The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0°C and that the outer surface of the skin is at 30.0°C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m2 and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person's skin?

Video transcript

Hey everyone today, we're dealing with the problem about thermal conductivity. So we're being told that a food container has a total area of 0.242 m squared, A warm liquid is placed in the container with a temperature of 42°C. And once thermal equilibrium has been established, the temperature of the outer wall of the container was found to be 35°C. We're also told that the container has a thickness of 2.4 mm and loses heat to the surroundings at a rate of 191 watts. With all this information, we're being asked to find the thermal conductivity of the material that the container is made of. So for this problem, we have a few values. We have the heat flow rate, the heat flow rate. We have the highest temperature recording. We have the lowest temperature recorded. We have the thickness and we have the area of the container. And we're being asked to find the thermal conductivity. So let's write all this out. This is 191 watts. Is the Heat. Low rate. The highest temperature recorded Was 42°C. The lowest was 35°C. The thickness of the container was 2.4 millimeters. But we want this in meters because we want our S. I. Units. So we just multiply this by we have one m for every to the third millimeters. So our mm will cancel out and we'll be left with 2.4 times 10 to the negative 3rd m. The area of the container is 0.24, two meters squared. And with this, we can use the formula, we can use the formula H is equal to K. A delta T over L. Where delta T. Is the change in temperature between the highest and lowest points. However, we have all the values except for K, which is thermal conductivity and what we're looking for. So let's rearrange this to solve for K. And by doing so we get that K is equal to H. L over a delta T. So substituting in our values, we can then say that K is equal to 191 watts, multiplied by a thickness of 2.4 times to the negative three m negative three m divided by an area of 0.242 m two. And the the change in temperature should be in kelvin but the difference in temperature between centigrade and kelvin will be the same. So the change in temperature and let's write this in red. The change in temperature will be equal to the highest temperature minus the lowest temperature, Which is 45 degrees Celsius minus 35 degrees Celsius, which equals seven degrees Celsius, which is the same thing as seven degrees kelvin Which is the same thing as the change in 7° Kelvin, I should say, or seven Kelvin. So we'll write that here and the change in temperature is seven Kelvin uppercase K. Not to be confused with the thermal conductivity. So finally calculating our final value, we get that K is equal to 0.271 Hoops 271. What per meter kelvin? Which gives us answer choice. D. I hope this helps. And I look forward to seeing you all in the next one.