# CHM 130 Lab 4

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CHM130Lab 4

Calorimetry

Name:

Data Table: (12 points)

ALUMINUM METAL

Pre-weighed Aluminum metal sample mass (mmetal)

20.09 g

Temperature of boiling water and metal sample in the pot (Ti(metal))

dsdfa(Ti

99°C

Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water))

24°C

Maximum Temperature of water/metal in calorimeter after mixing (Tf)

28°C

LEAD METAL

Pre-weighed Lead metal sample mass (mmetal)

20.03g

Temperature of boiling water and metal sample in the pot (Ti(metal))

103°C

Temperature of cool water in the calorimeter prior to adding hot metal sample (Ti(water))

25°C

Maximum Temperature of water/metal in calorimeter after mixing (Tf)

26°C

IRON METAL

Pre-weighed Iron metal sample

mass

*…show more content…*

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.html. (5 points)

Mass = Density * Volume

Density = 0.9970479 g/ml

Mass = 0.9970479 g/ml * 75mL = 74.78 g

16. Use the equation: q = m(SH)ΔT to solve for the amount of heat gained by the water from metal. You have the mass of water from calculation #15, the specific heat of water is 4.184 J/g(oC), and the temperature change of water from calculation #13. (10 points) Q = 74.78 g * 4.184 J/g(oC) * 1° = 312.88 J/g(oC)

17. Use the equation: q = m(SH)ΔT to solve for the specific heat of the metal.

For q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change of the metal can be found in calculation #14. (10 points)

SH = q/(m* ΔT) = 312.88 / (19.99 * 75) = 312.88 / 1499.25 = 0.20869

18. Determine the percent error using the equation and knowing that the actual specific heat of iron is 0.450 J/g(oC): (10 points)

Percent Error = actual-experimental x 100

Actual

Percent Error: (0.450 – 0.209)/ (0.450) * 100 = 53.56%

Follow-Up Questions:

19. The First Law of Thermodynamics states that heat lost by one object is gained by another?

Mass = Density * Volume

Density = 0.9970479 g/ml

Mass = 0.9970479 g/ml * 75mL = 74.78 g

16. Use the equation: q = m(SH)ΔT to solve for the amount of heat gained by the water from metal. You have the mass of water from calculation #15, the specific heat of water is 4.184 J/g(oC), and the temperature change of water from calculation #13. (10 points) Q = 74.78 g * 4.184 J/g(oC) * 1° = 312.88 J/g(oC)

17. Use the equation: q = m(SH)ΔT to solve for the specific heat of the metal.

For q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change of the metal can be found in calculation #14. (10 points)

SH = q/(m* ΔT) = 312.88 / (19.99 * 75) = 312.88 / 1499.25 = 0.20869

18. Determine the percent error using the equation and knowing that the actual specific heat of iron is 0.450 J/g(oC): (10 points)

Percent Error = actual-experimental x 100

Actual

Percent Error: (0.450 – 0.209)/ (0.450) * 100 = 53.56%

Follow-Up Questions:

19. The First Law of Thermodynamics states that heat lost by one object is gained by another?