Brackenholt Heat Stroke Plan: A Lab Comparing Four Cooling Methods

Answer: Depends on the group's readings; the ice-water tub (D) should rank first and still air (A) last.
Each rate is the minute 0 temperature minus the minute 10 temperature, divided by 10 minutes, and should be reported in °C per minute. In most groups bottle D cools much faster than any other bottle, while bottle A falls by only a few degrees over the 10 minutes. Bottles B and C usually fall between A and D, and their order depends on the fan speed, how cold and how damp the towel was, and how much of each ice bag touched the bottle, so accept either order for B and C when it matches the student's calculated rates. Check that the ranking follows from the rates and not from a guess.

Answer: Depends on the group's readings; the rate from minute 0 to minute 2 should be clearly larger than the rate from minute 8 to minute 10.
In most groups the early rate is several times the late rate. Heat flows from the bottle into the ice water at a rate that depends on the temperature difference between them. At the start the bottle is far warmer than the ice water, so heat leaves quickly; by minute 8 the difference is much smaller, so heat leaves more slowly and the temperature falls fewer degrees each minute. Accept any explanation that ties the slowdown to the shrinking temperature difference. If a group's late rate is not smaller, check whether its tub ran out of ice or whether readings were taken without stirring.

Answer: 5 °C per minute.
The temperature fell from 44.0 °C to 24.0 °C, a drop of 20.0 °C, over 4 minutes, and 20.0 ÷ 4 = 5.0 °C per minute. Written as an average rate of change, the value is -5.0 °C per minute, because the temperature decreased. Accept either form when the student states that the bottle was cooling.

Answer: 1.25 °C per minute.
The temperature fell from 14.0 °C to 9.0 °C, a drop of 5.0 °C, over 4 minutes, and 5.0 ÷ 4 = 1.25 °C per minute. This late rate is one-fourth of the early rate from question 3 (5.0 ÷ 1.25 = 4), even though the method never changed, because by minute 8 the bottle was much closer to the temperature of the ice water.

Answer: 6.5 °C.
The differences between the bottle and the 4.0 °C ice water are 44.0 - 4.0 = 40.0 °C, 24.0 - 4.0 = 20.0 °C, 14.0 - 4.0 = 10.0 °C, and 9.0 - 4.0 = 5.0 °C. The amount removed changes from one interval to the next (20.0, then 10.0, then 5.0 °C), so the change is not linear. The difference is multiplied by the same factor, one-half, over every 4-minute interval, and equal factors over equal intervals describe exponential decay. Continuing the pattern, the difference at minute 16 is 5.0 × 0.5 = 2.5 °C, so the bottle reads 4.0 + 2.5 = 6.5 °C. A student who subtracts another 5.0 °C and predicts 4.0 °C has treated the change as linear; the bottle keeps approaching the temperature of the ice water without reaching it in this pattern.

Answer: The tub surrounds nearly all of the bottle with moving cold water, while the ice packs touch only small patches of it.
Heat leaves the bottle by conduction into whatever touches it. Water conducts heat far better than air and can take in a large amount of heat, and the melting ice absorbs heat while staying at its melting point, which keeps the ice water cold. In the tub, cold water touches almost the whole surface of the bottle, and stirring carries warmed water away and brings colder water in. The ice bags touch only two narrow strips through plastic and small air gaps, so most of bottle C's surface still loses heat only to the room air. Bottle B lost heat mainly by evaporation: each bit of water that evaporates from the towel takes heat with it, and the fan carries away the moist air so that more water can evaporate. The towel also started cold, so it absorbed some heat directly in the first minutes. This is the same process by which evaporating sweat cools the skin. Accept answers that name contact area, the difference between water and air in carrying heat, and the effect of stirring; strong answers also explain evaporation for bottle B.

Answer: Accept any two differences that are correctly explained.
An athlete's body keeps producing heat, especially right after hard exercise, while the bottle only loses heat, so a person cools more slowly than the bottle's numbers suggest. Blood carries heat from deep in the body out to the skin, so how fast the core cools depends on circulation as well as on the cooling method. An athlete has many times the mass of a 450 mL bottle, so every method takes much longer in minutes, which means the lab supports comparing methods but not predicting treatment times. The neck, armpits, and groin are only a small share of the body's skin, which makes the small contact area of ice packs an even larger drawback on a person than on the bottle. A person in a tub must also have the head supported above the water and must be watched by a trained adult the whole time. Accept any two differences with a correct statement of how each one changes the result or its use.

Answer: A tub of ice water ready before every hot-weather practice, with 911 called and cooling started on site.
Strong recommendations cite a rate, such as bottle D's 10-minute rate compared with bottle C's, or the trainer's 5 °C per minute early rate from question 3, and then list the steps of the plan: call 911 at once; start cooling on site right away rather than waiting for the ambulance; remove the helmet and as much equipment and clothing as can be taken off quickly; place the athlete's trunk and limbs in the ice water with the head supported above the surface; stir the water and keep an adult with the athlete the whole time; and do not give anything to drink to an athlete who is confused or drowsy. Trained medical staff, such as the athletic trainer or emergency responders, decide when cooling can stop. If a tub is not possible, rotating ice-water towels and ice bags over as much skin as possible is the backup. Also accept plans that add preparation steps for the fictional Brackenholt field, such as filling the tub and stocking ice before practice on hot days, and prevention steps such as water breaks and gradual acclimatization. Do not accept a plan that relies on ice packs alone or delays cooling until the athlete reaches the hospital.