Calculate the equilibrium temperature of the water.

Question

A 46.0g sample of aluminum, which has a specific heat capacity of 0.897·J·g−1°C−1, is dropped into an insulated container containing 100.0g of water at 60.0°C and a constant pressure of 1atm. 

The initial temperature of the aluminum is 6.6°C.

Assuming no heat is absorbed from or by the container, or the surroundings, calculate the equilibrium temperature of the water.

Be sure your answer has the correct number of significant digits.

I tried to calculate the equilibrium alone, but it doesn't seem to be correct, I think I might be messing up with sig figs. If I could see the process done to answer this to be better in the future, that would be great!

Answer 1

They are essentially asking for Tf (final temperature). Recall that heat gained by Al must equal heat lost by water.

Heat gained by Al is q = mC∆T = (46.0 g)(0.897 J/gº)(Tf - 6.6)

Heat lost by H2O is q = mC∆T = (100.0 g)(4.184 J/gº)(60.0 - Tf)

Setting these 2 equations equal and solving for Tf, we have...

(46.0 g)(0.897 J/gº)(Tf - 6.6) = (100.0 g)(4.184 J/gº)(60.0 - Tf)

41.26Tf - 272.3 = 25,104 - 418.4Tf

459.66Tf = 25,376

Tf = equilibrium temperature = 55.2ºC (3 sig. figs.)

Or actually, since temperature of Al has only 2 sig. figs. you might report it as 55ºC