Observations
Mass of Fe metal:100 MG
Volume of water in Calorimeter: 100 ML
Mass of water in Calorimeter (use 1g per ml): 100G
Initial Temperature of metal: 20
Initial Temperature of water: 20
Final Temperature (both metal and water): 100
Determine the heat gained by the water (show your work):
Q water =
Q Fe =
Determine the specific heat of the metal (show your work):
C Fe =
Determine the atomic weight of your metal:
Look up the actual atomic weight on your periodic table. Determine your percent error using the equation from assignment 2.04 (show your work).
Mass of Al metal:
Volume of water in Calorimeter:
Mass of water in Calorimeter (use 1g per ml):
Initial Temperature of metal:
Initial Temperature of water:
Final Temperature (both metal and water):
Determine the heat gained by the water (show your work):
Q water =
Q Al =
Determine the specific heat of the metal (show your work):
C Al =
Determine the atomic weight of your metal:
Look up the actual atomic weight on your periodic table. Determine your percent error using the equation from assignment 2.04 (show your work).
I did part of it and i have no idea how to do the rest. Please help..
To determine the heat gained by the water in the first part of the question, we can use the formula:
Qwater = mwater * cwater * ΔT
where:
Qwater is the heat gained by the water
mwater is the mass of water in grams (convert from mL using the density of water, which is 1g/mL)
cwater is the specific heat capacity of water, which is approximately 4.18 J/g°C
ΔT is the change in temperature of the water, which is the final temperature minus the initial temperature.
Given:
Mass of water in Calorimeter = 100g
Specific heat capacity of water (cwater) = 4.18 J/g°C
Initial Temperature of water = 20°C
Final Temperature = 100°C
Calculations:
mwater = 100g
ΔT = 100°C - 20°C = 80°C
Qwater = 100g * 4.18 J/g°C * 80°C
Qwater = 33440 J
So, the heat gained by the water is 33440 Joules.
To determine the heat gained by the Iron (Fe) metal, we can use the same formula as above:
QFe = mFe * cFe * ΔT
where:
QFe is the heat gained by the Iron (Fe) metal
mFe is the mass of Iron (Fe) metal in grams (convert from MG using the conversion factor 1MG = 1x10^9 g)
cFe is the specific heat capacity of Iron (Fe), which is approximately 0.449 J/g°C (lookup value)
ΔT is the change in temperature of the metal, which is the final temperature minus the initial temperature.
Given:
Mass of Fe metal = 100 MG = 100x10^9 g (convert from MG to grams)
Specific heat capacity of Fe (cFe) = 0.449 J/g°C (lookup value)
Initial Temperature of metal = 20°C
Final Temperature = 100°C
Calculations:
mFe = 100x10^9 g
ΔT = 100°C - 20°C = 80°C
QFe = 100x10^9 g * 0.449 J/g°C * 80°C
QFe = 3.592x10^12 J
So, the heat gained by the Iron (Fe) metal is 3.592x10^12 Joules.
To determine the specific heat of the metal (Fe), we can rearrange the formula as follows:
cFe = QFe / (mFe * ΔT)
where:
cFe is the specific heat capacity of Iron (Fe)
QFe is the heat gained by the Iron (Fe) metal
mFe is the mass of Iron (Fe) metal in grams (convert from MG using the conversion factor 1MG = 1x10^9 g)
ΔT is the change in temperature of the metal, which is the final temperature minus the initial temperature.
Given:
Mass of Fe metal = 100 MG = 100x10^9 g (convert from MG to grams)
Heat gained by the Fe metal (QFe) = 3.592x10^12 J
ΔT = 80°C
Calculations:
cFe = 3.592x10^12 J / (100x10^9 g * 80°C)
cFe = 4.49 J/g°C
So, the specific heat capacity of the Iron (Fe) metal is 4.49 J/g°C.
To determine the atomic weight of the metal, we need more information about the metal used in the experiment. Please provide the name or symbol of the metal so we can look up its atomic weight on the periodic table and calculate the percent error using the formula from assignment 2.04.
To determine the heat gained by the water, you can use the formula:
Q = m * c * ΔT
Where:
Q = heat gained or lost
m = mass of the substance
c = specific heat capacity of the substance
ΔT = change in temperature
For the water:
m = mass of water = 100 g
c = specific heat capacity of water (approximately 4.18 J/g°C)
ΔT = final temperature - initial temperature = (100°C - 20°C) = 80°C
Q water = m * c * ΔT
= 100 g * 4.18 J/g°C * 80°C
= 33440 J
So, the heat gained by the water is 33440 Joules.
Next, to determine the specific heat of the metal, you can use the formula:
Q metal = m metal * c metal * ΔT
You already have the values for Q metal and ΔT from the previous part. You can rearrange the equation to solve for c metal:
c metal = Q metal / (m metal * ΔT)
For the Fe metal:
Q metal = -Q water (since the heat lost by the metal is gained by the water)
= -33440 J
m metal = mass of Fe metal = 100 MG (1 MG = 1 million grams = 1 x 10^9 g)
ΔT = final temperature - initial temperature = (100°C - 20°C) = 80°C
c Fe = -33440 J / (100 x 10^9 g * 80°C)
= -4.18 x 10^-11 J/g°C
Note: The negative sign indicates that the metal lost heat.
To determine the atomic weight of the metal, you'll need more information about the metal. Please provide the name or any other characteristics of the metal so that I can help you find its atomic weight and calculate the percent error.