# Calometry Lab

Topics: Heat, Energy, Thermodynamics Pages: 3 (587 words) Published: May 18, 2013
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. Now, the...