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Calculate the rate of heat conduction out of the human body assuming that the core internal temperature is 37.0 degree Celsius, the skin temperature is 34.0 degrees Celsius, the thickness of the fatty tissues between the core and the skin average is 1.00 centimeters, and the surface area is 1.40 meters squared.
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The thermal conductivity of human fatty tissues is 0.200 watts per meter degrees Celsius.
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Okay, so in this question, we’ve been asked to calculate the rate of heat conduction out of the human body, and we’ve got to make a few assumptions.
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Firstly, we can assume that the core internal temperature is 37.0 degrees Celsius.
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Secondly, we assume that the skin temperature is 34.0 degrees Celsius.
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Next, we’re told to assume that the thickness of fatty tissues between the core and the skin average is 1.00 centimeters, and the surface area is 1.40 meters squared.
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Finally, we’ve been told that the thermal conductivity of human fatty tissues is 0.200 watts per meter degrees Celsius.
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So essentially what we have is a human being.
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Let’s draw a diagram.
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This is a diagram of the most average human being in the world.
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We’ll call him Joe because now he’s an average Joe.
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So why is he the most average person in the world?
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Well in the question, we’ve been told that the thickness of the fatty tissues between the core and the skin average is 1.00 centimeters.
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So if we say that this outer layer, the orange layer, is his skin, and this inner part here we assume this to be the core so the orange part is the skin the inner part is the core, then the thickness of the fatty tissue between the skin and the core on average is 1.00 centimeters.
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Now of course on a real human body, there are areas where the fatty tissue is thicker and there are areas were the fatty tissue is thinner.
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However, on average the thickness of the fatty tissue is 1.00 centimeters.
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So the assumption that we’re making is that Joe over here has got 1.00 centimeters of thickness of fatty tissue all around his body.
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This means we can keep things simple.
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And we don’t need to worry about the regions on his body where the fat is thicker and the regions where it’s thinner because even if we consider all of these complications, it would average out to give us the same result.
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So anyway, let’s start labelling some of the stuff we’ve been given at the question.
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We know that Joe’s inner core temperature is 37.0 degrees Celsius, and we know that his skin temperature is 34.0 degrees Celsius.
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We’ve also been told that the surface area of his skin is 1.40 meters squared.
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And finally, we know that the thermal conductivity of human fatty tissues is 0.200 watts per meter degrees Celsius.
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We can use all of this information to work out the rate of heat conduction out of the human body.
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To do this, we can recall that this is the equation we’re looking for.
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𝑄 is the heat conducted out of the material that we’re studying.
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𝑡 is the time taken for this heat conduction to occur.
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𝜅 is the thermal conductivity of the material that we’re studying.
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𝐴 is the surface area of that material.
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Δ𝑇 is the temperature difference across the material, so the difference in temperature between one side of the material and the other.
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And finally, 𝑑 is the thickness of that material.
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Now in this case, the material that we’re talking about is the fatty tissue found all over Joe’s body.
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And we’re trying to find out the rate of heat conduction.
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But we’ve just said that 𝑄 is the heat transferred and 𝑡 is the time taken for this heat transfer to occur.
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So 𝑄 divided by 𝑡 must be the rate of heat transfer; it’s the amount of heat transferred divided by the time taken or the amount of heat transferred per unit time.
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This is the same as the rate of heat transfer.
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In other words, we’re trying to find out what the left-hand side of the equation is.
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And luckily for us, we already know everything on the right-hand side of the equation.
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The thermal conductivity, well that’s 𝜅; we’ve already been told this.
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We also know the surface area of the material that we’re studying, which in this case is the surface area of the skin because the skin surrounds the fatty tissue.
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And this surface area is 1.40 meters squared.
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And for Δ𝑇, well we know the difference in temperature between one side of the material and the other because we know the core temperature and the skin temperature.
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So we can work out the difference in temperature by subtracting the skin temperature from the core temperature.
