Lab5 - exp 5 PDF

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Enthalpy and Specific Heat PRE-LAB QUESTIONS 1. Water and steam are both 100ºC when water is boiling, but a burn from steam is worse than a burn from the water. Hypothesize why this is the case.

When water changes from liquid to gas it requires energy (the heat of vaporization). So, when steam hits the skin it goes from gas to liquid phase and condenses on the skin (phase change). In this process it releases a lot of energy causing the burn to be much more severe. When boiling water comes in contact with the skin it does release some amount of energy where it would decrease its temperature (transferring heat to the skin) but would not have to go through a phase change. Because of no phase change the injury is smaller and the burn is less severe.

2. A 10-g ice cube, initially at 0ºC, is melted in 100 g of water that was initially 20ºC. After the ice has melted, the equilibrium temperature is 10.93 ºC. Calculate:

a. The total heat lost by the water (the specific heat for water is 4.186 J/g·°C). Mass of water = 100g Specific heat for water = 4.186 J/g·°C Temperature change = 20°C – 10.93°C = 9.07°C

Q = Mass × C (specific heat) × T (Change in temp) Q = 100 g × 4.186 J/g°C × 9.07°C Q = 3796.702 Q = 3796.7 J b. The heat gained by the ice cube after it melts (the specific heat for ice is 2.093 J/g·°C). Mass of ice = 10g Specific heat for ice = 2.093 J/g·°C Temperature change = 20°C – 10.93°C = 9.07°C

Q = Mass × C (specific heat) × T (Change in temp) Q = 10 g × 2.093 J/g°C × 10.93°C

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Enthalpy and Specific Heat Q = 228.7649 Q = 228.8 J

c. The heat it took to melt the ice (Hint: It takes 334 J of heat energy to melt 1 g of ice). Hfusion = 334 J/g Mass of ice = 10 g

Q = Mass of ice × Hfusion Q = 10 g × 334 J/g Q = 3340 J

3. Inside a calorimeter is 100 g of water at 39.8ºC. A 10-g object at 50ºC is placed inside the calorimeter. When equilibrium has been reached, the new temperature of the water and metal object is 40ºC. What type of metal is the object made from? Hint: Use Table 1 in the Introduction for reference.

Mass of water = 100 g Initial temperature of water = 39.8ºC Mass of unknown metal = 10 g Final temperature = 40ºC Temperature change in water = 40ºC - 39.8ºC = 0.2 ºC Temperature change in metal = 50ºC - 40ºC = 10 ºC

mwater ·Cwater ·ΔTwater = mmetal ·Cmetal ·ΔTmetal 100 g × 4.186 J/g°C × 0.2 = 10 × C of metal x 10 4.186 J/g°C × 0.2 = C of metal 0.8372 J/g°C = C of metal (The specific Heat) Metal with specific heat closest to is Aluminum (0.902 J/g°C )

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Enthalpy and Specific Heat EXPERIMENT 1: SPECIFIC HEAT OF A METAL Data Sheet Table 1. Mass Mass (g) 97.9 g (100ml) Water Unknown Metal Strip

10.9 g

Table 2. Temperature Data Time

Temperature (°C) Trial 1

Temperature (°C) Trial 2

Temperature (°C) Trial 3

Initial Temp Water (Ti (water))

23 °C

23 °C

24 °C

Initial Temp Metal (Ti (metal))

97 °C

97 °C

97 °C

1 minute

24.5 °C

25 °C

26 °C

2 minutes

24 °C

24.5 °C

25.5 °C

3 minutes

24 °C

24 °C

25 °C

4 minutes

23 °C

24 °C

25 °C

5 minutes

23 °C

24 °C

25 °C

Table 3. Specific Heat Calculations Calculation

Trial 1

Trial 2

Trial 3

Highest Water Temp

24.5 °C

25 °C

26 °C

1.5 °C

2 °C

2 °C

-72.5 °C

-72 °C

-71 °C

Water Specific Heat

4.186 J/g°C

4.186 J/g°C

4.186 J/g°C

Heat Water Absorbed (qwater = mwater• Cwater•

Q= 97.9 x 4.186 x1.5 =614.7 J

Q= 97.9 x 4.186 x 2 =819.6 J

Q=97.9 x 4.186 x 2 =819.6 J

ΔTwater = Thighest water – Ti (water)

ΔTmetal = Thighest water – Ti (metal)

