Pre-weighed Aluminum metal sample mass (mmetal) 50.0 grams
Temperature of boiling water and metal sample in the pot (Ti(metal))

dsdfa(Ti 90.0 C
Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 17.0 C
Maximum Temperature of water/metal in calorimeter after mixing (Tf) 47.0 C

LEAD METAL

Pre-weighed Lead metal sample mass (mmetal) 50.0 grams
Temperature of boiling water and metal sample in the pot (Ti(metal)) 95.0 C
Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water)) 17.0 C
Maximum Temperature of water/metal in calorimeter after mixing (Tf) 50.0 C

IRON…show more content… Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change of the metal can be found in calculation #8. (10 points)

12. Determine the percent error using the equation and knowing that the actual specific heat of lead is 0.130 J/g(oC): (10 points)
Percent Error = actual-experimental x 100
Actua
=|0.130-0.125|/0.125*100
=4%

Iron Sample:
13. Calculate the change in temperature for the water caused by the addition of the aluminum by subtracting the initial temperature of the water from the final temperature of the water. ΔT = Tf - Ti (5 points)

60.0C-18.0C=42.0C
14. Calculate the change in the metal’s temperature by subtracting the initial temperature of the iron from the final temperature of the iron. ΔT = Tf - Ti (5 points)
60.0-100.0C=-40.0C
15. Determine the mass of the water by using the water’s density (specific to the initial temperature) and the volume of the water. Remember, density = mass/volume. You can look up the density of the water at your specific temperature at http://www.ncsu.edu/chemistry/resource/H2Odensity_vp.htmﾧl. (5 points)

Mass=density*volume=0.9985986*75=75g

16. Use the equation: q = m(SH)ΔT to solve for the amount of heat gained by the water from metal. You have