Specific Heat & Temperature Changes - Video Tutorials & Practice Problems

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1

concept

Specific Heat & Temperature Changes

Video duration:

6m

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Hey guys. So in earlier videos, we talked about temperature in this video, I'm gonna show you a related idea of thermodynamics called heat and the role that he plays in changing the temperature of an object. We're also gonna talk about something called the specific heat. And I'm gonna show you a very straightforward equation that you absolutely need to know. Let's check this out here. So before we go any further, I want to talk about heat and temperature. These are two words that we use sort of interchangeably in everyday life. So we say things like I can't stand the heat, it's unbearable. What we actually mean is that we can't stand the temperature aware of what we are. But these things have very specific definitions of physics that you need to know. Let's talk about temperature. We've actually already talked about temperature before, given by the letter T. It's kind of just the measure measure of the kinetic energy of the molecules that make up a substance. Kind of a measure of how hot or cold something is or feels. So the idea here is that these two substances, and if these molecules are moving faster, they're going faster. They have more kinetic energy and therefore the temperature is higher, like 20 degrees Celsius versus 10 for a colder object. So, that's temperature. Heat on the other hand, is a little bit different. Heat is given by the letter Q. And what he is is it's a transfer of energy. It's a transfer of energy between two materials that is caused by a difference in temperature. So here's the idea these two substances over here. Both have their own individual temperatures. One is at 20 and then one is at 10. So, those temperatures are different. And so because of that difference in temperature, it's gonna cause energy to be transferred between these two materials. That's what he is. So, the analogy that I like to use is that temperature is like kinetic energy, or just energy in general. An object can have energy. And heat, on the other hand, is kind of like work. An object can't have work. Work is something that is transferred or done between two objects, and that's exactly what heat is transferred. That's caused by temperature difference. Now, what's important about heat is that it has a direction heat always flows from hotter two colder substances. So, this heat that gets transferred, this Q has a direction it always flows from hotter to colder. It's basically always goes from high to low temperature. Just like a lot of things in physics go from sort of high to low. It's the same idea here. And they're basically they're going to continue exchanging this energy heat is going to continue flowing from the hotter to the colder until they reach what's called thermal equilibrium. So, you can imagine that these two things, one that 21 is at 10 when you mix them together, eventually they'll reach an equilibrium temperature that is 15 degrees Celsius. Just making that up here. And then what happens is that when they reach this equal the Q. Is equal to zero. There's no more heat transfer anymore. All right, So now that we know what heat is. Let's talk more specifically about what heat does. Because you're gonna need to know the equation that relates this heat with the temperature change. So when a material absorbs heat like plus Q. Or loses heat minus Q. It's gonna change temperature. So the idea here is that if you took water and heated it from 0 to 10°C, you had to put in some heat in order to do that in order to increase its temperature. But you can also change objects phase. Now, we're only gonna talk about temperature in this video. We're gonna talk about phase in a later one. All you need to know for now is that phase is kind of like solids, liquids and gasses is like the states of matter. And if you input energy, you can also change the phase of an object. So, for example, if you stuck ice in a microwave, you would change the temperature that ice. But also at a certain point you're gonna melt the ice into water. It's gonna change from solid to liquid. Now, we're going to talk specifically about the change in temperature equation over here, the equation for this is Q equals M. C. Times delta T. This is sometimes referred to as the M. Cat equation because if you imagine that this delta sign sort of grows legs. That kind of looks like the word M. Cat. Right? So the M. Is the mass of the object. That DELTA T. Is the change in the temperature. And this variable this letter C. Here is something called the specific heat of the material. Here are the units for the specific heat. And one way you can think about the specific it is, it's kind of like a measure of thermal inertia. Remember that inertia is like a resistance to change this thermal inertia is the resistance of changes in temperature. You can think about it is how hard it is to change temperature from any heat that is gained or lost if you ever need this C. Variable there is constant. It's going to be given to you because it's basically just has to do with the material that is involved. And the idea here is that if the sea is higher for higher values of C, it's more difficult to change temperature for the same amount of heat basically if you're see is higher than you need more Q. In order to change in objects temperature. Alright, so here's actually some examples. So I've actually got some substances down here with some pre some common ones with their specific heats. So for example copper, iron and lead all have specific heats that are sort of in the hundreds but ice and water have specific heats that are in the thousands. So they're much much higher than metals. This kind of makes sense if you take an iron skillet and you put it on the stove, it's gonna heat up very, very quickly. But then the water that you put into the pot pot or something like that to boil it, it takes a long time for it to change temperature. Alright, so let's go ahead and get to our example here. So we've got um the heat required to raise the temperature of 50 g of water from 40 to 55. So we've got the mass which is equal to 0.5. And we've got the specific heat, let's see actually we're going from a to temperature that's 10 knots equals 40 degrees and to a T. Final of 55 degrees Celsius. We also know that we're dealing with water. So the specific heat of water that we're gonna use this is C. W. I'm just gonna call this is 41 86 that's the these are the units right here, just in case you need them. So with that being said, how much heat do we need? Well this is just Q. Equals M. C. Times delta T. So we have em we have the C. For water and all we need to do is just figure out the change in the temperature. So we were told that we're going from 40 over to 55 degrees Celsius. So the idea here is does delta T actually equal 15? And the answer is yes, because remember Delta T in Celsius is the same thing as delta T in kelvin. So if you're calculating the deltas, it actually doesn't matter if you were to convert these things to kelvin, the difference would still be 15. So it actually doesn't matter which one you plug in. So the idea here is we have Q equals this is 0.05 times 41 86. And then we're gonna multiply this by 15 and what you should you should get 3140 jewels. So thats how much heat it takes in order to raise the temperature of this amount of amount of water by 15°C. That's that's it for this one. Guys, let me know if you have any questions.

