Designing a Hand Warmer PDF

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Designing a Hand Warmer Purpose of Experiment: Research and design an effective hand warmer that is inexpensive, non-toxic, and safe for the environment.

Background: Most commercial hand warmers consist of a plastic package containing a solid, and an inner pouch filled with water. When the hand warmer pack is activated, the solid dissolves in water and produces a large temperature change in the form of heat. In this laboratory experiment, the heat of solution will be calculated from the data for six salts in order to determine the most viable and safe hand warmer. The heat of solution is the enthalpy change associated with the dissolution of a substance in a solvent. In order to carry out experiments involving heats of solution and other enthalpy changes, an insulated vessel called a calorimeter must be used. The process of a solute dissolving in water may either release heat into the resulting aqueous solution or absorb heat from the solution, but the amount of heat exchanged between the calorimeter and the outside surroundings should be minimal. A hand warmer should not  be endothermic, meaning it should not absorb heat from its surroundings. A cold pack is endothermic at room temperature; a hand warmer/hot pack is exothermic, or heat-releasing, at room temperature. The temperature change must also be recorded. The temperature change is used to calculate the total amount of heat released. The molar heat of solution tells us how many kilojoules of heat is released per mole of a salt. Using the molar heat of solution, we can compare it with the costs for the salt to determine which one is most cost-effective. However, because the purpose of this experiment also includes the objective

Pre-Laboratory Questions:

1. When chromium(II) chloride, CrCl2, is dissolved in water, the temperature of the water decreases. a. Is the heat of solution exothermic or endothermic? i. The heat of solution is endothermic -- the solution absorbs heat from the surroundings. b. Which is stronger -- the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond strength of CrCl2 and intermolecular forces between water molecules? Explain. i. The attractive forces between water molecules and chromium and chloride ions is faster, because the bond contains intramolecular forces, and the other ions contains intermolecular forces (which aren’t as strong as intramolecular forces). 2. A solution was formed by combining 25.0 g of solid A with 60.0 mL of distilled water, with the water initially at 21.4 degrees Celsius. The final temperature of the solution was 25.3 degrees Celsius. Calculate the heat released as the solid dissolved, qsoln, assuming no heat loss to the calorimeter (see Equation 1). a. Q = mc(ΔT) b. Q = 85 * 4.18 * (25.3 - 21.4) c. Q = 85 * 4.18 * 3.9 d. Q = 1,385.67 J e. Qsoln = -1,385.67 J

3. In Question 2 above, the calorimeter was found to have a heat capacity of 8.20J/℃. If a correction is included to account for the heat absorbed by the calorimeter, what is the heat of solution, qsoln? a. qcal = ΔT(℃) * Ccal(J/℃) b. qcal =  (25.3-21.4) * 8.20 c. qcal = 31.98 J d. qsoln = -(qaq + qcal) e. qsoln = -(1385.67 J + 31.98 J) f. qsoln = -1417.65 J

4. The solid in Question 2 was aluminum sulfate, Al2(SO4)3. Calculate the molar heat of solution, ∆Hsoln, for aluminum sulfate. Hint: The units for molar heat of solution are kilojoules per mole (kJ/mol). First determine the heat released per gram of solid. a. Molar heat = (qsoln / mass) * molar mass b. Molar heat = (-1417.65 J / 25) * (26.982 * 2 + 32.06 * 3 + 16 * 12) c. Molar heat = -19401.68 J d. Molar heat = -19.40 kJ 

Salts: Ammonium chloride, NH4Cl, 5 g Calcium chloride, anhydrous, CaCl2, 5 g Sodium acetate, NaCH3CO2, 5 g Sodium chloride, NaCl, 15 g Lithium chloride, LiCl, 15 g Sodium carbonate, Na2CO3, 15 g 

Procedure:  Part A. Heat Capacity of the Calorimeter 1. Set up a calorimeter consisting of two nested polystyrene cups in a ring clamp attached to a support stand. 2. (Optional: Place a magnetic stirrer below the calorimeter.) 3. Lower the ring clamp until the bottom of the cup just sits on the surface of the magnetic stirrer (see Figure 1). 4. Measure 100.0 mL of distilled water in a 100-mL graduated cylinder and transfer the water into the calorimeter. 5. Add a magnetic stirring bar to the calorimeter,and set the bar spinning slowly. If a magnetic stirrer is not available, use a stirring rod. Do not remove the stirring rod from the calorimeter. 6. Measure and record the initial temperature of the water. 7. Heat 125 mL of distilled water to 60-70 C  in a 250-mL beaker. 8. Using heat-resistant gloves, measure 100.0 mL of the hot water in a 100-mL graduated cylinder. 9. Measure and record the temperature of the hot water. 10. Immediately pour the hot water into the room temperature water in the calorimeter.

