T_f=22072.6`

`T_f=26.1`

Therefore, the final temperature of the solution is 26.1 degrees Celsius

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