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A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in $-1.00 ^\circ\textrm{C}$ water. The walrus’s internal core temperature is $37.0^\circ\textrm{C}$, and it has a surface area of $2.00 \textrm{ m}^2$. What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood?
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Walrus on ice. (credit: Captain Budd Christman, NOAA Corps)
Walrus on ice. (credit: Captain Budd Christman, NOAA Corps)
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Final Answer

$10 \textrm{ cm}$

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OpenStax College Physics Solution, Chapter 14, Problem 37 (Problems & Exercises) (2:23)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We have a walrus with an internal temperature which we call T2 since this is where the thermal energy is coming from and it's going towards this other body which is T1 namely the sea water which is at negative one degree Celsius and T2 the internal body temperature is that of a mammal which is 37 degree Celsius and there is this insulating layer of blubber and we have to figure out what the thickness is in this question. Now we're told that the rate of heat transfer is a 150 watts and so that's Q over t in this rate of heat transfer due to conduction formula and the total area of the walrus is two square meters and the thermal conductivity of this material is that of fatty tissue without too much blood vessels and I think that is the wording that it said. Yeah fatty tissue without blood. So we have to solve this for d and we multiply both sides by d and divide both sides by this fraction Q over t and so we are gonna think of Q over t as a single thing with we're dividing by so on this side we multiply by d and we divide by Q over t and then Q over t cancels here and the d cancels over here. If this looks a bit strange, you might prefer instead of dividing by Q over t I often like to multiply by its reciprocal so we can multiply by the reciprocal of t over Q and now clearly the Q's and t's cancel. So, d is thermal conductivity times the area multiplied by the difference in temperatures divided by the rate of heat transfer. So that’s 0.2 Joules per second per meter per Celsius degree thermal conductivity for fat multiplied by the area two square meters surface area of the walrus multiplied by the difference in temperature which is 37 degree Celsius minus negative one degree Celsius and keeping in mind that that makes a plus, minus and a minus, divide by 150 Watts and that gives ten centimeters must be the thickness of the fatty tissue insulating the walrus.