2

Problem

Problem

You are given a sample of an unknown metal. You weigh the sample and find that its weight is 29.4N. You add 1.25×10^{4} J of heat energy to the sample and find that its temperature increases from 52°C to 70°C. What is the specific heat of this unknown metal?

A

23.6 J/(kg⋅K)

B

231.5 J/(kg⋅K)

C

14.3 J/(kg⋅K)

D

1.20×104 J/(kg⋅K)

3

example

Heating Cup of Water

Video duration:

3m

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Everybody. So welcome back here. Let's take a look at this problem here. So we have a cup of water that we're gonna put on a hot plate, which is rated at 200 watts. Now, all of this energy from the hot plate is going into the water as heat energy. So let me just draw this real quick here. Imagine I have, you know, this is like my hot plate, it's like a little, you know, sort of like a stove or whatever, like a portable stove. And I've got this little cup of water that's here on the inside like this. So I've got my couple of water. And basically what happens is this hot plate is going to supply energy into this water as heat energy. So there's gonna be some cue that transfers in now. What we wanna do is take and we want to figure out how long it takes for the water to warm from 10 to 90°C here. So which variable is that we're actually looking for delta little T. That's time. Remember, don't confuse this with capital delta T. Because that's a change in temperature. So how do we figure out what this DELTA T. Is our Q equals M. Cat equation doesn't have time in it. But we're actually gonna use another equation that we saw earlier in physics because we have something here that's measured in units of power, which is wants. So remember that the power equation is just P equals W over delta T. That was the works supply over change in time. Now, remember work is also just energy. So another way to sort of say this is just the amount of energy that something requires or takes divided by the change in time. So this is really where my delta T. Comes from. So if I want to solve for this DELTA T. Here, all I'm gonna do is just trade places with the power of variable. So they're just gonna swap places and my delta T. Just becomes the energy divided by power. Right? So this is units of jewels and this is units of jewels per seconds. When you divide them, you're just gonna end up, you're gonna end up with seconds. Alright, so what's going on here then? What's the amount of energy that I need? Well again, if you think about this, the hotplate delivers all of this heat energy into the water and all of it goes into warming the water from 10 to 90 degrees Celsius. So really what happens is the E. In my power equation is actually just being transferred into the water as heat energy. It's just Q. So this E. Here just becomes Q. So my delta T. Equation just becomes Q. Overpay. Now, what happens is my delta T. I have an equation for Q. Remember the specific heat equation is just M. C. Times delta T divided by the power. So that's just what that becomes. And so now I can start head go ahead and plug everything in. So I've got the mass of the water, that's 0.3 kg, that's 0.3 then I need the specific heat of water which is 86. That's just here in this table right here. So I've got 41 86. Now, it's the change in the temperature. Remember that change in the temperature could be either in Celsius or kelvin, it doesn't matter which one you use, If I'm going from 10 to 90, that means that the change in temperature, that's big. T is equal to 80. Alright, so this is just gonna be 80 over here. And then finally, just divide by the power, I'm told that the electric hot plate here is rated for 200 watts. So this hot plate here is just p equals 200. So this is just gonna come 200 over here. And then what you end up with is you end up with one times 10 to the fifth when you work out the numerator divided by 200 you'll end up with delta T. Is equal to 500 seconds and that's just about eight and 8.3 repeating minutes. Alright, so we'll take about eight and roughly 8.5 minutes in order to warm this water from 10, which is like room temperature to about 90 which is almost boiling. Alright, so that's your final answer. Let me know if you guys have any questions for this

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