A 23.5g aluminum block is warmed to 65.9°C and plunged into an insulated beaker containing 55.0g water initially at 22.3°C. The aluminum and the water are allowed to come to thermal equilibrium.

Assuming that no heat is lost, what is the final temperature of the water and aluminum?

25.97oC

Explanation:

Heat lost by aluminum = heat gained by water

M (Al) x C (Al) x [ Temp (Al) - Temp (Al+H2O) ] = M (H2O) x C (H2O) x [ Temp (Al+H2O) - Temp (H2O) ]

Where M (Al) = 23.5g, C (Al) = specific heat capacity of aluminum = 0.900J/goC, Temp (Al) = 65.9oC, Temp (Al+H2O) = temperature of water and aluminum at equilibrium = ?, M (H2O) = 55.0g, C (H2O) = specific heat capacity of liquid water = 4.186J/goC

Let Temp (Al+H2O) = X

23.5 x 0.900 x (65.9-X) = 55.0 x 4.186 x (X-22.3)

21.15 (65.9-X) = 230.23 (X-22.3)

1393.785 - 21.15X = 230.23X - 5134.129

230.23X + 21.15X = 1393.785 + 5134.129

251.38X = 6527.909

X = 6527.909/251.38

X = 25.97oC

So, the final temperature of the water and aluminum is = 25.97oC