Results for Calorimetry SD
Specific Heat of a Metal
Unknown number?
1
Specific Heat of a Metal
a) Mass of water from Trial 1? (Show work with short verbal descriptions of numbers. For example, 23.00 g (mass of calorimeter and water) - 3.00 g (mass of calorimeter) = 20.00 g)
b) Mass of metal from Trial 1? (Show work with short verbal descriptions of numbers.)
c) delta T (water) from Trial 1? (Show work with short verbal descriptions of numbers.)
d) delta T (metal) from Trial 1? (Show work with short verbal descriptions of numbers.)
a) Mass of water from Trial 1 = 42.25 g (mass of calorimeter and water) - 3.25 g (mass of empty calorimeter) = 40.00 g
b) Mass of metal from Trial 1 = 22.40 g (mass of unknown metal sample)
c) delta T (water) from Trial 1 = 26.0 degrees C (final temperature of the water and metal after mixing) - 22.0 degrees C (initial temperature of the water) = -73.5 degrees C
d) delta T (metal) from Trial 1 = 26.0 degrees C (final temperature of the water and metal after mixing) - 22.0 degrees C (initial temperature of the metal) = 4.0 degrees C
Specific Heat of a Metal
a) q (water) from Trial 1? (Show work with short verbal descriptions of numbers.)
b) Specific heat of metal from Trial 1? (Show work with short verbal descriptions of numbers.)
a) q (water) from Trial 1 = 40.00 g (mass of water) x (4.184 J/gC) x -73.5 degrees C (temperature change of the water) = -12300.96
b) Specific heat of metal from Trial 1 = - (40.00 g (mass of water) x (4.184 J/gC) x -73.5 degrees C (temperature change of the water)) / 22.40 g x 4.0 degrees C (temperature change of the metal) = 0.407 J/gC
Specific Heat of a Metal
Calculate the average specific heat of your unknown metal sample. Show all work with proper units.
Average specific heat of unknown metal sample = N/A (we did not do a Trial 2)
Heat of Solution
Answer the following questions:
a) Unknown number?
b) What is the mass of water used? (Show work with short verbal descriptions of numbers.)
c) What is the mass of the solid used? (Show work with short verbal descriptions of numbers.)
a) Unknown number = 3
b) What is the mass of water used = 42.92 g (mass of calorimeter and water) - 3.25 g (mass of empty calorimeter) = 40.00 g
c) What is the mass of the solid used = (42.92 g (mass of calorimeter and water) - 23.15 g (mass of unknown sample)) + (45.05 g (mass of another full unknown sample) - 41.75 g (mass of unknown sample)) = 23.07 g (we added more unknown sample to reach above 20.00 g of unknown sample as instructed)
Heat of Solution
Answer the following questions:
a) delta T for the solution?
b) q solution for the reaction (Assume the specific heat of the solution is the same as that of water (4.184 J/g/°C)? (Show work with short verbal descriptions of numbers.)
c) q unknown? (in joules)
d) Enthalpy of solution, per gram of solute?(in joules/g)(Show work with short verbal descriptions of numbers.)
e) Average delta H if multiple trials were performed.
a) delta T for the solution = 17.0 degrees C (final temperature of the water plus unknown salt) - 99.5 degrees C (initial temperature of the water) = -5.0 degrees
b) q solution for the reaction = (62.74 g (mass of water and salt) x (4.184 J/gC) x -5.0 degrees C (temperature change of the solution) = -1312.5208 J
c) q unknown = - (62.74 g (mass of water and salt) x (4.184 J/gC) x -5.0 degrees C (temperature change of the solution) = 1312.5208 J
d) Enthalpy of solution, per gram of solute = (1312.5208 J (q unknown) / 23.07 g (mass of unknown sample)) x (1 kj / 1000 mol) = 57 J/g
e) Average delta H if multiple trials were performed = N/A
Heat of Neutralization
Show the following for Trial 1
a) Change in temperature? (Show work with short verbal descriptions of numbers.)
b) q solution? (Show work with short verbal descriptions of numbers.)
c) q reaction?
d) Enthalpy of neutralization (delta H) (qreaction mole of H+ and OHminus reacting)? (Show work with short verbal descriptions of numbers.)
e) Average delta H if multiple trials were performed
a) Change in temperature = 32.5 degrees C (final temperature of the solution) - 21.95 degrees C (average temperature of the HCl and NaOH solutions) = 10.55 degrees C
b) q solution = 1.02 g/mL (density) x (25.0 mL + 25.0 mL (volume of HCl and NaOH solutions)) x (4.184 J/gC) x 10.55 degrees C (temperature change of solutions) = 2251.2012
c) q reaction = - (1.02 g/mL (density) x (25.0 mL + 25.0 mL (volume of HCl and NaOH solutions)) x (4.184 J/gC) x 10.55 degrees C (temperature change of solutions)) = -2251.2012
d) Enthalpy of neutralization (delta H) (qreaction mole of H+ and OHminus reacting) = (-2251.2012 J (q reaction) / 0.05 mol (mole of H+ and OHminus reacting)) x (1 kj / 1000 J) = -45 kj/mol
e) Average delta H if multiple trials were performed = N/A
Your instructor will give you instructions as what to do with this box. It may be used for bonus points, or for taking away points (for example late submissions) or for error analysis if something went wrong during the lab. If no instructions have been provided to you, please leave this box blank.
5% = 8.25