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!
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)
Sure, I can help you with calculating the specific heat of aluminum.
First, let's gather the necessary information:
Mass of the aluminum (Mm) = 6.128g
Mass of the water (Mw) = 25 mL
Temperature of the water in the calorimeter (Tiw) = 23.6 °C
Temperature of the metal in the water bath (Tim) = 100 °C
Final temperature (Tf) = 28.4 °C
Now, let's use the formula you mentioned: Q = M x Cs x ΔT
Where:
Q = heat absorbed or released (in Joules)
M = mass of the substance (in grams)
Cs = specific heat of the substance (in J/g°C)
ΔT = change in temperature (in °C)
To find the heat transferred (Q), we'll use the following equation:
Q = (Mw x Cs x ΔTw) + (Mm x Cs x ΔTm)
Where:
ΔTw = Tf - Tiw
ΔTm = Tf - Tim
Now, let's calculate ΔTw and ΔTm:
ΔTw = Tf - Tiw = 28.4 °C - 23.6 °C = 4.8 °C
ΔTm = Tf - Tim = 28.4 °C - 100 °C = -71.6 °C
Since ΔTm is negative, it means that heat is being released by the aluminum.
Now, substitute the values into the heat equation:
Q = (Mw x Cs x ΔTw) + (Mm x Cs x ΔTm)
You want to find the specific heat of aluminum (Cs), so rearrange the equation:
Cs = Q / [(Mw x ΔTw) + (Mm x ΔTm)]
Now, plug in the values you have:
Cs = Q / [(Mw x ΔTw) + (Mm x ΔTm)]
Note: To find Q, you may need additional information such as the heat capacity of the calorimeter or the heat capacity of the water.
Once you have the result, compare it to the known value for the specific heat of aluminum (0.897 J/g°C). If your calculated value is significantly different, there may be some errors in your measurements or calculations during the lab.
I hope this helps! Let me know if you have any further questions.