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Current time:0:00Total duration:11:39

Video transcript

in a previous video we talked about thermal expansion when you heat things up atoms jiggle more and they move farther away from each other causing the whole material to expand but in this video we're gonna try and answer questions like how much copper expands on heating or how much aluminum expands on heating so how do you give a number to how much something expands well it turns out that there is one single quantity that can tell you how much something expands and that quantity is called as linear expansion expansion coefficient coefficient and by the end of this video we'll familiarize ourselves with this scary-looking term alright so let's start with an example imagine someone gives you a gold wire it's so imagine here is a gold wire and let's say it has a length say it has a length L and it's sitting on on your desk at some temperature T now you decide to heat this thing up so you have a world wire and you heat it up and then it heats up it's temperature rises and the whole thing expands and you can see it expanding so as you heat it up we will see that the wire starts expanding like this okay let's put that thing over here this is the expanded wire and since the temperature has increased now the new temperature let's say is T plus delta T delta T represents the change in temperature how much the temperature increased and now we'll compare the length of this wire to the previous length now this is how long the wire was initially so this would be L and due to the expansion this is the extra length so let's write that down this would be L this is L and this this region is that extra length Delta L and the big question now is what does Delta L depend on let's think about this we the wire expanded because we increase the temperature so we could say that more you increase the temperature more the wire will expand in other words you could say that more the value of delta T more the value of Delta L right well experiments support this in fact it turns out that for small values of Delta L for small values of Delta L its even found that Delta L is proportional to delta T proportional to delta T and what I mean by this is that if you double the value of Delta P then Delta L will also double and you may be wondering what do you mean by small values of Delta L how small is small well as long as the expansion is much smaller than the original length in other words in other words as long as Delta L is much smaller than L this will hold true not like what I've shown in this diagram over here Delta L is considerably large I have exaggerated over here in fact but we will see in reality Delta L is very tiny compared to L itself okay all right anything else that Delta L depends on hmm turns out there's one more thing that'll depend on Delta L turns out to be also proportional to the original length and you may be like whoa wait a second why would Delta L depend on L well if you think about this it makes sense imagine that we started out with the wire of not length L but length 2 L think about this imagine we had a wire gold Huff length 2 L and imagine again we heated up and increased temperature by delta T we could assume that this wire is made up of two wires each of length L right we could do that and since we know that each wire or each section of the wire is going to expand by Delta L we will see that this section will expand by Delta L and this one will expand by Delta L as well therefore the total expansion do you agree would be 2 Delta L so if you double the length of the wire the expansion doubles and if you triple it this will triple so in other words Delta L or the expansion length is proportional to the original length of the wire so if we put these two together we could now say Delta L is proportional to L times delta T and we could now replace this proportionality with an equal to sign by putting a constant over here and that constant can be used any electric and abuse for that over here we're going to use alpha L as the symbol for the constant times L delta T and this expression now is going to help us calculate how much something expands on heating and this constant alpha L is what we call as the linear expansion coefficient so let me just put her for the short form over here I won't write it again so linear expansion coefficient this is a linear expansion coefficient all right now let's get to know I'll file a little bit better so let's put this thing a little bit aside okay the first thing we'll do is figure out the units of alpha L we can do that by dividing both sides by L delta T so we'll isolate alpha L so if you divide both sides by L delta T the right hand side would be alpha L which I'm writing over here and the left hand side would be Delta L divided by L delta T divided by L delta T and from this we can now figure out the units of this let's see Delta L has the units of well it's length so it's meters L is also linked that's also meters and delta T temperature well the standard unit is Kelvin so this cancels which means we end up with 1 over Kelvin or Kelvin inverse that's the unit of alpha L the linear expansion coefficient for example for example for gold for gold it turns out that our final values I just looked it up is equal to approximately 14 times 10 to the minus 6 minus 6 Kelvin inverse and this might seem a little bit weird as to what do we even mean by 14 times 10 to the minus 6 Kelvin inverse well what it means is if you took one meter long wire one meter long gold wire and if you increase this temperature by one Kelvin then notice L becomes 1 delta T becomes 1 which means Delta L becomes equal to alpha L and therefore in our case the expansion in the gold wire this expansion extra length the expansion the length would be this number 14 times 10 to the power minus 6 meters this is meters because this is meters so the Alpha L the linear expansion coefficient tells us how much a 1 meter long wire when increased the temperature by one Kelvin would expand that's what this number really is and it doesn't have to be meters this could have been 1 feet if this was 1 feet then this would be 14 times 10 to the power minus 6 feet so whatever unit of length you take same unit of Delta L you end up with so more the value of alpha L more the material expands on heating or more sensitive the material is to changes in temperature so now let's look at some values of alpha L of some typical materials so here it is we have all five values in terms of 10 to the power minus 6 Kelvin inverse gold is 14 times 10 to the power minus 6 Kelvin inverse a look at copper it's 17 which means copper expands more on heating compared to gold aluminium expands even more and look at glass its value is just 4 times 10 to the minus 6 Kelvin indoors which means glass doesn't expand as much as these metals do all right I want to end this video with a simple numerical here it is so we have a gold wire which has a length of 50 meters at Kelvin we have to calculate the change in its length when the temperature is increased to 400 Kelvin our file value is given it's the same thing it's gold I want you to pause this video and see if you can try to figure this out yourself try to do it intuitively all right time for solution we know that alpha L is given to us is 14 times 10 to the minus 6 Kelvin inverse which means if we have one meter long gold wire and increases temperature by one Kelvin then the expansion would be 14 times 10 to the minus 6 meters but we have 50 meter long wire whoo that means now if we had expired increase the temperature by 1 Kelvin the Delta would be 50 times this number that makes sense for 1 meter would be so much for 50 meters would be 50 times that much so it would be 14 times 10 to the minus 6 Kelvin inverse multiplied by 50 50 meters but this would be the expansion in the length for 1 Kelvin rise all right for one killing pork elements 1 Kelvin rice but how much given rise do we have we're increasing the temperature from 300 Kelvin to 400 Kelvin well how much is the change in the temperature here whew that's a hundred Kelvin rise if this is the expansion for 1 Kelvin rise 400 Kelvin rise well we just multiplied by 100 so multiply 100 Kelvin and this is because Delta L is proportional to delta T you should discuss that and so now we just have to plug this in let's see how much we get we get 14 times 50 times 100 is 5,000 5,000 times 10 to the minus 6 and notice the kelvin inverse and kelvin cancels and we end up with meters of course because delta l should be in terms of meters all right so if you multiply this five fours are 2500 570 thousand times 10 to the minus 6 meters and that is equal to let's just shift this a little bit okay that's equal to 10 power minus 6 so let's just shift decimals 1 2 3 4 5 6 which means we get about zero seven meters and that's how much the wire expands on heating in this example and notice we have just used the same expression that we derived over here notice Delta L equals alpha L times L times delta T so even without the expression if you just if you just understand what alpha L really means if you just understand this concept then it's very easy to figure out or solve problems like this