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Save on Textbooks! Other   Other Homework Help ( 742 ) Math   Math- Algebra 1 ( 1628 )   Math- Algebra 2 ( 807 )   Math- Calculus ( 280 )   Math- Geometry ( 398 )   Math- Other ( 1004 )   Math- Precalculus ( 299 )   Math- Trigonometry ( 190 ) Biology   Biology- Animal ( 105 )   Biology- Cell ( 378 )   Biology- Ecological ( 63 )   Biology- General ( 414 )   Biology- Microbiology ( 69 ) Business   Business- Accounting ( 390 )   Business- Finance ( 290 )   Business- Other ( 212 ) Chemistry   Chemistry- Biochem ( 111 )   Chemistry- Organic ( 95 )   Chemistry- Other ( 768 ) Economics   Economics- Macroeconomics ( 519 )   Economics- Microeconomics ( 361 ) Essay Service   Essay Correction ( 48 )   Essay Writing ( 225 ) History   History- World ( 459 )   US History- Post 1877 ( 278 )   US History- Pre 1877 ( 205 ) Language   English ( 442 )   Foreign Languages ( 80 )   World Literature ( 117 ) Physics   Physics- Electricity, Magnetism ( 158 )   Physics- General ( 905 )   Physics- Mechanical, Heat, Sound ( 234 )   Physics- Wave, Quantum Physics ( 59 ) z Medical Questions Your Question 1. The problem statement, all variables and given/known data 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. 2. Relevant 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) 3. 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! Please Login to Answer the Question LOGIN

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