The heat gained by the water in reaching equilibrium equals the heat lost by the metal. Assume that you know the specific heat of water is Cwater = 1.00 cal/g*C.
Cmetal*6.128*(100-28.4) = Cwater*25*(28.4-23.6)
Cmetal = 1.00*25*4.8)/[6.128*71.6]
= 0.274 cal/(g*C)
Need help with my lab report for specific heat of unknown metal!
I need to Find the Specific Heat of Al! I know its Q= M x Cs x Change in Temp
I am getting a such a high number am just confused if its right.
Here is what I got in Lab:
Aluminum
Mass of Metal (Mm): 6.128g
Mass of water (Mw) : 25 mL
Temp of water in calorimeter: 23.6 C
Temp of metal in water-bath: 100 C
T Final: 28.4 C
PLS helppp.... i am stuck on this!
1 answer
1 answer