Question

To a sample of water at 23.4oC in a constant pressure calorimeter of negligible heat capacity is added a 12.1 g piece of aluminium whose temperature is 81.7oC. If the final temperature of water is 24.9 oC, calculate the mass of the water in the calorimeter. Ans:98.6g

-I know that The specific heat of aluminum is 0.900 J/g · °C
- ΔT Al is 24.9°C − 81.7°C = −56.8°C
- ΔTwater and ΔTcalorimeter are both 24.9°C − 23.4°C = 1.5°C.
-The specific heat of water is 4.184 J/g · °C.
But I tried using m=q/sΔt. I'm really stuck,can anyone help me?

Answer 1

Ok I can help you out. You were on the right track .. use \[-q _{metal} = q _{H2O}\] to find you're q of the water. So q of the metal = 12.1g(mass) x -56.8 degree celsius(delta T) x .900 J/g degree celsius (Cs of the metal) ... so now you have -618.552 which gives you +618.552 as you q for water.
Now plug that into q = Cs x M x Delta T so you have 618.552 = 4.184 x M x 1.5
Then divide 618.552/(4.184 x 1.5) = 98.55 g :)

Answer 2

Ohhh!!! I see it now! Thank you so much!! :D

Answer 3

No problem at all :)