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Enthalpy and Specific Heat Calculation ∆Twater) Heat Lost by Metal (-qmetal=qwater) Metal Specific Heat Cmetalr=qmetal / ( mmetal• ∆Tmetal )

Trial 1

Trial 2

Trial 3

-614.7 =614.7

-819.6 =819.6

-819.6=819.6

C=-819.6/(10.9x-72) C= 1.044 J/g°C

C=-819.6/(10.9x-71) C=1.059 J/g°C

C=-614.7/(10.9x-72.5) C= 0.7779 J/g°C

Photo Documentation Paste photos of steps 8 and 12 for your first trial below:

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Enthalpy and Specific Heat

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Enthalpy and Specific Heat

Experiment 1 Post-Lab Questions 1. Why is ΔTmetal < 0?

ΔTmetal < 0 because metal will lose energy and water will gain that energy when unknown metal is submerged in the water in this experiment.

4. Why is ΔTwater > 0 ?

ΔTwater > 0 because water would gain the energy released from the metal when the unknown metal is submerged in water in this experiment.

5. What is the average specific heat capacity of the unknown metal for the three trials? Show your work. C= (0.7779 + 1.044 + 1.059) / 3

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Enthalpy and Specific Heat C =0.9603 J/g°C

6. What is the unknown metal in this experiment? Use Table 1 in the Introduction for reference.

The unknown metal in this experiment is aluminum. The specific heat of aluminum is 0.902 J/g°C which is closest to my answer 0.9603 J/g°C.

7. A metal sample weighing 43.5 g at a temperature of 100.0 °C was placed in 39.9 g of water in a calorimeter at 25.1°C. At equilibrium, the temperature of the water and metal was 33.5°C. d. What was ΔT for the water? (ΔT = Tfinal - Tinitial)

ΔT = T(final) – T(initial) ΔT = 33.5 °C - 25.1 °C ΔT = 8.4 °C

e. What was ΔT for the metal?

ΔT = T(final) – T(initial) ΔT = 33.5 °C - 100 °C ΔT = -66.5 °C

f.

Using the specific heat of water (4.186 J/g°C), calculate how much heat flowed into the water.

Q = Mass × C (specific heat) × T (Change in temp) Q = 39.9 x 4.184 x 8.4 Q = 1402.31 K

g. Calculate the specific heat of the metal.

Mass of water = 39.9 g Mass of metal = 43.5 g

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Enthalpy and Specific Heat Specific heat of water = 4.186 J/g°C Temperature change in water = 8.4 °C Temperature change in metal = - 66.5 °C

mwater ·Cwater ·ΔTwater + mmetal ·Cmetal ·ΔTmetal =0 (39.9) * (4.184) * (66.5) + (43.5) * (c₁) (8.4) = 0 1402.3 - 2892.8c₁ = 0 c₁ = 1402.3 / 2892.8 c₁ = 0.4848 J / (g-C) specific heat of the metal

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Enthalpy and Specific Heat EXPERIMENT 2: ENERGY IN FOOD Data Sheet Table 4. Calorie Testing Measurement

Mass of Cracker in Food Holder (g)

Water Temperature (°C)

Initial

2.0 g

24 °C

Final

1.1 g

35 °C

Difference

0.9 g

11 °C

Photo Documentation Paste photos of step 5 below

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Enthalpy and Specific Heat

Post-Lab Questions 1. Calculate the thermal energy absorbed by the water from the burning cracker according to the equation below:

q = 0.9 x 4.186 x 11 q = 41.4 J 8. Is there a relationship between the final mass of the cracker and an increase in temperature of the water inside of the can? Explain your experiment observations.

Yes, there is a relationship between the final mass of the cracker and an increase in temperature of the water inside of the can. They are inversely proportional to each other where one (the temperature of water) would increase and the other (mass of cracker) would decrease.

9. Based on the design of the experiment and your estimated amount of error, would you have designed this experiment differently? Explain why or why not and, if applicable, how you would design it differently.

I would add in some sort of stand so I do not have to hold the can with the help of a glass rod myself. I would add an electronic thermometer which would show more precise temperature reading. By doing these additions I think I would produce a more accurate temperature reading.

10. Does toasting bread change the number of calories in the bread (i.e., does a slice of toast have more/fewer calories than a slice of bread)? Explain your answer using the data in Table 4.

No, toasting bread does not change the number of calories in the bread. The toasting of the bread might affect the mass of the bread when compared to slice bread, but the calories will remain the same. This is because toasting will only remove the water in the bread and not the calories.

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