Inducing Therapeutic Hypothermia
Sarah Byce
Herrington
CHM 201-10
5 November 2012
Byce 2
Statement of Question
How many instant cold packs will it take to induce therapeutic hypothermia on a patient
who has suffered from a heart attack?
Background
For this lab, instant ice packs were cut open; the ammonium nitrate crystals were mixed
with water to determine the amount of heat given off. This would help calculate the number
of ice packs needed to cool the body to therapeutic hypothermia. To do these calculations, a
basic understanding of the heat equation is needed (Appendix: Equation 1).
Procedures
In this experiment, 100mL of tap water was poured into a plastic graduated cylinder. The
temperature was taken of this water while in the cylinder. The entire contents, minus the
water bag, of the instant cold pack were weighed. The total weight of the ammonium nitrate
crystals was then divided by three to get three trials (Appendix: Calculation 1). A third of
ammonium nitrate was then weighted out and poured into a Styrofoam coffee cup. The
coffee cup was used to keep in as much as the heat as possible as opposed to a glass beaker.
The water from the graduated cylinder was poured into the coffee cup. A thermometer was
immediately placed in the coffee cup while the mixture was continuously being stirred. Once
the thermometer temperature leveled off, the temperature was recorded (Appendix: Table 1)
and the experiment was repeated two more times. After all the trials were completed
calculations needed to be done to calculate the heat given off by the entire ice pack.
Data and Observations *see Appendix
Calculations *see Appendix
Analysis and Discussion
1. The energy change in Joules when one gram of ammonium nitrate is dissolved:
𝑡𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 1 𝑏𝑎𝑔 (𝐴𝑝𝑝𝑒𝑛𝑑𝑖𝑥: 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2)
10,460 𝐽
=
= 260.78 𝐽
𝑡𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑎𝑚𝑚𝑜𝑛𝑖𝑢𝑚 𝑛𝑖𝑡𝑟𝑎𝑡𝑒
40.11 𝑔
2. The energy change in Joules of the entire cold pack using 100 mL of water (Appendix:
Equation 2):
(3,514.56 𝐽) + (3,430.88 𝐽) + (3,514.56 𝐽) = 10,460 𝐽
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3. The number of our cold packs needed to induce therapeutic hypothermia on the patient:
𝑏𝑜𝑑𝑦 ℎ𝑒𝑎𝑡 (𝐽)(𝐴𝑝𝑝𝑒𝑛𝑑𝑖𝑥: 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 2) 840,000 𝐽
=
=≈ 80 𝑐𝑜𝑙𝑑 𝑝𝑎𝑐𝑘𝑠
ℎ𝑒𝑎𝑡 𝑜𝑓 𝑜𝑛𝑒 𝑐𝑜𝑙𝑑 𝑝𝑎𝑐𝑘 (𝐽)
10,460 𝐽
4. The cold packs on the plane have 200 grams of ammonium nitrate, because of this less
cold packs would be needed:
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑜𝑢𝑟 𝑎𝑚𝑚𝑜𝑛𝑖𝑢𝑚 𝑛𝑖𝑡𝑟𝑎𝑡𝑒 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 𝑎𝑚𝑚𝑜𝑛𝑖𝑢𝑚 𝑛𝑖𝑡𝑟𝑎𝑡𝑒
=
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑐𝑜𝑙𝑑 𝑝𝑎𝑐𝑘
𝑋
40.11𝑔
200 𝑔
=
… 40.11𝑋 = 2,092,000 … 𝑋 = 52,156.57 𝐽
10,460 𝐽
𝑋
𝑏𝑜𝑑𝑦 ℎ𝑒𝑎𝑡 (𝐽)(𝐴𝑝𝑝𝑒𝑛𝑑𝑖𝑥: 𝐶𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 2)
840,000 𝐽
=
≈ 16 𝑐𝑜𝑙𝑑 𝑝𝑎𝑐𝑘𝑠
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑝𝑙𝑎𝑛𝑒 𝑖𝑐𝑒 𝑝𝑎𝑐𝑘
52,156.57 𝐽
5. When compared to the class data, when sorted by amount of energy change for one gram
of ammonium nitrate (Appendix: Table 2), our data was on the high end of the scale.
There was only one group that had a higher value for energy being released. When the
data was sorted by the number of our cold packs needed, our data was fairly in the middle
(Appendix: Table 3). From this sort, it was clear to see the lower the energy change for
ammonium nitrate the more ice packs needed to induce hypothermia. It was also clear to
see a direct relationship between the number of plane cold packs needed compared to
the number of your cold packs needed, which only makes sense, the less of your cold
packs, the less plane cold packs needed. There were a couple outliers however; one
group said there should be 61 cold packs where another said 1,043 packs were needed.
6. Some of the things that could contribute to the calculation differences are inconsistent or
incorrect measuring. There were different procedures for determining energy change and
number of cold packs. A major difference was the size of the ammonium nitrate crystals;
some had white crystals where others had small brown balls.
7. Some things that could improve the experiment would be, having more cold packs to test.
If this was done, my group wouldn’t have had to divide by three, we could have used an
entire packet. After doing the experiment, I realized we should have measured the
amount of water in the pack, to know for sure how the cold pack would have reacted.
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This could also affect how cold the pack gets. Maybe the more water the colder or
warmer the ice pack gets.
Conclusion
In conclusion, it was determined, from our procedure, it takes about 261 J to cool one gram
of ammonium nitrate. It would take 80 of our cold packs to induce therapeutic hypothermia
and about 16 plane cold packs to induce the hypothermia. Based on the class data, the
greater the energy change the greater the number of cold packs needed. The goal for this
experiment was met when the conclusion, or understanding, about how many cold packs
would be needed to induce hypothermia as well as how this is happening. This lab taught
that when water is added to ammonium nitrate it becomes colder and there is an energy
change between these two substances. This concept is important for understanding
temperature regulation, as well as how a cold pack works. The experiment went smoothly,
but could be improved upon. It may have been better to get a few cold packs to work with so
know a better idea of the energy change.
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Appendix A: Tables, Calculations, and Equations
Equation 1: Determining Heat
𝑞 = 𝑚𝑐∆𝑇
q= heat (Joules); m= mass (grams); c= specific heat (J/g*°C); ∆T= change in temperature (°C)
Calculation 1: Trial Weights
𝑡𝑜𝑡𝑎𝑙 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑖𝑐𝑒 𝑝𝑎𝑐𝑘 (𝑔)
𝑇𝑟𝑖𝑎𝑙 𝑊𝑒𝑖𝑔ℎ𝑡 =
3
40.11𝑔
= ≈ 13.37𝑔
3
Table 1: Individual Data
Equation 2: Total Energy Change
𝑇𝑜𝑡𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦 𝐶ℎ𝑎𝑛𝑔𝑒 = 𝑇𝑟𝑖𝑎𝑙 𝑂𝑛𝑒 + 𝑇𝑟𝑖𝑎𝑙 𝑇𝑤𝑜 + 𝑇𝑟𝑖𝑎𝑙 𝑇ℎ𝑟𝑒𝑒
Calculation 2: Heat Removed from the Body
𝑞 = 𝑚𝑐∆𝑇
q= ?; m= 60kg; c= 3.5 J/g*°C; ∆T= (37°C-33°C)
𝑞 = (60,000𝑔)(3.5 𝐽/𝑔 ∗ °𝐶)(37°𝐶 − 33°𝐶) = 840,000 J = 840 kJ
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Table 2: Sorted Class Data- Energy Change
Table 3: Sorted Class Data- Number of Our Cold Packs Needed