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RC
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Heat Transfer - Example 2

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 second example video looking at heat transfer methods. Now in this video, we're going to consider boiling a pan of water. Simplify the problem. We're going to say that we have a cold reservoir that is the water which is equal to 100 degrees Celsius, and then our hot reservoir being the burner underneath is going to be 300 degrees Celsius. Given some volume of water inside the pot and the latent heat of vaporization, we're almost everything we need. But we do need some properties of the pot itself. First of all, if we say it's made of stainless steel, it gives us the thermal conductivity here. And then we also will need a width that is to say, how thick is the bottom of the pot? Because this is where all the heat is coming from. So we need to worry about this length, and then we need to worry about the area of the bottom, the assuming a a circular pot or a cylindrical shape with a radius equal to seven centimeters. We should have all the information that we need in order to solve this problem. So first of all, we want to remember that since we're just trying to boil off all of this water, the amount of heat we need is going to be M times LV, where m is the mass of water that we have. We don't know the mass, but we do know the volume. And if you recall row times V is just a xga good as mass. And so row is going to be 1 kg her leader for water multiplied by one half leader. And so our masks is going to be one half kilogram. Okay, so we have our Q determined. But what else we know is that we have a de que de ti which we're going to write as some amount of heat that's exchange over some amount of time is going to be equal to K times A over l multiplied by the delta T now delta T here remember, isn't a final temperature and initial temperature. It's going to be the hot reservoir temperature minus the cold reservoir temperature. Now, the reason I said we were simplifying this problem is because recall that this temperature will not change since all the energy is going into boiling the water Okay, so we have everything we need here, except what we need is the time there's our delta t right here. So we can do is solve for Q in this equation, which will give us Delta T multiplied by K A over l Times T h minus t c and then set it equal to this queue up here because they are the same. Que remember that we are giving heat from the burner into the water. So the the amount of heat that goes into vaporization is the amount of heat transferred into the pot. Okay, so here's our equation. All we need to do is solve for Delta t so Delta T is going to be equal to We've got our length times mass times the latent heat of vaporization divided by K times, a times th minus TC. And now we confined the amount of time it will take in order to boil a half a liter of water

RC
University of North Carolina at Chapel Hill
Top Physics 101 Mechanics Educators
Christina K.

Rutgers, The State University of New Jersey

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