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.