11. 12. 13. 14.

Insert the thermometer, and stir the water. Record the mixing temperature Tmix after 20 seconds. Empty the calorimeter and dry the inside. Calculate the calorimeter constant, Ccal.

 Part B - Calorimetry Procedure  MgSO2(s) + H2O (l) → Mg2+(aq) + SO42-(aq) 1. Measure 100.0 mL of distilled or deionized water in a 100-mL graduated cylinder and transfer to the calorimeter. 2. Measure and record the initial temperature of the water. 3. Measure 5.0012 g of anhydrous magnesium sulfate in a weighing dish. 4. Put a magnetic stir bar or stirring rod into the calorimeter and slowly stir the water. 5. Quickly add the anhydrous magnesium sulfate to the calorimeter and insert the thermometer. 6. Monitor the temperature and record the highest or lowest temperature reading. 7. Calculate the molar heat of solution for magnesium sulfate. Include the correction due to the heat capacity of the calorimeter.  Part C 1. Measure 40.0 mL of dH2O in a 100-mL graduated cylinder and transfer to the coffee-cup calorimeter. 2. Measure and record the initial temperature of the water. 3. Measure 5.00 g of (1. NH4Cl), (2. CaCl2), and (3. LiCl) in a weighing dish. 4. Put a magnetic stir bar or stirring rod into the calorimeter and slowly stir the water. 5. Quickly add the anhydrous magnesium sulfate to the calorimeter and insert the thermometer. 6. Monitor the temperature and record the highest temperature or lowest temperature reading. 7. Repeat steps one to six for the remaining two salts. 8. Calculate the molar heat of solution for the salts, and include the correction due to the heat.     

 

Results:   SET A: NH4Cl Laboratory Results

CaCl2

LiCl

Mass of Salt (grams) 5.00 grams

5.00 grams

5.00 grams

Volume dH2O (mL)

40.0 mL

40.0 mL

40.0 mL

Initial Temperature (C)

22.0 C

22.5

22.6

Highest/Lowest Temperature (C)

13.0 C

37.5

33.6

Change in Temperature (C)

-9° C

15° C

11° C

Molar Heat of Solution with Ccal

23.4 KJ/mol

-80.7 KJ/mol

-22.6 KJ/mol

Set B: Laboratory Results

NaCl

NaCH3OO

Na2CO3

Initial temp (o C)

22.0 ºC

22.8 ºC

22.8 ºC

Mass of salt (g)

5.0835 g

5.0197 g

5.0171 g

Highest/lowest temperature (o C)

20.4 ºC

26.7 ºC

29.0 ºC

Change in

-1.6 ºC

3.9 ºC

6.2 ºC

temperature (o C) Volume of water (mL)

40.5 mL

40.0 mL

40.0 mL

Molar heat of solution

4.2 kJ/mol

-14 kJ/mol

-30 kJ/mol



Calculations:  Calorimeter Constant qwater =  200 * 4.18 * (39.1 - 40.15) = -877.8 J qcal = 877.8 J Ccal = (877.8)/(39.1-22.8) = 53.85 J/°C   Molar Heat of Solution for NH4Cl  qrxn = qcalorimeter + qsoln qrxn = (CΔT) + (mCΔT) qrxn = 53.8 J/C * (-9) + 45g * 4.184 J/g°C * (-9) qrxn = 2178 J  (2178 J)/0.093 mol = 23419 J/mol = 23.42 kJ/mol of heat released.     Molar Heat of Solution for CaCl2  qrxn = qcalorimeter + qsoln qrxn = (CΔT) + (mCΔT) qrxn = 53.8 J/C * (15) + 45g * 4.184 J/g°C * (15)

qrxn = 3631.2 J  -3631.2 J/0.045 mol = -80693 J/mol = -80.7 kJ/mol of heat released.   Molar Heat of Solution for LiCl  qrxn = qcalorimeter + qsoln qrxn = (CΔT) + (mCΔT) qrxn = 53.8 J/C * (11) + 45g * 4.184 J/g°C * (11) qrxn = 2622.8 J  -2622.8 J/0.118 mol = -22567 J/mol = -22.6 kJ/mol of heat relased.  Cost of Solution for LiCl     Repeat the above calculations for the rest of the salts.

