In a calorimetry experiment in the lab, you were able to record the following data:
Mass of water in a calorimeter = 90.0 g
Initial temperature of water in calorimeter = 22.0o C
Mass of unknown metal = 45.0 g
Initial temperature of hot metal = 100.0o C
Final temperature of water after you place hot metal in calorimeter filled with water = 28.0o C
Based on the above data, what is the Cp of the metal?
(Cp H2O = 4.184 J/g oC)
q = massmetal x Cpmetal x delta Tmetal.
q = massH2O x CpH2O x delta TH2O.
Use equation 2 first to calculate q, then use equation 1 (using q from the first calculation) to calculate Cp metal. Post your work if you get stuck.
How do i determine the cpH20 and the delta Tmetal?
Cp for H2O is given in the problem. Last line in your post.
Use equation 2 first. You know mass H2O, you know Cp H2O, and you know delta T. Calculate q. Plug that q into equation 1. Now you have q, mass metal, and delta T metal. Calculate Cp metal.
sorry, didn't even see that. how does 0.6973 sound?
sounds ok. I suppose that's J/g
Thanks alot Dr.Bob!
To calculate the specific heat capacity (Cp) of the metal, you can use the formula:
Q = m * Cp * ΔT
Where:
Q is the heat transferred (in Joules),
m is the mass of the substance (in grams),
Cp is the specific heat capacity (in Joules per gram per degree Celsius), and
ΔT is the change in temperature (in degrees Celsius).
First, calculate the heat transferred to the water using the formula:
Qwater = mwater * Cpwater * ΔTwater
Where:
mwater is the mass of the water (90.0 g)
Cpwater is the specific heat capacity of water (4.184 J/g oC)
ΔTwater is the change in temperature of the water (Final temperature - Initial temperature = 28.0o C - 22.0o C = 6.0o C)
Qwater = 90.0 g * 4.184 J/g oC * 6.0o C
Next, calculate the heat transferred from the metal to the water using the formula:
Qmetal = mmetal * Cpmetal * ΔTmetal
Where:
mmetal is the mass of the metal (45.0 g)
Cpmetal is the specific heat capacity of the metal (unknown)
ΔTmetal is the change in temperature of the metal (Final temperature - Initial temperature = 28.0o C - 100.0o C = -72.0o C)
Qmetal = 45.0 g * Cpmetal * -72.0o C
Since energy is conserved, the heat transferred from the metal to the water is equal to the heat transferred to the water.
Qwater = Qmetal
Now, you can equate the two equations:
90.0 g * 4.184 J/g oC * 6.0o C = 45.0 g * Cpmetal * -72.0o C
Solving for Cpmetal:
Cpmetal = (90.0 g * 4.184 J/g oC * 6.0o C) / (45.0 g * -72.0o C)
Cpmetal ≈ -0.260 J/g oC
Note: The Cp value of the metal is negative because the metal lost heat to the water, causing its temperature to decrease.