Calometry Lab
Cal: Metal: | Aluminum | Zinc | Iron | Copper | Mass of metal: | 27.776 g | 41.664 g | 34.720 g | 41.664 g | Volume of water in the calorimeter: | 26.0 mL | 26.0 mL | 26.0 mL | 26.0 mL | Initial temperature of water in calorimeter: | 25.3 °C | 25.3 °C | 25.3 °C | 25.3 °C | Temperature of hot water and metal in hot water bath: | 100.5 °C | 100.5 °C | 100.5 °C | 100.5 °C | Final temperature reached in the calorimeter: | 31.6 °C | 34.8 °C | 33.1 °C | 34.5 °C |
Part I:
Part II: Metal: | Metal A | Metal B | Metal C | Mass of metal: | 15.262 g | 25.605 g | 20.484 g | Volume of water in the calorimeter: | 24.0 mL | 24.0 mL | 24.0 mL | Initial temperature of water in calorimeter: | 25.2 °C | 25.3 °C | 25.2 °C | Temperature of hot water and metal in hot water bath: | 100.3 °C | 100.3 °C | 100.3 °C | Final temperature reached in the calorimeter: | 27.5 °C | 32.2 °C | 28.0 °C |
Part 1&2:
Part I:
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
qwater = m × c × ΔT.
We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
qwater = m × c × ΔT
m = mass of water = density x volume = 1 x 26 = 26 grams
ΔT = T(mix) - T(water) = 38.9 - 25.3 = 13.6
q(water) = 26 x 13.6 x 4.18
q(water) = 1478 Joules
SPECIFIC HEAT: qmetal = -205 J = 15.363 g X c X (27.2 - 100.3 C) c = 0.183 J/gC
PART2. Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation. q(water) = - q(metal) q(metal) = - 1478 Joules q(metal) = m × c × ΔT m = 27.776 g
ΔT = T(mix) - T(metal)
ΔT = 38.9 - 100.5 = - 61.6
C = q(metal) / m x ΔT
C = -1478 / (-61.6 x 27.776 )
C = 0.864 J / (g × °C)
Part 3:
1&2:
For #1 theres a specific heat of 0.864 J / (g × °C) and that is closest to the specific heat of aluminum. So, for this experiment, let's call your metal aluminum.
Part I:
Part II: Metal: | Metal A | Metal B | Metal C | Mass of metal: | 15.262 g | 25.605 g | 20.484 g | Volume of water in the calorimeter: | 24.0 mL | 24.0 mL | 24.0 mL | Initial temperature of water in calorimeter: | 25.2 °C | 25.3 °C | 25.2 °C | Temperature of hot water and metal in hot water bath: | 100.3 °C | 100.3 °C | 100.3 °C | Final temperature reached in the calorimeter: | 27.5 °C | 32.2 °C | 28.0 °C |
Part 1&2:
Part I:
1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation
qwater = m × c × ΔT.
We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
qwater = m × c × ΔT
m = mass of water = density x volume = 1 x 26 = 26 grams
ΔT = T(mix) - T(water) = 38.9 - 25.3 = 13.6
q(water) = 26 x 13.6 x 4.18
q(water) = 1478 Joules
SPECIFIC HEAT: qmetal = -205 J = 15.363 g X c X (27.2 - 100.3 C) c = 0.183 J/gC
PART2. Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation. q(water) = - q(metal) q(metal) = - 1478 Joules q(metal) = m × c × ΔT m = 27.776 g
ΔT = T(mix) - T(metal)
ΔT = 38.9 - 100.5 = - 61.6
C = q(metal) / m x ΔT
C = -1478 / (-61.6 x 27.776 )
C = 0.864 J / (g × °C)
Part 3:
1&2:
For #1 theres a specific heat of 0.864 J / (g × °C) and that is closest to the specific heat of aluminum. So, for this experiment, let's call your metal aluminum.