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(a) What is the rate of heat conduction through the 3.00 -cm-thick fur of a large animal having a $1.40-{m}^{2}$ surface area? Assume that the animal's skin temperature is $32.0^{\circ} {C},$ that the air temperature is $-5.00^{\circ} {C},$ and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?

Part $(a) : 39.7 \mathrm{W}$Part $(\mathrm{b}) : 819 \mathrm{kcal}$

Physics 101 Mechanics

Chapter 14

Heat and Heat Transfer Methods

Thermal Properties of Matter

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

Hope College

Lectures

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for this question were active. Asked to find the rate of heat conduction through the fur of a large animal. So they give us in the question that the thickness of the fur is three centimeters, which is your 30.3 meters. They also tell us that the area of the fur is 1.4 square meters. The temperature of the skin is 32 degrees Celsius and the temperature of the air, which is where the heat is going, is minus five degrees Celsius. They tell us that the thermal conductivity of the skin is the same as that of air. So we look that up. Kay Air equals 0.23 Jules per seconds times, meters, times, degree Celsius. And in this case, we want to use the thermal conductivity equation for the rate of heat conduction, which is Q equals K times, eh? Times t to minus t one over D. Okay, so part eh asks us to find Q, so he's gonna plug in K to your 0.23 Jules per seconds times meters times degree Celsius turns a just 1.4 square meters. Have the change in temperature so that's 32 degrees Celcius, minus negative five degree Celsius. This becomes positive. This is just adding divided by D, which is your 0.3 meters. So our degrees Celsius cancels out with this addition up top, the meter squared cancels with the meter's in the bottom. We are left with 39.7 Jules per second, so that's our answer for part A. This is the same as 39.7 watts, by the way. So for Part B were asked to find how much this animal would have to eat to maintain this rate of heat loss. So in a day we have from part A. We have jewels for seconds. We want to find out how many seconds or in a day. So one day is 24 hours, which is 60 minutes times. Well, it's already like this, 24 hours times 3600 seconds per hour, which you should memorize. You use a lot and we get 8.64 times 10 to the four seconds. So to get the heat in Jules, we just take from part a 39.7 jewels per second, times the seconds in a day, which we just found to be a 0.64 times to another four seconds, which is the same as 3.43 times, 10 to the six jewels. But typically, when we're talking about food, we want to know what's the energy and kill a cow's. So we need to look up the conversion from Jules to kilocalories. So that's times 2.39 times 10 to the minus four kilocalories per Jules, she gives us our final answer of 819 kilocalories.

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