Calculating thickness of insulation, rate of T rise and fall

[Brownstone]

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/m²hr, and the usual muscle efficiency (20%) applies. They are completely covered by clothing with a coefficient of thermal conductivity of 0.36 Cal-cm/m²hr °C. Their total surface area is 1.7m², 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/m²hr. 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) = mc'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.
calories burned with 20% efficiency:
20/100 = x/500
x = 100 Cal = q/t

taking c' into account:
(q/t) = mc'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.

 

[haruspex]

The metabolic rate of walking is 140 Cal/m2hr, and the usual muscle efficiency (20%) applies.

I'm not sure you have interpreted this correctly. (And do you mean 140 Cal/m²hr, or just Cal/hr?)

calories burned with 20% efficiency:
20/100 = x/140
x = 28 Cal = q/t

The energy consumed gets turned into useful work done plus waste heat. The 20% efficiency implies 20% is the useful work done and 80% goes as heat. The heat is the part that goes into keeping the walker warm.

 

[haruspex]

q/t = ((k ⋅ A) / L)(TCORE - TAIR)
28 = ((0.36 ⋅ 1.7) / L)(37 - (-10))

Are you sure it's the core temperature you need here?

 

[Brownstone]

I'm not sure you have interpreted this correctly. (And do you mean 140 Cal/m²hr, or just Cal/hr?)

The energy consumed gets turned into useful work done plus waste heat. The 20% efficiency implies 20% is the useful work done and 80% goes as heat. The heat is the part that goes into keeping the walker warm.

So I should've used the 80% instead of the 20%?

 

[haruspex]

So I should've used the 80% instead of the 20%?

I think so.

 

[Brownstone]

"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)."

I'd already tried that.

 

[haruspex]

"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)."

I'd already tried that.

I hadn't seen that because it was in the section on part b, which I'd not got to. But it makes no sense to me that the walker is generating four times the heat and it barely makes any difference to the answer. Using the 80% what do you get for part a? And you have not answered post #3, where I point out another probable error, but of less significance.