Experiment 22 Lab Report 5l2r42
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Hot and Cold Pack Lab Report Andrew Nguyen Collin Teng Lab Section #22 March 21st, 2016
Introduction Instant hot and cold packs are used to help heal injuries at a faster rate. These packs are used often by athletes. Cold packs are used to reduce the temperature from the injury, which reduces the blood flow. This process is called vasoconstriction. Hot packs transfer heat from the pack to the injury, which increases the blood flow to the injury. As more blood flows to the injury more oxygen and nutrients is transferred to the injury, which helps treat the injury. We are given task is to use the most effective and precise analysis techniques to determine which salts would be the most effective to use within a hot and cold pack. We will be given ammonium nitrate, calcium chloride, lithium chloride, and potassium chloride. With the given material, we will use the calorimeter process to determine which salt is the most effective. After determining which salt is most effective, the task is to determine the cost of producing the hot and cold pack. Then find the salt that can be made from less than 5 dollars and is most efficient. There is one technique that will be employed in this experiment to find the data. That technique is the calorimeter test. Using this technique we will be finding which salts would be the best at creating hot packs or cold packs. We must use the calorimeter test to determine changes of temperature when each salt is added. When you find this you can then determine the calorimeter constant using the following equation: Q calorimeter=C calorimeter∗ΔT calor i meter
which is also equal to
(m warm water∗C warm water∗ΔT warm water )( mcool water∗C cool water∗ΔT cool water ).
In order to determine if the compounds are endothermic or exothermic we must take the q dissolution
and divide that by the moles of the substance that we used in the experiment.
That number will represent the
Δ H , if the
Δ H is positive the compound is endothermic
and if it is negative then it is exothermic. With these equations the questions proposed by the company should be easily answered. Experimental
● ● ● ●
Necessary Reagents and Chemicals: Ammonium nitrate Calcium chloride Lithium chloride Potassium chloride
● ● ● ● ● ● ● ● ● ●
Necessary Equipment: MeasureNet temperature probe 2 Styrofoam Cups and a Lid Magnetic stirrer and stir bar Rubber stopper Temperature probe Wire Ring Stand Utility Clamp Test tube 150 mL or 250 mL Beaker
Calorimeter Procedure The calorimeter procedure involves two parts, part A involves finding the calorimeter constant.You must first set up the MeasureNet, you must calibrate the system by making an ice bath. The setup for this part of the lab involves the placing the two styrofoam cups inside each other to serve as a makeshift calorimeter. The makeshift calorimeter will be placed on the stirring plate and the temperature probe will be inserted into the opening of the lid on the styrofoam cups (calorimeter). The temperature probe will be attached to the ring stand using the utility clamp and will also be attached to the measure net system in order to record the data. The actual experiment involves taking 40-50 grams of water and placing it inside the calorimeter. Using a magnetic stir bar turn the stirring plate to a medium low setting, this will stir the solution. Then using a hot plate heat up 60 mL of water inside a beaker. Once the hot water has been heated 45-
60 degrees Celsius above the room temperature, extract 50 mL of water using a graduated cylinder and press start on the MeasureNet system to begin recording data. Make sure to save the graphs. Part B involves doing the same as part A but use one gram of the designated salts instead of the hot water, and use 25 mL of water instead of 40-50 mL of water. Once all this is done, use the graphs to determine if the equation is exothermic or endothermic and compare the data to the prices to find the optimal salt.
