Change the chapter
Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is $1.40 \textrm{ m}^2$?
Question by OpenStax is licensed under CC BY 4.0.


Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 14, Problem 39 (Problems & Exercises) (1:37)

Sign up to view this solution video!


0 votes with an average rating of .

Calculator Screenshots

OpenStax College Physics, Chapter 14, Problem 39 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. A person is covered in wool clothing that has a thickness of 2 centimeters which is 2 times 10 to the minus 2 meters and they are losing heat at a rate of 50 watts and the surface area of this person is 1.4 square meters and since they are wearing wool we can put a thermal conductivity of 0.04 Joules per second per meter per Celsius degree. Now our goal is to figure out what is the temperature difference that must exist in order for this rate of heat transfer given a certain thickness area and thermal conductivity. So, our formula for rate of heat transfer through conduction is this. So Q over t equals the thermal conductivity times the area times change in temperature divided by thickness. And so we will solve for Delta T by multiplying both sides by d over KA and all those factors cancel on this side and we are left with delta T isolated and we then switch the sides around so that the unknown is on our left. So delta T equals d over KA times Q over t. So Q over t is a single number 50 watts, that’s the rate of heat transfer, and we multiply that by the thickness of 2 times 10 to the minus 2 meters divided by the thermal conductivity for wool 0.04 and then divide also by 1.4 square meters and this gives 20 Celsius degrees must be the temperature difference between the person side of the clothing in the outside.