1757 Words8 Pages

CHM130

Lab 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…*

You have the mass of water from calculation #9, the specific heat of water is 4.184 J/g(oC), and the temperature change of water from calculation #7. (10 points)

Mass of water: 74.8g

Specific heat of water is 4.184 J/g(oC)

Change in temp. of water: 1°C

Q = 74.8 * 1 * 4.184 = 312.96 J/g(oC)

11. 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 #8. (10 points)

Mass of Sample: 20.03 g

Change in temperature of metal: 77°C

Q = 312.96 J/g(oC)

SH = q/(m* ΔT):

SH = (312.96) / (20.03 * 77) = 312.96 / 1542.31 = 0.2029°C

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

Actual

Percent Error: (0.130 – 0.203)/ (0.130) * 100 = -56.15% error

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)

ΔT = 26°C – 25°C = 1°C

14. Calculate the change in the metal’s temperature by subtracting the initial temperature of the iron from the final

Lab 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

You have the mass of water from calculation #9, the specific heat of water is 4.184 J/g(oC), and the temperature change of water from calculation #7. (10 points)

Mass of water: 74.8g

Specific heat of water is 4.184 J/g(oC)

Change in temp. of water: 1°C

Q = 74.8 * 1 * 4.184 = 312.96 J/g(oC)

11. 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 #8. (10 points)

Mass of Sample: 20.03 g

Change in temperature of metal: 77°C

Q = 312.96 J/g(oC)

SH = q/(m* ΔT):

SH = (312.96) / (20.03 * 77) = 312.96 / 1542.31 = 0.2029°C

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

Actual

Percent Error: (0.130 – 0.203)/ (0.130) * 100 = -56.15% error

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)

ΔT = 26°C – 25°C = 1°C

14. Calculate the change in the metal’s temperature by subtracting the initial temperature of the iron from the final

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