Results
Initial Temperature of Trial 1: 18.5522 Initial mass: 45 g Final Temperature Trial 1: 41.3166 Temp of Hot Water: 80.00 Grams of hot water: 50g
Initial Temperature of Trial 2: 19.5143 Initial mass: 45 g Final Temperature Trial 2: 47.5390 Temp of Hot Water: 85.00 Grams of hot water: 50g
q calorimeter = −qwarm water - q cool water
q=mC Δ T
Trial 1:
J q= (50 g*4.184 *( 41.3166 C−80.00 C ))= g∗C -8092.57 q=(45g*4.184
Trail 2: q= (50 g*4.184
J *( 47.5390 C−85.00 C ))= g∗C
-7836.8412 J *(41.3166-18.5522 C))=4286.08 g∗C
q calorimeter =-(-8092.57)-(4286.08)= 3806.49 Calorimeter Constant=
3806.49 J J =167.21 22.7644 C C Average Calorimeter Constant=
q=(45g*4.184
J *( 47.5390 -19.5143 C))=5276.49 g∗C
q calorimeter =-(-7836.8412)-(5276.49)= 2560.35 Calorimeter Constant=
2560.35 J J =91.36 28.0247 C C
167.21+91.36 = 129.29 2
J C
Initial Temperature of Trial 1: 25.0344 Initial mass: 45 g Final Temperature Trial 1: 28.4660 Grams of salt: 1g
Initial Temperature of Trial 2: 19.6048 Initial mass: 45 g Final Temperature Trial 2: 22.4608 Grams of salt: 1g
−qdissolution = msolution∗C solution∗Δ T solution )+( C calorimeter∗Δ T calorimeter ¿ ¿
Trial 1: −qdissolution C))+ ( 129.29
J 46 g∗4.184 ∗¿ = (28.4660 C-25.0344 g∗C ¿
J ∗(28.4660 C−25.0344 C) ¿ = -1104.13 J C
Moles of LiCl: Change in Heat: kJ mol
1mol 1.000 g * = .0236 mol 42.39 ❑ −1104.13 J 1 kJ * =-46.79 .0236 mol 1000 J
Trial 2: −qdissolution = 46 g∗4.184 129.29
J ∗(22.4608C−19.6048 C) )+( g∗C ¿
J ∗(22.4608 C−19.6048 C)¿ = -918.93 C
Moles of LiCl: Change in Heat:
1mol 1.000 g * = .0236 mol 42.39 ❑ −918.93 J 1 kJ kJ * =-38.94 .0236 mol 1000 J mol
Average Change in Heat:
Cost of Lithium Chloride:
kJ −46.79+(−38.94) → exothermic (hot pack) =-42.87 mol 2 1.000 gave r age g =.000996 *90 mL *23.32 C mL∗C 45 mLaverage∗22.32 Caverage ΔT
=2.09 g 2.09 g *
$ 65.00 = $ .27salt + $ .73 labor +$ .36capital +$ .19 plastic bag = $1.55 Hot Pack 500 g
Initial Temperature of Trial 1: 20.4422 Initial mass: 45 g Final Temperature Trial 1: 22.9409 Grams of salt: 1g
Initial Temperature of Trial 2: 19.9749 Initial mass: 45 g Final Temperature Trial 2: 22.7206 Grams of salt: 1g
Trial 1: −qdissolution =
Trial 2: −qdissolution =
46 g∗4.184
46 g∗4.184
129.29
J ∗(22.9409C−20.4422 C) )+( g∗C ¿
J ∗(22.9409−20.4422)¿ = -806.58 C
Moles of Ca Cl2 : mol
1 mol 1.000 g * = .009011 110.98 g ❑
( 129.29
J ∗(22.7206−19.9749) )+ g∗C ¿
J ∗(22.7206−19.9749)¿ = -883.44 C
Moles of Ca Cl2 : mol
1 mol 1.000 g * = .009011 110.98 g ❑
Change in Heat:
−806.58 J 1 kJ * =-89.51 .009011 mol 1000 J
kJ mol
Change in Heat:
−883.44 J 1 kJ * =-98.04 .009011 mol 1000 J
kJ mol
Average Change in Heat:
Cost of Calcium Chloride:
kJ −89.52+(−98.04) → exothermic (hot pack) =-93.78 mol 2 1.000 g average g =.00110 *90 mL *20.21 C mL∗C 45 mLaverage∗20.21 Caverage ΔT
=2.00 g 2.00 g *
$ 31.70 = $ .13salt + $ .73labor + $ .36 capital + $ .19 plastic bag = $1.41 Hot Pack 500 g
Initial Temperature of Trial 1: 19.5830 Initial mass: 45 g Final Temperature Trial 1: 18.4221 Grams of salt: 1g
Initial Temperature of Trial 2: 19.9876 Initial mass: 45 g Final Temperature Trial 2: 18.7666 Grams of salt: 1g
+
J −qdissolution = 46 g∗4.184 g∗C ∗¿ (18.4221-19.5830)) ¿
