 ## Physics

Learn the toughest concepts covered in Physics with step-by-step video tutorials and practice problems by world-class tutors

20. Heat and Temperature

# Heat Transfer

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concept

## Introduction to Heat Transfer 7m
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concept

## Conduction 10m
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Hey, guys, in this video, we're gonna talk about conduction in more detail. While we've talked about conduction in the qualitative sense, the conceptual sense we haven't used any equations to describe conduction specifically, how quickly can he be conducted from one object to another? All right, that's what we're gonna focus on this video. Let's get to it. Remember that conduction is the transfer of heat through direct contact. Okay. Conduction is the most common type of heat transfer you're gonna encounter in your studies and your introductory physics courses. That's why it conduction. It was basically the only type of heat transfer that we've seen up to this point. Okay. When studying Kalorama tree, all he transfers where via conduction. Okay. And that was another point that I made when you put two objects in thermal isolation together in contact, the heat transfers always going to be conduction. Okay, well, we're interested in is how rapidly he could be conducted from ah, hot substance to a cold substance. Right? It always goes from hot to cold. And we're gonna get to that later on when we cover the second law of thermodynamics. But we want to know how quickly. This happens how long it takes the happen. Okay. Materials have a natural allowance for heat flow, known as the thermal conductivity given by Okay, okay. It's how easily they allow heat to be transferred quickly through them. Okay, The larger the thermal conductivity, the faster heat is conducted. Okay. So materials with a high thermal conductivity are called thermal conductors and material with low thermal conductivity are called thermal insulate er's. All right. Now, when dealing with heat, we talk often about a heat current. Okay, The current for the heat is just how rapidly the heat is moving per second. Okay, so it's just Q over Delta T. We've seen problems before That says heat was entering at 95 joules per second. That was the heat current. Okay, How much energy per second. Okay, so the conduction current is the heat current four conduction. Alright, let me minimize myself. We have to substances here. We have one at a hot temperature, one of the high temperature which we just call hot, and one at a low temperature, which was just called Colt. And a connection between the two. This is the conducting material. This is the conductor, okay, And the conductor is described by three things. It's got a cross sectional area. It's got a length and not written here. It has a conductivity, those air the three aspects that described the conductor. Besides that, you also have the temperature off the hot substance and the temperature of the cold substance, which have nothing to do with the conductor. Those air about these systems, the conduction current through the conductor is going to be given by K times a times the hot temperature minus the cold temperature over L. Okay, this is a very important equation, and the units are gonna be jewels per second because it's just the amount of heat transferred per second. All right, there are a few important consequences of this equation. First, the conduction current, like I said, is the rate at which heat is conducted through surface. There was substance. Okay, I explained that let's move past that. The heat conducted would then just be given by H Times Delta t. Okay, as long as H is a constant, If h is not a constant, then you couldn't just multiply it by the amount of time because H might change as that time goes on, if you knew the average conduction current, you could multiply it by the amount of time and find the total heat transfer. But this equation right here, typically Onley works if h is a constant. Okay, now notice H should not be a constant. Okay, The conduction current should absolutely change as the hot substance became colder because it's releasing heat and the cold substance becomes hotter. So naturally this is gonna drop, and this is gonna go up. That's what happens as he goes from the hot substance to the cold substance, so H should not be a constant. The conduction current will be constant if the hot and cold substances are what we call reservoirs. Like a reservoir of water. A reservoir of water is a giant source of water. Okay, what a reservoir is for anything, and we use it a lot in thermodynamics is a reservoir is an infinite source or sink of heat. That means that it can absorb and release an infinite amount of heat without changing its temperature one bit. Okay, that's what it means to be a reservoir. So if we look at our conduction current equation imagine now that the hot objects in the cold objects were reservoirs and the conductor and the conductor was connected between the two reservoirs. Then, no matter how much he went through the conductor, the temperature of the reservoirs would never change. That's the point of being a reservoir. It's an infinite source, so it can produce as much heat as it wants. It's an infinite sink so it can absorb as much heat as it wants, all without leading Thio any change in temperature. So if this substance and this substance here, where reservoirs in the conduction current through the conductor would in fact be a constant okay, and that's an important point to make because you'll probably see reservoirs quite a bit in thermodynamics. All right, let's do an example. Ah, hot reservoir at 100 degrees Celsius is connected to a cold reservoir at zero degrees Celsius by a 15 centimeter long piece of iron with the 150.5 square meter cross section. How much heat crosses the piece of iron and five seconds, and then it gives us the thermal conductivity of iron. Okay, so we're talking about how much heat in some amount of time, so we know that we need to use Q equals H Delta T. Okay. And we know that h the conduction current is K a th minus TC over l. We're told that the hot source and the cold source are actually reservoirs in this problem a hot reservoir and a cold reservoir. So the conduction current is gonna be constant. Let's calculate that the thermal conductivity of iron is 5 and the units of Watts per meter Kelvin R s I units, the cross sectional area is 0.5 The hot reservoir is 100 degrees Celsius minus zero degrees Celsius. Okay, Now, because this is a change in temperature, this is a difference in temperature, right? You have a hot minus a cold, even though there's no delta there because there's a change in temperature. We can simply leave this in degrees Celsius because that change in Celsius is equivalent to a change in Kelvin. And we do need Kelvin because if you notice the S I unit right here, is Kelvin okay? Divided by the length. And we're told that it's a 15 centimeter long piece of iron. So this is 150.15 m, plugging all of that in the heat. Current is 26 50 watts. Okay, Jules, per second is what I gave is the units for conduction current, but a jewel per second is just a watt. So most of the time conduction current is given in watts. Okay, Now, we confined the total heat transferred and were a perfectly allowed to use this equation because since the hot source in the cold source our reservoirs, their temperatures don't change and therefore h doesn't change. So this is 26 50 times. We were asked for it in five seconds. Okay. And so this is 13,250 jewels or 1.33 Kill the jewels. Like I said, typically, like thio give these units and kill a jewels. Because most of the problems with this is not it's 13.3, not 1.33 13.3 killing jewels. Because most of these heats are large enough to be represented as kill a jewels and to largely represented his jewels. Alright, that wraps up our discussion on the conduction current and conduction in specific. Alright, Thanks for watching guys
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Problem

A cubic Styrofoam cooler containing ice on a hot day is shown in the following figure. The thickness of each wall of the cooler is 15 mm, with a side length of 1 m. If it is 40°C outside, how long will 2 kg of ice last in the cooler? Assume that during the melting process, the temperature inside the cooler remains at 0°C and that no heat enters from the bottom of the cooler. Note that the latent heat of fusion for water is 334 kJ/kg and the thermal conductivity of Styrofoam is 0.033 W/mK. 4
concept 13m
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