- 06 Jul '10 23:35So you are leaving a store with a cold bottle of soda, one liter at it is at 5 degrees C.

The outside air is 50 degrees C, 122 degrees F, very hot day indeed. There is no wind. So you think, this may be like the running in the rain problem.

If I walk at a normal pace, and have to cover 200 meters, does the soda remain colder if I walk at a normal pace or if I run as fast as I can? Which technique gets me the coldest soda when I get to my nice air conditioned car?

Of course it's a eco unfriendly bottle about 0.5 mm thick, so the heat gets inside very quickly. It has a temperature gauge inside that registers in one tenth of a second in response to temperature changes and is sensitive to 1/1000 of a degree C(there are thermo probes that sensitive in fact). So which way will register the lowest temperature when you get to your car. If you walk slowly, you make 1 meter per second and if you sprint, you make 10 meters per second. - 07 Jul '10 01:28

Wouldn't you just be "pushing" the heat in at a faster rate?*Originally posted by AThousandYoung***It seems obvious to me that the answer would be to run. First of all you spend less time in the heat; second of all it creates a breeze as you run through the air.**

It seems obvious. Why would you think walking would work better? - 07 Jul '10 16:17

The "breeze" you are creating would have a heating effect, not a cooling one.*Originally posted by AThousandYoung***It seems obvious to me that the answer would be to run. First of all you spend less time in the heat; second of all it creates a breeze as you run through the air.**

It seems obvious. Why would you think walking would work better? - 07 Jul '10 18:20Interesting question. Thoroughly extraneous (is anyone really going to sit down and do a heat transfer calculation for a 200 m sprint with a pop can?), but interesting nonetheless.

As I recall from my heat transfer course, energy will can transferred to the can in up to 3 ways:

a) conduction;

b) convection;

c) radiation.

In this case, there is no conduction (we assume the hand/can system is adiabatic), so that leaves us with convection and radiation. The rate of heat transfer due to convection "C" into the can is given by:

C = hA(Tb-Ts)

where:

h = convective heat transfer coefficient (typically 10-100 W/m2K for air)

A = surface area (assume 1 m2)

Tb = bulk fluid temperature = 323 K

Ts = surface temperature = 278 K

When walking we will assume h = 10 W/m2K, and when running we will assume h = 50 W/m2K. When walking, the rate of convective heat transfer is then:

C = 10 * 1 * (323 - 278) = 450 W

When running, we have:

C = 50 * 1 * (323 - 278) = 2250 W

That's quite a range! Moving on, the rate of heat transfer to the can due to radiation "R" is given as (as best as I can recall...someone fact check this equation please!):

R = e * s * A * (Tb^4 - Ts^4)

where:

e = emissivity coefficient (assume e = 1 for the shiny can)

s = Stefan-Boltzmann constant = 5.67 x 10-8 W/m2K4

A = area (assume 1 m2)

Tb = bulk fluid temperature = 323 K

Ts = surface temperature = 278 K

Therefore, we have:

R = 1 * 5.67x10-8 * 1 * (323^4 - 278^4) = 279 W

Quite a bit less! Now, the total time for the trip when walking is 200m / 1 m/s = 200 s, and 200m / 10 m/s = 20 s when running. Assuming the can stays roughly the same temperature for the entire trip (and calculates its own change in temperature only at the end of the trip ), the heat transfer to the can "Q" during the walk is:

Q(walk) = (C+R)dt = (450 + 279)*200 = 145.8 kJ

During the run we have:

Q(run) = (2250 + 279)*20 = 50.6 kJ

So it seems that**running**is indeed the best option. - 07 Jul '10 18:38

One question about the radiative equation, you assume a surface area of one meter squared. Is that supposed to be the surface area of the can? Since we specified a one liter can, that seems way out of line with the actual surface area which would probably be more like 1/10th of a square meter, maybe a bit more.*Originally posted by PBE6***Interesting question. Thoroughly extraneous (is anyone really going to sit down and do a heat transfer calculation for a 200 m sprint with a pop can?), but interesting nonetheless.**is indeed the best option.[/b]

As I recall from my heat transfer course, energy will can transferred to the can in up to 3 ways:

a) conduction;

b) convection;

c) radiation.

In this case, there is no co ...[text shortened]... = (2250 + 279)*20 = 50.6 kJ

So it seems that [b]running - 07 Jul '10 18:47

I only assumed 1 m2 because it makes the calculations that much easier. If I had kept the equations in their algebraic forms until the end, it would be easier to see that both terms contain the same "A". Changing this parameter would affect the magnitude of the answers, but the ratio between them would be the same. If we use 0.1 m2, then the answers would be 14.8 kJ and 5.1 kJ, respectively. This is more in line with what you'd expect, considering the heat capacity of the soda in a 335 mL soda can is somewhere in the neighbourhood of 1.5-2 kJ/K (representing temperature increases of about 8 C and 3 C, respectively).*Originally posted by sonhouse***One question about the radiative equation, you assume a surface area of one meter squared. Is that supposed to be the surface area of the can? Since we specified a one liter can, that seems way out of line with the actual surface area which would probably be more like 1/10th of a square meter, maybe a bit more.** - 07 Jul '10 18:48

Heheh...be my guest! (It should be at least as straightforward as the other calculations, but now you're introducing heat transfer resistance from hand to can, can to soda, worrying about the Biot number for the soda, etc...just try and make as many simplifying assumptions as you can!).*Originally posted by iamatiger***Say you have an inefficient metabolism and your hand gets very hot when you run?** - 07 Jul '10 20:15 / 2 edits

I did say 'bottle'. I said that because when I walk around with a bottle of soda I hold it by the neck which is usually above the soda line just to minimize shielding effects of your hand for this problem. Your hand of course would be at about 38 Degrees C and the air at 50 would create a bit of thermal shield.*Originally posted by PBE6***Heheh...be my guest! (It should be at least as straightforward as the other calculations, but now you're introducing heat transfer resistance from hand to can, can to soda, worrying about the Biot number for the soda, etc...just try and make as many simplifying assumptions as you can!).** - 08 Jul '10 14:53The thing I like about these engineering-type problems is that they require judgement in choosing the correct model before any math gets done. Since there are numerous ways to model these situations, to varying degrees of accuracy, the problem is still fun to solve even after someone else has already posted a solution.

As a follow-up, why doesn't someone come up with a model that involved conduction from the hand to the bottle, with T(hand) constant at 38 C and a contact area of 40 in2? You'll have to make some sort of estimate for the heat conduction coefficient, but the key is making a reasonable argument to support your choice. Checking the final answer to see if it's reasonable is an important step as well. If it isn't, maybe changing some of your assumptions will help. The possibilities are endless!