( 129.29
J ∗(18.4221−19.5830) ¿ = 373.52 C
Moles of KCl: Change in Heat:
1mol 1.000 g * = .01341 mol 74.5513 g ❑ 373.52 J 1 kJ kJ * =27.85 .01341 mol 1000 J mol
Average Change in Heat:
+
J −qdissolution = 46 g∗4.184 g∗C ∗¿ (18.7666-19.9876)) ¿
( 129.29
J ∗(18.7666−19.9876)¿ = 392.86 C
Moles of KCl: Change in Heat:
1mol 1.000 g * = .01341 mol 74.5513 g ❑ 392.86 J 1 kJ kJ * =29.29 .01341 mol 1000 J mol
27.85+ 29.29 kJ → endothermic (cold pack) =28.57 2 mol
Cost of Potassium Chloride:
1.000 g average g =.00112 *90 mL *19.79 C mL∗C 45 mLaverage∗19.79 Caverage ΔT
=2.00 g 2.00 g *
$ 28.19 = $ .11salt +$ .73 labor +$ .36capital +$ .19 plastic ba g = $1.39 Cold Pack 500 g
Initial Temperature of Trial 1: 19.9470 Initial mass: 45 g Final Temperature Trial 1: 18.3374 Grams of salt: 1g
−qdissolution = 46 g∗4.184∗¿ (18.3374-19.9470 ))+ ¿ J ( 129.29 ∗(18.3374−19.9470)¿ = 517.895 C Moles of
NH 4 (NO3 ) :
1mol 1.000 g * =. 80.052 g ❑
01249 mol
Initial Temperature of Trial 2: 19.8505 Initial mass: 45 g Final Temperature Trial 2: 18.3356 Grams of salt: 1g
−qdissolution = 46 g∗4.184∗¿ (18.3356-19.8505 ))+ ¿ J ( 129.29 ∗(18.3356−19.8505)¿ = 487.425 C Moles of
NH 4 ( NO3 ) :
1mol 1.000 g * =. 80.052 g ❑
01249 mol
Change in Heat:
517.90 J 1 kJ kJ * =41.47 .01249 mol 1000 J mol
Average Change in Heat:
487.43 J 1 kJ kJ * =39.03 .01249 mol 1000 J mol
41.47+39.03 kJ → endothermic (cold pack) =40.25 2 mol
Cost of Ammonium Nitrate:
2.00 g *
Change in Heat:
1.000 g average =.00112*90 mL *19.90 C =2.00 g 45 mLaverage∗19.90 Caverage ΔT
$ 26.20 = $ .10salt + $ .73labor + $ .36 capital + $ .19 plasticbag = $1.38 Cold Pack 500 g
Discussion
In this experiment we were suppose to find the best salt to use in the cold and hot packs that the companies provided for us. To do this we used calorimetry which gave us the data to find if the reaction was exothermic or endothermic, to find the amount that was needed to reach a temperature of 0 for the endothermic reactions or a temperature of 65 degrees, and to find the final cost for those salts. The calorimeter technique that was used in the experiment was used to find the change in temperature of a system when a certain amount of the salt is added. One mistake that could have been made during the process include contamination of the salts. Another mistake that should be outlined is that some of the salt may have not gone into the calorimeter which would have made the “one gram” measurement more accurate. Conclusion In this experiment, various salts were tested for effectiveness and cost. The salt that was concluded to be the most effective as a cold pack was ammonium nitrate because it had the highest effectiveness for the lowest price. The cost of one ammonium nitrate packed came to be about $1.38 where the next lowest salt was potassium chloride and that came out to be $1.39. These prices reflected the cost of the amount of the material used to reach 0 degrees celsius. For the hot packs the most effective one came to be Calcium Chloride. The price of this hot pack to reach the asking temperature of 65 degrees was $1.41. The next lowest priced hot pack was Lithium Chloride at $1.55.