       Discussion: The purpose of this lab was to research and design an effective hand warmer that is expensive, non-toxic, and safe for the environment. We calculated the calorimeter constant, and performed a laboratory experiment to determine the molar heat of solution for six salts. Our calculations, compared with the cost and the MSDS, revealed that the optimal hand warmer would have to use the solution LiCl, lithium chloride, to

make the most efficient and safe hand warmer. However, at a hefty cost of $2.93 per mole of LiCl, it is the most expensive hand warmer to make. (68.30 $/kg * 1kg/1000g) * 42.394 g/mol = $2.93 per mole of Lithium Chloride The second option would be CaCl2, which is much less expensive. At this point, it would rather be a question of cost versus safety of salt used.    

  Error Analysis:  There were several possible instances of error in this lab. These experimental errors could have affected the final result and conclusion based off inaccurate values. The accuracy of the thermometer could be a given instance of experimental error, given that an incorrect temperature reading could alter all the data gathered, and all the data calculated after the experiment, including Molar Heat of Solution (and therefore the cost calculation for the salts). However, given that the thermometer that was used in the experiment could only read the temperature of a substance/solution up to one decimal point, the heat that was lost to the surroundings could be too insignificant to really count as a possible source of error. However, if this experiment was repeated with a thermometer that could read up to two or three, or even more decimal points, the accuracy of the thermometer could definitely be an instance of experimental error.  

Picture of thermometer used in the experiment:

     Not all of the salt may have been dissolved when taking the temperature reading. This would mean that there could be some extra energy not accounted for, which would affect the data and calculations, and also the final result for the most efficient and non-toxic hand warmer. However, again, given that the thermometer used in this experiment only records one decimal point, if an extremely minimal amount of salt was left undissolved, the data still would not be affected by this error. A thermometer that is more precise could be affected by smaller amounts of undissolved salt. The calorimeter was also not closed, meaning that heat leaking to the surroundings could have been another possible source of error. However, not a lot of heat may have been leaked to the surroundings, because the temperature measurement was taken relatively quickly (approximately five minutes on average). Again, the accuracy of the thermometer is also a deciding factor if this source of error is negligible or not. 

AP Chem Review Questions:  Review the following data from a calorimetry experiment to determine the heat of fusion of ice. After shaking off any excess water, several ice cubes were added to 99 g of warm water contained in a calorimeter. The initial temperature of the

warm water was 46.8 C. The ice-water mixture was stirred until the temperature reached a stable, minimum value, which was 1.1 C. Any unmelted ice remaining at this point was immediately and carefully removed using tongs, and the mass of the water in the calorimeter was measured—154 g.  1. Use the energy equation to calculate the amount of heat in joules released by the warm water as it cooled. a. qhot = 99 g * 4.184 J/g°C * (-45.7°C) b. qhot = -18929 J c. qhot ≈  -19000 J 2. Based on the Law of Conservation of Energy, what amount of heat was absorbed by the ice as it melted? a. (-1) * -19,000 = 19,000 J 3. Determine the amount of energy absorbed per gram of ice as it melted. a. 154g - 99g = 55g b. 19000 J / 55 g = 344.16 J/g ≈ 340 J/g 4. Calculate the heat of fusion (the heat required to melt ice) in units of kilojoules/mole. a. Heat of Fusion = (340 J/g * 18.02 g/mol) * (1kJ/1000J) b. Heat of Fusion = 6.127 kJ/mol c. Heat of Fusion = 6.1 kJ/mol 5. The literature value for the heat of fusion of ice is 6.02 kJ/mole. What is the percent error for the experimentally determined heat of fusion? a. Percent error = (6.1-6.02) / 6.02 * 100% b. Percent error = 1.33% 6. When a mixture of ice and water originally at 0 C is heated, the temperature remains constant (within experimental error) until all of the ice melts. Explain what happens to the heat energy that is absorbed during this time while the temperature does not change. a. The heat energy that is absorbed is used to keep the temperature of the ice constant as it melts. During this time, all of the heat energy is used in order to change the state from a solid to a liquid. The temperature remains constant until all of the ice has melted into the liquid state, as the picture on the next page shows. (Note that the

picture should be flipped horizontally - it should go from vapor to solid, not solid to vapor. b. .

c. .  d. . e. . Conclusion: The stated purpose of this lab, which is to design an effective hand warmer that is expensive, non-toxic, and safe for the environment, was met. I carried out this experiment and calculated the molar heat of solution for my salts, and used the MSDS to determine the most viable and effective hand warmer....