Thermal conductivity for mountain climber

[Vanessa23]

Homework Statement

A mountain climber wears a goose down jacket 3.28 cm thick with total surface area 1.10 m2. The temperature at the surface of the clothing is -19.3°C and at the skin is 36.0°C. Determine the rate of heat flow by conduction through the jacket assuming it is dry and the thermal conductivity, k, is that of down.
Part 2:
Determine the rate of heat flow by conduction through the jacket assuming the jacket is wet, so k is that of water and the jacket has matted down to 0.462 cm thickness.

Homework Equations

thermal conductivity of goose down is .025 J/(mKs)
thermal conductivity of water is .561W/(mKs)

phi dot = KA (delta T/delta X)

K is goose down (.025 j/smk)
A is area
T is temperature
x is distance heat flows

Part 2
The rate of heat flow= (the rate of heat flow you found in
part 1) x (k water/ k goose down) x (thickness of goosedown/
thickness of wet jacket)

The Attempt at a Solution

.025*(.0328*1.10)*(36+19.3/.0328)

I get the rate is 1.52 W, yet it is the wrong answer. If I had that answer I could use it to get part 2 where I would do the following
rate=rate1*(.561/.025)*(.0328/.00462)

Please let me know what I am doing wrong! Thank you!

 

[Shooting Star]

Why is the distance between the two surfaces in the numerator? Delta_x should be in the bottom.

 

[ogb p]

I would like to clarify that the given information and equations seem to be correct. However, the approach for calculating the rate of heat flow through the jacket may need to be revised.

For part 1, the correct equation to use would be phi dot = (K * A * delta T) / delta X, where K is the thermal conductivity of goose down, A is the surface area of the jacket, delta T is the temperature difference between the outer and inner surfaces of the jacket, and delta X is the thickness of the jacket.

Using the given values, the rate of heat flow would be calculated as follows:

phi dot = (0.025 J/(mKs) * 1.10 m2 * (36.0°C - (-19.3°C))) / 0.0328 m

= 1.45 W

For part 2, the approach is correct, but the calculation for the rate of heat flow in part 1 needs to be used. The equation for part 2 would be:

rate = rate1 * (k water / k goose down) * (thickness of goose down / thickness of wet jacket)

= 1.45 W * (0.561 W/(mKs) / 0.025 J/(mKs)) * (0.0328 m / 0.00462 m)

= 1.45 W * (22.44) * (7.09)

= 226.5 W

Therefore, the rate of heat flow through the wet jacket would be 226.5 W.

I hope this helps clarify the approach and calculations for this problem. It is also important to note that the given values and equations may not account for all variables and assumptions, so the results may vary slightly.