Bibliography Stanton B., Zhu L., Atwood C. (2010) . Experiments in General Chemistry Featuring MeasureNet 2nd Edition.
Introduction Instant hot and cold packs are used to help heal injuries at a faster rate. These packs are used often by athletes. Cold packs are used to reduce the temperature from the injury, which reduces the blood flow. This process is called vasoconstriction. Hot packs transfer heat from the pack to the injury, which increases the blood flow to the injury. As more blood flows to the injury more oxygen and nutrients is transferred to the injury, which helps treat the injury. We are given task is to use the most effective and precise analysis techniques to determine which salts would be the most effective to use within a hot and cold pack. We will be given ammonium nitrate, calcium chloride, lithium chloride, and potassium chloride. With the given material, we will use the calorimeter process to determine which salt is the most effective. After determining which salt is most effective, the task is to determine the cost of producing the hot and cold pack. Then find the salt that can be made from less than 5 dollars and is most efficient. There is one technique that will be employed in this experiment to find the data. That technique is the calorimeter test. Using this technique we will be finding which salts would be the best at creating hot packs or cold packs. We must use the calorimeter test to determine changes of temperature when each salt is added. When you find this you can then determine the calorimeter constant using the following equation: Q calorimeter=C calorimeter∗ΔT calor i meter
which is also equal to
(m warm water∗C warm water∗ΔT warm water )( mcool water∗C cool water∗ΔT cool water ).
In order to determine if the compounds are endothermic or exothermic we must take the q dissolution
and divide that by the moles of the substance that we used in the experiment.
That number will represent the
Δ H , if the
Δ H is positive the compound is endothermic
and if it is negative then it is exothermic. With these equations the questions proposed by the company should be easily answered. Experimental
● ● ● ●
Necessary Reagents and Chemicals: Ammonium nitrate Calcium chloride Lithium chloride Potassium chloride
● ● ● ● ● ● ● ● ● ●
Necessary Equipment: MeasureNet temperature probe 2 Styrofoam Cups and a Lid Magnetic stirrer and stir bar Rubber stopper Temperature probe Wire Ring Stand Utility Clamp Test tube 150 mL or 250 mL Beaker
Calorimeter Procedure The calorimeter procedure involves two parts, part A involves finding the calorimeter constant.You must first set up the MeasureNet, you must calibrate the system by making an ice bath. The setup for this part of the lab involves the placing the two styrofoam cups inside each other to serve as a makeshift calorimeter. The makeshift calorimeter will be placed on the stirring plate and the temperature probe will be inserted into the opening of the lid on the styrofoam cups (calorimeter). The temperature probe will be attached to the ring stand using the utility clamp and will also be attached to the measure net system in order to record the data. The actual experiment involves taking 40-50 grams of water and placing it inside the calorimeter. Using a magnetic stir bar turn the stirring plate to a medium low setting, this will stir the solution. Then using a hot plate heat up 60 mL of water inside a beaker. Once the hot water has been heated 45-
60 degrees Celsius above the room temperature, extract 50 mL of water using a graduated cylinder and press start on the MeasureNet system to begin recording data. Make sure to save the graphs. Part B involves doing the same as part A but use one gram of the designated salts instead of the hot water, and use 25 mL of water instead of 40-50 mL of water. Once all this is done, use the graphs to determine if the equation is exothermic or endothermic and compare the data to the prices to find the optimal salt.
