A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.2 cm thick on the inside wall surface. The wood has k = 0.080 W/m K, and the Styrofoam has k = 0.027 W/m K. The interior surface temperature is 19.0°C, and the exterior surface temperature is -10.0°C. (a) What is the temperature at the plane where the wood meets the Styrofoam?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use. In order to solve this problem, a shipping container is to be thermally insulated. The architect uses 2.5 centimeters graduated cork to insulate the container from the outside. On the inside, the container is coated with a 2.0 centimeter thick layer of felt K for the graduated cork and felt are 0.048 watts per meters multiplied by Kelvin and 0.040 watts per meter multiplied by Kelvin. Respectfully, the interior temperature is 18.0 °C. While the exterior temperature is negative 8.0 °C determine the temperature at the interface of the materials assume the materials are in contact. So that's our end goal. So ultimately, we are asked to figure out what the temperature at the interface of the materials is. That's our final answer that we're trying to solve for. We're also given some multiple choice answers. They're all in the same units of degrees Celsius. So let's read them off to see what our final answer might be. A is 13.1. B is 5.27 C is 16.7 and D is 7.60. So first off, let us recall and use the equation for the rate of heat flow. And let's call this equation one. And the rate of heat flow R is equal to capital Q divided by T which is equal to lowercase K which is our thermal conductivity multiplied by capital A multiplied by capital T subscript H. So th minus TC all divided by capital L. So now we need to rearrange equation one to solve for capital R divided by capital A which is the rate that heat flows through the container's walls. So when we do that, we will find that capital R divided by A which we're gonna call this equation two is equal to K multiplied by TH minus TC, all divided by capital L. So now we need to start solving for T and in, and in order to solve for T, we need to set equation one equal to. So we need to set both equations equal to each other. But by two equations is we need to set using equation one, we need to use equation one for the quark hour later. So we need to use for the outer layer which is quark. So that's gonna be our first equation and we need to set it equal to our second equation which is gonna be focusing on the felt layer. So let's make a quick little note here off to the side that we're gonna use F is gonna denote the felt layer and we're gonna use C is going to equal the quark layer. So we don't get confused. So anything subscripts F will mean that we're dealing with the felt anything subscript C means we're dealing with the quark layer. OK. Now that we have that in mind, so now we need to solve for T. So let's start setting our equation for that considers the quark outer layer and then the inner layer which is felt. So with that in mind, we know that KC multiplied by a multiplied by capital T minus TC divided by LC is equal to KF multiplied by a multiplied by th minus T divided by Awesome. So we need to note that K which is the thermal conductivity of the material will be different for both felt and cork. And we also need to note that R and R divided by A will be equal for the wall. And L in this case is the thickness of the material. And T which is our final answer that we're trying to solve for is the temperature at the boundary between these two different materials. So it's the temperature at the boundary between the cork and the felt materials. So now we need to simplify our equation. So let us first divide both sides by A. And when we do that, we will find that multiplied by KC multiplied by T minus TC is equal to LC multiplied by KF multiplied by T minus T, which we can simplify to multiplied by KC, multiplied by T minus multiplied by KC, multiplied by TC is equal to LC multiplied by KF multiplied by T. Oh, let's make that look nicer. So KF multiplied by TH minus LC multiplied by KF multiplied by T. OK. So we need to keep simplifying. So LF multiplied by KC multiplied by T plus LC multiplied by KF multiplied by T equals LC multiplied by KF multiplied by TH plus LF multiplied by KC multiplied by TC. And if you're wondering why in the world are we doing? This is because ultimately, we're trying to isolate and solve for T. So we have to keep simplifying and rearranging things until we get T by itself. So simplifying once again. So LF multiplied by KC plus LC multiplied by KF multiplied by T is equal to LC multiplied by K F multiplied by T H plus LF multiplied by KC multiplied by TC. And finally getting T by itself, we get that T is equal to LC multiplied by KF multiplied by TH plus LF multiplied by KC multiplied by TC. And this is all divided by LF multiplied by KC plus LC multiplied by KF. OK. So we need to quickly note and let's make a little note. Here in blue, we need to note that delta T is equal to th minus TC. So therefore, we are dealing with a change in temperature equation. And we can also note that delta T will have the same equation result if we use units of degrees Celsius or Kelvin. So it doesn't matter if we use degrees Celsius or Kelvin units. So we still provide the same answer using our delta T equation. So to keep things simple, let's just use degrees Celsius units. So now at this stage, we could write down all of our known variables. So we know that LC is equal to 2.5 centimeters. But let's use dimensional analysis really quick to convert centimeters to meters. So in 1 m, there's 100 centimeters. So that means it's equal to 0.025 m. And then we know using similar methods that would be 0.0 2 0 m. And we know that our teh which is like the warmer temperature is equal to 18.0 °C. And we know that tec is equal to negative is that's our cold temperature. It is negative 8.0 °C. So now we can finally solve for T, which is our final answer by plugging in all of our known variables back into equation three. So plugging in all of our known variables back into equation three, we will find that T is equal to 0.025 m multiplied by 0.040. And its units are watts per meters multiplied by degrees Celsius multiplied by 18.0 °C plus aren't the sea 0.0 to 0 m multiplied by 0.040. And its units are watts per meters multiplied by degrees Celsius. And this is also multiplied by negative 8.0 °C. And this is all divided by 0.025 m multiplied by oh, and don't forget our second parenthesis up top here and then going back to the bottom multiplied by 0.0 40 watts per meters multiplied by degrees Celsius plus 0.020 m multiplied by 0.0 48. And its units are watts per meters multiplied by degrees Celsius. Awesome. So let me plug that into a calculator. We will find that T is equal to 5.27 °C when we round to two decimal places. And that's it. That's our final answer. Hooray. We did it. So looking at our multiple choice answers at the top, our final answer has to be the letter B 5.27 °C. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.