Part I Data and Observations: Initial temperature of hot metal: Volume of water in calorimeter: Initial temperature of water in calorimeter: Mass of...
Part I Data and Observations:
| Initial temperature of hot metal: | 100 ° C |
| Volume of water in calorimeter: | 26 ml |
| Initial temperature of water in calorimeter: | 25 ° |
| Mass of metal: | 27.776 g |
| Metal name: | Aluminum |
| Final Temperature in calorimeter: | 38.9 ° C |
Part I Calculations:
1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J/ (g · cC), and the density of water is 1.00 g/mL.
q = m * c * Δt
m=26g
c=4.18
Δt= 38.9 – 25 = 13.9
26 * 4.18 * 13.9 = 1511 joules
2. Calculate the specific heat of the metal.
q surroundings = - q system
q = m × c × Δt
c = qmetal / (massmetal * temp change of metal)
c=Q/m*temp change
Q=1511 joules
m=27.776g
Δt =100-38.9=61.1
c=1511/(27.776 *61.1)
c=1511/1697
c=0.890 J.g-1.oC-1
Part II Data and Observations:
| Initial temperature of hot metal: | 100 °C |
| Volume of water in calorimeter: | 24 ml |
| Initial temperature of water in calorimeter: | 25 °C |
| Unknown Letter (A, B, or C): | A |
| Mass of metal: | 25.605 |
| Final Temperature in calorimeter: | 29.1 °C |
Part II Calculations:
1. Calculate the energy change (q) of the surroundings (water). 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.
q = m × c × Δt (Use values for water)
m = 24 ml
c=4.18
Δt = 29.1 – 25 = 4.1
24 * 4.18 *4.1 = 411.31 joules
2. Calculate the specific heat of the metal.
q surroundings = - q system
q = m × c × Δt
c = qmetal / (massmetal * temp change of metal)
c=Q/m*temp change
Q=411 joules
m=25.61g
Δt = 100-29.1 = 70.9 oC
c = 411.3 J / (25.61 gx 70.9 oC) = 0.226 J/(g.oC)