Results
Initial Temperature of Trial 1: 18.5522 Initial mass: 45 g Final Temperature Trial 1: 41.3166 Temp of Hot Water: 80.00 Grams of hot water: 50g
Initial Temperature of Trial 2: 19.5143 Initial mass: 45 g Final Temperature Trial 2: 47.5390 Temp of Hot Water: 85.00 Grams of hot water: 50g
q calorimeter = −qwarm water - q cool water
q=mC Δ T
Trial 1:
J q= (50 g*4.184 *( 41.3166 C−80.00 C ))= g∗C -8092.57 q=(45g*4.184
Trail 2: q= (50 g*4.184
J *( 47.5390 C−85.00 C ))= g∗C
-7836.8412 J *(41.3166-18.5522 C))=4286.08 g∗C
q calorimeter =-(-8092.57)-(4286.08)= 3806.49 Calorimeter Constant=
3806.49 J J =167.21 22.7644 C C Average Calorimeter Constant=
q=(45g*4.184
J *( 47.5390 -19.5143 C))=5276.49 g∗C
q calorimeter =-(-7836.8412)-(5276.49)= 2560.35 Calorimeter Constant=
2560.35 J J =91.36 28.0247 C C
167.21+91.36 = 129.29 2
J C
Initial Temperature of Trial 1: 25.0344 Initial mass: 45 g Final Temperature Trial 1: 28.4660 Grams of salt: 1g
Initial Temperature of Trial 2: 19.6048 Initial mass: 45 g Final Temperature Trial 2: 22.4608 Grams of salt: 1g
−qdissolution = msolution∗C solution∗Δ T solution )+( C calorimeter∗Δ T calorimeter ¿ ¿
Trial 1: −qdissolution C))+ ( 129.29
J 46 g∗4.184 ∗¿ = (28.4660 C-25.0344 g∗C ¿
J ∗(28.4660 C−25.0344 C) ¿ = -1104.13 J C
Moles of LiCl: Change in Heat: kJ mol
1mol 1.000 g * = .0236 mol 42.39 ❑ −1104.13 J 1 kJ * =-46.79 .0236 mol 1000 J
Trial 2: −qdissolution = 46 g∗4.184 129.29
J ∗(22.4608C−19.6048 C) )+( g∗C ¿
J ∗(22.4608 C−19.6048 C)¿ = -918.93 C
Moles of LiCl: Change in Heat:
1mol 1.000 g * = .0236 mol 42.39 ❑ −918.93 J 1 kJ kJ * =-38.94 .0236 mol 1000 J mol
Average Change in Heat:
Cost of Lithium Chloride:
kJ −46.79+(−38.94) → exothermic (hot pack) =-42.87 mol 2 1.000 gave r age g =.000996 *90 mL *23.32 C mL∗C 45 mLaverage∗22.32 Caverage ΔT
=2.09 g 2.09 g *
$ 65.00 = $ .27salt + $ .73 labor +$ .36capital +$ .19 plastic bag = $1.55 Hot Pack 500 g
Initial Temperature of Trial 1: 20.4422 Initial mass: 45 g Final Temperature Trial 1: 22.9409 Grams of salt: 1g
Initial Temperature of Trial 2: 19.9749 Initial mass: 45 g Final Temperature Trial 2: 22.7206 Grams of salt: 1g
Trial 1: −qdissolution =
Trial 2: −qdissolution =
46 g∗4.184
46 g∗4.184
129.29
J ∗(22.9409C−20.4422 C) )+( g∗C ¿
J ∗(22.9409−20.4422)¿ = -806.58 C
Moles of Ca Cl2 : mol
1 mol 1.000 g * = .009011 110.98 g ❑
( 129.29
J ∗(22.7206−19.9749) )+ g∗C ¿
J ∗(22.7206−19.9749)¿ = -883.44 C
Moles of Ca Cl2 : mol
1 mol 1.000 g * = .009011 110.98 g ❑
Change in Heat:
−806.58 J 1 kJ * =-89.51 .009011 mol 1000 J
kJ mol
Change in Heat:
−883.44 J 1 kJ * =-98.04 .009011 mol 1000 J
kJ mol
Average Change in Heat:
Cost of Calcium Chloride:
kJ −89.52+(−98.04) → exothermic (hot pack) =-93.78 mol 2 1.000 g average g =.00110 *90 mL *20.21 C mL∗C 45 mLaverage∗20.21 Caverage ΔT
=2.00 g 2.00 g *
$ 31.70 = $ .13salt + $ .73labor + $ .36 capital + $ .19 plastic bag = $1.41 Hot Pack 500 g
Initial Temperature of Trial 1: 19.5830 Initial mass: 45 g Final Temperature Trial 1: 18.4221 Grams of salt: 1g
Initial Temperature of Trial 2: 19.9876 Initial mass: 45 g Final Temperature Trial 2: 18.7666 Grams of salt: 1g
+
J −qdissolution = 46 g∗4.184 g∗C ∗¿ (18.4221-19.5830)) ¿
( 129.29
J ∗(18.4221−19.5830) ¿ = 373.52 C
Moles of KCl: Change in Heat:
