LAB IV SPECIFIC HEAT CAPACITY DETERMINATION
DATA SHEET
The metal sample suspected of being aluminum is warmed and then submerged into water, which is near room temperature. The final temperature of the water and the metal is given below. The specific heat capacity of water is 4.18 J/g·°C. Calculate the specific heat capacity of the metal based on the data below. Remember heat lost = heat gained.
Type of metal used: Al
Trial 1
Trial 2
Trial 3
Mass of metal, g
2.746 g
2.750 g
2.900 g
Mass of water, g
15.200 g
15.206 g
15.201 g
Initial Temp. of Water, °C
24.7 °C
24.6 °C
24.5 °C
Initial Temp. of Metal, °C
72.1 °C
72.2 °C
71.9 °C
Final Temp of Water & Metal, °C
26.3 °C
26.2 °C
24.7 °C
ΔT for water, °C
1.6 °C
1.6 °C
0.2 °C
ΔT for metal, °C
-45.8 °C
-45.9 °C
-47.2 °C
Specific heat capacity of metal, J/g·°C
0.390 J/g·°C
0.390 J/g·°C
0.402 J/g·°C
Average specific heat capacity, J/g·°C
0.394 J/g·°C (use two significant figures due to ΔT of water)
The Law of Conservation of Energy states that the total amount of energy in a system is constant. So, for this experiment, the thermal energy flows as heat from the hotter substance (metal) to the cooler one (water in the cups). The mathematical translation of this statement is then: - Heat lost by metal = Heat absorbed by water - qmetal = qwater - (mass of metal) x (spec. heat of metal) x (ΔT of metal) = (mass of H2O) x (spec. heat of H2O) x (ΔT of H2O)
The specific heat of water is 4.18 J/g·°C. The final temperature is the same for both metal and water because they reached equilibrium. ΔT of metal = Final T for both - Initial T of the metal (boiling water T) ΔT of H2O = Final T for both – Initial T of the H2O in the cups