Any idea how to do this project. Thermal Insulator Rules: Container size: 500 ml

Question

Any idea how to do this project. Thermal Insulator Rules: Container size: 500 ml water bottle (bring two bottles). Insulator materials: items that existed before 1912: (No Styrofoam, no wetsuits, no plastic, no boats, etc). Other restrictions: if you can't wear it and still swim, then you can not use it. Level III Thermodynamics: increase the time that heat drops 37 degree C to 20 degree C by 25%. Winner, the best insulator (the longest time to reach the 37 degree C to 20 degree C heat loss) Level III Optional. Stay true to the original concept of the project. (If you can't wear it and swim, it shouldn't be used in your project).

Explanation / Answer

For heat transfer across an object of material with heaqt conductivity, k, surface area = A, thickness t is
Q' = kA(dT)/t [ dT is the temperature difference across material] [ Q' is rate of heat transfer]
so, to increase the time by which heat transfered across dT = (37 - 20) = 17 C by 25 p.c.
for same value of t , dT and A, k has to be increased by 25 pc
hence we need to find materias whith physical property of that of a cloth except, it should have decreased constant of thermal conductivity by (1 - 1/1.25)*100 = 20 pc
so, from tables of thermal conductivity, something like carbon fibre cloth would be anice choice to have a decreased coefficeint of thermal conductiviy.

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