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And finally, we know 𝑑, which is the thickness of the material, which in this case is the thickness of the fatty tissue.
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So it looks like we can plug stuff into our equation.
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But before we do that, there are a couple of things that we need to work out.
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Firstly, like we said, we need to work out Δ𝑇 as we said earlier Δ𝑇 is the difference in temperature between one side of the material that we’re looking. at and the other side of the material.
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In our case then, Δ𝑇 is equal to 37.0 degrees Celsius, which is the core temperature, minus 34.0 degrees Celsius, which is the skin temperature.
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And this gives us Δ𝑇 is equal to 3.0 degrees Celsius.
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Secondly, we also know that 𝑑 is equal to 1.00 centimeters, the thickness of the fatty tissue.
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However, we don’t want this in centimeters.
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We want it in the standard unit of meters.
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So we need to remember that a centimeter is the same as 10 to the power of negative two meters.
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So 1.00 centimeters is the same as 1.00 times 10 to the power of negative two meters.
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And at this point, we can stop plugging stuff into our equation.
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But hold on a second!
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Why didn’t we convert Δ𝑇 into Kelvin?
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If we’re converting everything to standard units, then why are we okay with leaving Δ𝑇 in degrees Celsius?
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Well the reason for this is twofold.
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Firstly, whenever we’re looking at differences in temperature, the difference between two temperatures in degrees Celsius is the same as the difference between the same temperatures in Kelvin.
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We can quickly show where this is true.
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When we calculated Δ𝑇, we said that it was equal to 37.0 degrees Celsius minus 34.0 degrees Celsius.
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What we could have done is converted each of these quantities into Kelvin.
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And that would give us 37.0 plus 273 Kelvin minus 34.0 plus 273 Kelvin.
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But essentially, what ends up happening is we’ve got a positive 273 here and a minus 273 here.
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Those just cancel out.
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So Δ𝑇, regardless of whether we working in Kelvin or degrees Celsius, will always be the same.
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Once again we get a Δ𝑇 value 3.0 Kelvin.
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And this is the same as 3.0 degrees Celsius here.
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So differences in temperature are the same in Kelvin as they are in degrees Celsius even though the temperatures themselves are not the same.
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The second reason for staying in degrees Celsius for Δ𝑇 is that our value for thermal conductivity has been given to us in watts per meter degrees Celsius.
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And we need to stay consistent in our units more than anything.
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So we’ll just stick with degrees Celsius.
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And at this point, let’s stop plugging stuff.
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In the right-hand side of the equation, it looks something like this: we’ve got the value of 𝜅; we’ve got the values of 𝐴, Δ𝑇, and 𝑑.
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And if we consider the units quickly, we’ve got this degrees Celsius cancelling with this degrees Celsius, and we’ve got this meter squared cancelling with one and two powers of meters as well.
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So all we’re left with is a unit of watts.
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And that makes sense.
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The unit of watt is a unit of power.
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And power is the rate of energy transfer.
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And we know that heat has the same units as energy.
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So heat transferred divided by time or the rate of heat transfer will have a unit of watts, which means dimensionally we’re doing okay.
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So we can evaluate the fraction to give us an answer of 84.0 watts.
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And we’re giving this answer to three significant figures because all of the values that we’ve been given in the question have been given to three significant figures.
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But hang on a minute!
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Didn’t we say earlier that Δ𝑇 was equal to 3.0 degrees Celsius?
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But that’s two significant figures, isn’t it?
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Well yes it is two significant figures, but the values that we used to calculate Δ𝑇, that’s 37.0 degrees Celsius and 34.0 degrees Celsius, have been given to three significant figures.
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So we could have just plugged those values into our calculation rather than calculating Δ𝑇 first.
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And hence our final answer has to be to three significant figures.
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And so our final answer is that the rate of heat conduction out of the human body is 84.0 watts to three significant figures.