- #1

Brownstone

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## Homework Statement

A.[/B] Someone is walking outside when the temperature of the air is -10°C. The metabolic rate of walking is 140 Cal/m

^{2}hr, and the usual muscle efficiency (20%) applies. They are completely covered by clothing with a coefficient of thermal conductivity of 0.36 Cal-cm/m

^{2}hr °C. Their total surface area is 1.7m

^{2}, and this area can be used as the area through which heat is being conducted away from the body. Their skin temperature is the same as the inside surface of the clothing, 35°C. The temperature of the outside of the clothing is -10°C. It is dark outside. Their mass is 75 kg. Ignore heat lost to breathing. How thick does the down have to be to maintain a constant internal body temperature of 37°C?

**B.**Say the person in Part A increases their exertion so that their metabolic rate rises to 500 Cal/m

^{2}hr. The conditions are the same as described above with one exception: assume that the clothing is only 0.1 cm thick. At this thickness, they are definitely losing heat to the environment. This heat loss is the only means they have to cool themselves (humidity too high inside the clothes to evaporate sweat, ignore heat lost to breathing). At what rate will their temperature rise/fall? Use that the specific heat of the body, on average is 8.3 X 10

^{-4}Cal/g °C.

## Homework Equations

(q/t) = ((k

**⋅**A) / L)(T

_{CORE}- T

_{AIR})

where

q = heat

t = time

k = thermal conductivity of clothes

A = surface area

L = thickness of down required to maintain 37 °C body temperature

T

_{CORE}= body temperature

T

_{AIR}= air temperature

(q/t) =

*m*c'ΔT

where

q = heat

t = time

*m*= mass

c' = specific heat capacity

ΔT = change in temperature

## The Attempt at a Solution

A.[/B]

__calories burned with 20% efficiency:__

20/100 = x/140

x = 28 Cal = q/t

__q/t formula to find L:__

q/t = ((k

**⋅**A) / L)(T

_{CORE}- T

_{AIR})

28 = ((0.36 ⋅ 1.7) / L)(37 - (-10))

solving for L:

L = ((0.36 ⋅ 1.7) / 28)(37 - (-10))

L = 1.02739 cm

**The above number seems reasonable for a jacket thickness.**

B.

B.

__calories burned with 20% efficiency:__

20/100 = x/500

x = 100 Cal = q/t

__taking c' into account:__

(q/t) =

*m*c'ΔT

100 = 75 ⋅ 8.3⋅10

^{-4}⋅ ΔT

solving for ΔT:

ΔT = 100 / (75 ⋅ 8.3⋅10

^{-4})

ΔT = 1606.43

*Maybe I'm bad at visualizing, but 1606.43 ΔT seems ridiculously large. Also, would I continue like this, or should I have substituted ΔT with the actual temperatures I've been given (37-(-10))?*__finding t:__

q/t = ((k

**⋅**A) / L)(T

_{CORE}- T

_{AIR})

100/t = ((0.36 ⋅ 1.7) / L)(37 - (-10))

solving for t:

t = ((100 ⋅ 0.1) / (0.36 ⋅ 1.77 ⋅ (37 - (-10)))

t = 0.347657

**At this point, I'm a bit confused as to whether this t makes sense conceptually. I can plug and chug (can I?), but I'd like to know what it is I'm doing. It's getting muddy here.**__rate of rise and fall of T:__

ΔT / t = 1606.43 / 0.347657

ΔT / t = 4620.74

**This seems too high. I'm not even sure what they mean by rate of rise and fall of T, but this seems way too high.**

__________________________

After this, I thought that I should've instead taken 80% instead of 20%, since I think I'm accounting for heat, so I tried that. Those calculations gave me ΔT = 6425.7, t = 1.39063, leading, again, to a rate of 4620.71 (that I should've seen).

**Please help.**