# Video: Calculating the Conductive Heating Power through a Boundary

A large animal has a 1.40 m² surface area. The animal’s surface is covered with 3.00-cm-thickness fur that has a thermal conductivity of 0.0230 W/m ⋅ °C. The animal’s skin is at a temperature of 32.00°C and the air temperature around the animal is −5.00°C. What is the rate of heat conduction through the fur? What daily intake of energy from food will the animal need to replace energy lost by heat conduction?

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### Video Transcript

A large animal has a 1.40 square meter surface area. The animal’s surface is covered with a 3.00-centimeter-thickness fur that has a thermal conductivity of 0.0230 watts per meter per degrees Celsius. The animal’s skin is at a temperature of 32.00 degrees Celsius and the air temperature around the animal is negative 5.00 degrees Celsius. What is the rate of heat conduction through the fur? What daily intake of energy from food will the animal need to replace energy lost by heat conduction?

We can call the rate of heat conduction through the fur 𝑑𝑄 𝑑𝑡. And we’ll call the daily intake of energy from food the animal will need capital 𝐸. In this exercise, we’re given lots of numerical information about the scenario. We’re told the thermal conductivity constant of the animal’s fur. We’re told the thickness of the fur. We’re told the total area of the fur. And we’re also told the animal’s temperature as well as the temperature of the air right around the animal.

Knowing all this, we want to solve for the time rate of change of heat through the animal’s fur to the outside. And we also want to solve for the energy the animal would need to intake to daily offset that energy loss due to heat. In general, if we had some sort of thermal barrier that will separate in two different temperatures, 𝑇 one and 𝑇 two, if this thermal barrier has an area we could call 𝐴 and a thickness we would call lower case 𝑡, then we could figure out how much heat flows across the barrier either from 𝑇 one to 𝑇 two or the opposite direction, depending on which one is higher, if we had one more piece of information about the material that makes up this thermal barrier.

There is a property of materials we’ve called 𝑘 which is the thermal conductivity constant. This value — which is in units of watts, which is energy per unit time divided by units of length times units of temperature difference — describes just how well a given material lets heat pass through. If we take our thermal conductivity constant, multiply it by the cross-sectional area of our barrier, then multiply that by the temperature difference between the two temperatures on either side of the barrier, making sure to take the absolute value so we have a positive result, and divide this by the thickness 𝑡 of our barrier, then overall we have a result for the rate at which heat passes across our thermal barrier, 𝑑𝑄 𝑑𝑡.

We can express this relationship in shorthand this way. And it’s this relationship, for how much heat per unit time passes our barrier, that we’ll use to solve for 𝑑𝑄 𝑑𝑡 for our animal. In terms of our given variables, 𝑑𝑄 𝑑𝑡 is equal to 𝑘 times 𝐴 times the absolute value of 𝑇 sub animal minus 𝑇 sub air, the temperature difference between the inside and outside the animal, all divided by capital 𝐿.

When we do plug in these values, our temperature difference ends up being 37.00 degrees Celsius. And we’re sure to convert our thickness of our animal’s fur into units of meters. Looking at the units in this expression, we see that meters in our denominator cancels out with square meters, the area of the animal’s fur. And that degrees Celsius also cancels out, leaving us with final units of watts which is in energy per unit time. To three significant figures, 𝑑𝑄 𝑑𝑡 is 39.7 watts. That’s how much power escapes the animal due to heat loss.

Next, we want to solve for the amount of food energy the animal would need to intake every day to offset this energy loss due to heat. To solve for capital 𝐸, we can start off by recalling that one watt is equal to a joule per second. Assuming a constant rate of heat loss throughout the day, this means that every second that passes, this animal loses 39.7 joules of energy due to heat. We multiply that result by 3600 seconds per hour to find the amount of energy lost per hour. And then we multiply that result by 24 hours in one day, to solve for the total amount of energy lost to heat in one day.

If we calculated out the product of these three fractions, that would give us the energy lost due to heat in one day. But that energy will be expressed in units called joules. A unit that may be more familiar when we speak of food intake is the unit of calories or kilocalories. One kilo calorie, which is often represented on food nutrition labels as one capital 𝐶 calorie, is equal to 4184 joules. So finally, we’ll multiply our fraction by one kilocalorie per 4184 joules to get a final answer in units of kilocalories, or, as we’re more used to see them, capital 𝐶 calories.

This result will be the total energy we want to solve for, capital 𝐸. When we calculated, we find it’s 820 kilocalories. This is how much food energy the animal would need to intake every day to offset energy lost due to heat.