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RC
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Heat Transfer - Example 1

Heat transfer is the exchange of thermal energy between physical systems. The rate of transfer is dependent on the temperature difference (thermal gradient) between the systems, and conductive, convective, and radiative properties of the material. Conduction is the transfer of energy between objects that are in direct physical contact. Convection is the transfer of energy between objects that are not in direct contact through the movement of fluids. Radiative heat transfer is the transfer of energy by radiation, which occurs due to a temperature difference between the systems.

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Video Transcript

welcome to our first example video. Looking at heat transfer techniques. In this video, we're going to consider conduction into a cold box, which could be a house if we wanted to call it that. So we have a cold box and then a hot reservoir on the outside. We have our hot reservoir at a temperature of 28 degrees Celsius. Are cool, are cool reservoir at a temperature of 20 degrees Celsius. And we're going to say that the walls and ceiling I'll have a R value of 20 watts per meter. Kelvin, what this does for us is when we come over to our d. Q t t equation, we can rewrite it in terms of the R value. Remember that R is equal to L over K, which means the Q t t is equal to a over our multiplied by th minus TC. So plugging things and then my my area is going to be the surface area of all of these different sides. So I have 12345 sides and I need toe add up all of the different surface areas. So I have two sides that are six by three so that's two sides that are 6 m by 3 m. Plus, we're going to have two sides that are 7 m by 3 m. So that's two sides times 7 m by 3 m and then finally we have the ceiling which is going to have dimensions of 6 m by 7 m. So there's our numerator divided by R, which is 20 lots per meter kelvin multiplied by the temperature difference, which is eight. Calvin. Remember, we want to stick with Calvin here if we can, though when it's a temperature difference, Degrees Celsius works just fine. So this is our rate of heat transfer. If we want to know the rate, though, at which temperature inside the boxes in clean increasing, we need to go back to our equation. Q equals N c Delta T. We use the Moller specifically in this case because we're dealing with a gas and it's a little easier to calculate. Remember, I have my ideal gas law PV equals and our T. So if I want to know the number of moles and they're assuming that we start out with the pressure off one atmosphere, we have a constant volume which I calculated we have, which I can be calculated a 6 m by 3 m by 7 m. Then we have a original temperature of 20 degrees Celsius so I can solve for the number of moles inside the house. And if we assume that it's airtight, then we can take this. And we know that Delta T is equal to Q over and see, and the rate of temperature increase will be d T. With respect to time is equal to one over and time see multiplied by de que d t, which we already have a value for So again, this is Mueller specific heat so you can plug in everything that you need here and look, you'll even notice that d t d t is actually going to be equal to is going to be proportional to a factor of temperature itself. So, um, this is how we can solve for the rate of temperature increase inside the box

RC
University of North Carolina at Chapel Hill
Top Physics 101 Mechanics Educators
Elyse G.

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University of Sheffield