A mountain climber wears a goose-down jacket 3.8 cm thick with...

Question

A mountain climber wears a goose-down jacket 3.8 cm thick with total surface area 0.95 m². The temperature at the surface of the clothing is -18°C and at the skin is 34°C. Determine the rate of heat flow by conduction through the jacket assuming (a) it is dry and the thermal conductivity k is that of goose down, and (b) the jacket is wet, so k is that of water and the jacket has matted to 0.50 cm thickness.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice prom together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, a deep sea diver jumps in the sea wearing a standard diving wet that has a thickness of 0.7 centimeters and a total surface area of 1.8 m squared. The surface temperature of the standard diving suit is negative two °C. And when it is on the skin, its temperature is 37 °C determine the heat rate flow by conduction through the wet suit assuming a it is dry. And the thermal conductivity K is that of a standard diving suit. Also, if the heat rate flow is calculated by conduction through the wet suit when B the suit is moist and the value of K is equal to that of water. Also consider if the suit has been compressed to a thickness of 0.35 centimeters. So that's our angle. Our angle is we're trying to solve for two separate conditions. So we're trying to figure out what the heat rate flow is by conduction through the wetsuit if it's dry. So that the diving suit is dry, that's our first condition. So we're trying to figure out what the heat rate of flow by conduction through the wetsuit when it's dry. And we're also trying to figure out what the heat rate of flow by conduction is if this diving suit is wet and that's it. We're trying to solve for these two separate conditions for the rate the heat rate of flow. Awesome. So now that we have an idea of what we're solving for, we can look at our multiple choice answers to see what our final answer pair might be. Let us note that our first answer is for the dry condition. And our second answer is for the wet condition and they're all in the same units of watts. So A is 250 15,000. B is 470 11,633 C is 501 and 11,000, 633. And finally, D is 501 and 20,000. Awesome. So first off, let us recall and use the equation to solve for the rate of heat flow by conduction, which states that the rate of heat flow, which is Q divided by T, where Q denotes the heat and T denotes the time is equal to K where K is the thermal conductivity multiplied by A where A is the surface area multiplied by T two minus T one, where T two minus T one is our change in temperature. So this is all divided by D where lowercase D is the thickness. We also need to note that the thermal conductivity of the diving suit is approximately 0.0 five watts per meter multiplied by Calvin. And the thermal conductivity of water is about 0.58 M. So 0.58 in its units are watts per meters multiplied by Kelvin. So now at this point, now that we know the thermal conductivity of water and the thermal conductivity of the diving, so we can start solving for part A officially. So with that in mind, we need to consider the dry diving suit condition first. So now at this point, we need to first convert the thickness value from centimeters to meters. So let us note that D or thickness is equal to 0.7 centimeters, which is also equal to, I'm just gonna perform the, the conversion for you. So it's 0.007 m. But you can use dimensional analysis or you can quickly look up this conversion. I've just given it to you to speed up our solving process. So now at this point, we can plug in all of our known variables to help us solve for Q divided by T, which is our final answer for part A. So Q divided by T is equal to zero point zero five watts per meter multiplied by Calvin multiplied by our surface area, which is one point 8 m squared, which this value is given to us in the problem itself multiplied by our change in temperature which is 37 °C minus negative two degrees Celsius. Awesome. So and then we need to divide everything by our thickness value which our thickness value as we should be able to recall is 0.007 m. Awesome. Therefore, when we plug into our calculator, we will find that Q of T or Q divided by T is equal to 501 watts. And that's it. You've solved for part A hooray, we did it. So now we could start solving for part B. So now we need to consider the wet diving suit condition. So now considering the wet or moist diving suit condition, we also need like our part A, we also need to convert our new thickness value from centimeters to meters. So we need to note that d in this particular case for this condition is 0.35 centimeters as given to us by the pro which is equal to 0.0035 m. So now at this point like part A, we need to plug in all of our known values to solve our rate of heat flow equation and solve for Q divided by T which is our final answer for part B. So Q divided by T is equal to 0.58 and its units are watts per meters multiplied by Calvin multiplied by our surface area, which once again is 1.8 m squared multiplied by our change in temperature which is 37 °C minus negative two °C, all divided by our new thickness value, which we should be able to recall that our thickness value is 0.00 3 5 m. So when we plug that into our calculator, we will find when we round to the nearest hole number, we will get 11,633 watts. And that's it. That's our answer for part B. Hooray. We did it. So looking at our multiple choice answers, the correct answer has to be the letter C dry is 501 watts and wet is 11,633 watts. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

Key Concepts

Heat Conduction

Thermal Conductivity

Insulation Thickness

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