1mol 1.000 g * = .01341 mol 74.5513 g ❑ 373.52 J 1 kJ kJ * =27.85 .01341 mol 1000 J mol
Average Change in Heat:
+
J −qdissolution = 46 g∗4.184 g∗C ∗¿ (18.7666-19.9876)) ¿
( 129.29
J ∗(18.7666−19.9876)¿ = 392.86 C
Moles of KCl: Change in Heat:
1mol 1.000 g * = .01341 mol 74.5513 g ❑ 392.86 J 1 kJ kJ * =29.29 .01341 mol 1000 J mol
27.85+ 29.29 kJ → endothermic (cold pack) =28.57 2 mol
Cost of Potassium Chloride:
1.000 g average g =.00112 *90 mL *19.79 C mL∗C 45 mLaverage∗19.79 Caverage ΔT
=2.00 g 2.00 g *
$ 28.19 = $ .11salt +$ .73 labor +$ .36capital +$ .19 plastic ba g = $1.39 Cold Pack 500 g
Initial Temperature of Trial 1: 19.9470 Initial mass: 45 g Final Temperature Trial 1: 18.3374 Grams of salt: 1g
−qdissolution = 46 g∗4.184∗¿ (18.3374-19.9470 ))+ ¿ J ( 129.29 ∗(18.3374−19.9470)¿ = 517.895 C Moles of
NH 4 (NO3 ) :
1mol 1.000 g * =. 80.052 g ❑
01249 mol
Initial Temperature of Trial 2: 19.8505 Initial mass: 45 g Final Temperature Trial 2: 18.3356 Grams of salt: 1g
−qdissolution = 46 g∗4.184∗¿ (18.3356-19.8505 ))+ ¿ J ( 129.29 ∗(18.3356−19.8505)¿ = 487.425 C Moles of
NH 4 ( NO3 ) :
1mol 1.000 g * =. 80.052 g ❑
01249 mol
Change in Heat:
517.90 J 1 kJ kJ * =41.47 .01249 mol 1000 J mol
Average Change in Heat:
487.43 J 1 kJ kJ * =39.03 .01249 mol 1000 J mol
41.47+39.03 kJ → endothermic (cold pack) =40.25 2 mol
Cost of Ammonium Nitrate:
2.00 g *
Change in Heat:
1.000 g average =.00112*90 mL *19.90 C =2.00 g 45 mLaverage∗19.90 Caverage ΔT
$ 26.20 = $ .10salt + $ .73labor + $ .36 capital + $ .19 plasticbag = $1.38 Cold Pack 500 g
Discussion
In this experiment we were suppose to find the best salt to use in the cold and hot packs that the companies provided for us. To do this we used calorimetry which gave us the data to find if the reaction was exothermic or endothermic, to find the amount that was needed to reach a temperature of 0 for the endothermic reactions or a temperature of 65 degrees, and to find the final cost for those salts. The calorimeter technique that was used in the experiment was used to find the change in temperature of a system when a certain amount of the salt is added. One mistake that could have been made during the process include contamination of the salts. Another mistake that should be outlined is that some of the salt may have not gone into the calorimeter which would have made the “one gram” measurement more accurate. Conclusion In this experiment, various salts were tested for effectiveness and cost. The salt that was concluded to be the most effective as a cold pack was ammonium nitrate because it had the highest effectiveness for the lowest price. The cost of one ammonium nitrate packed came to be about $1.38 where the next lowest salt was potassium chloride and that came out to be $1.39. These prices reflected the cost of the amount of the material used to reach 0 degrees celsius. For the hot packs the most effective one came to be Calcium Chloride. The price of this hot pack to reach the asking temperature of 65 degrees was $1.41. The next lowest priced hot pack was Lithium Chloride at $1.55.
Bibliography Stanton B., Zhu L., Atwood C. (2010) . Experiments in General Chemistry Featuring MeasureNet 2nd Edition.
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