Specific heat and latent leat of fusion and vaporization | Chemistry | Khan Academy

Let's get out of the special heat and Let's talk about the heat of melting and boiling. Let's say you have a container full of liquid and want to increase the temperature, will probably add heat. But how much heat do you need to add? There is a formula for this. Let's see what's going on here. It will depend on different things. First, how much is the temperature depends on what you want to increase. That is how much the temperature Want to increase? The more you want to increase you need to add a lot of heat. It also depends on how much material you have. In other words, from the mass of the liquid here. The more, the more to raise the temperature need a lot of heat. And it depends on one thing, from the specific heat capacity of a given material. Heat some materials from others more difficult.

And if the specific heat capacity of the material If high, to increase the temperature more coul heat will be needed. Let's be more specific. Suppose our math is water and its temperature is 20 degrees Celsius. Let's say a big pot There are two kilograms of water in it. Special heat capacity, you can look at them or the data. It turns out that the specific heat capacity of water 4186 coul section kilogram degrees Celsius.

And these units are special creates an idea about heat capacity. He says that if a kilogram of water is 1 degree Celsius 4186 coul energy is needed to heat. That's what you already say you need as much heat as you can. Water has a very high specific heat capacity. It is great without raising the temperature too much can hold a certain amount of energy. And let's say the question is from us raise the temperature to 50 degrees Celsius they want. How much to reach 50 degrees Celsius need to add heat? Well, we can come here. We need to find the heat we need to add, weight is 2 kilograms, that is, a special heat capacity of 2 kilograms, he is also 4186, tap temperature change. This also means that T is the initial output of the last T.

So what does the last T equal? T is the last 50 degrees Celsius, we want to get here, The output initially started with 20 degrees Celsius. Now we can find it by the amount of heat. And if you hit it all You will receive 251160 coul. Water temperature 30 degrees This is a very large energy to increase. Therefore, we use water mainly as a heat sink. A lot of heat can be given to the water and the temperature does not change much.

But this example is very simple. Let's look at the more difficult. Let's say you heat the cube from below rather than heating we just take hot metal. Suppose we have 0.5 kilograms of copper and we put it in the water. We preheated the copper and now we put it in the same bowl of water. And we want to know what they are will reach equilibrium temperature? Copper will cool and water will heat. Then when they reach the same temperature there will be a balance. What temperature is this? Well, we need to know a few things. I have already given the mass of copper. We need an initial temperature, initial temperature of copper.

I said we overheated it. Suppose it is 90 degrees Celsius. We also need to know the specific heat capacity of copper and its specific heat capacity is 387. And let's say the initial properties of water has. It was two pounds. The specific heat capacity of water always remains 4186. And let's say it's a temperature of 20 degrees Celsius starts with. That is, we know the equilibrium temperature, the equilibrium temperature at which they will reach about 20 to 90. We need to find out exactly what it will be. And the trick we will use here copper heat loss, and water is heat gain, How do we compare these temperatures? That the heat is not lost assuming they will be the same. We believe that heat is not lost, that is, you want him to occur on the calorimeter, indoors, in an environment that prevents heat from escaping. If the heat is not lost at this time, the heat received by the water will be equal to the heat lost by the copper. In essence, the heat from copper zero will be taken if you collect the heat added to the water, because one of them will be negative, the other is positive and their modulus will be equal.

How do we find them? We have a formula. I remembered that Q is equal to mc delta t, remember that I want it because it's like MCAT, i.e. mc, this delta resembles Aa, The total Q is equal to MCAT. Now I have to use a mass of copper, The weight of copper was 0.5 kilograms, hit 387 with a specific heat capacity tap temperature difference. I do not know the final temperature, this is normal. I will name this variable that I left blank. Let's end it with T I'll call it first. Started at 90 degrees Celsius, that is, let's go 90 degrees Celsius, plus heat from water, we use the same MCAT formula for this.

It weighs 2 kilograms The specific heat capacity is 4186. T end, I still don't know the T end, but I know the original Ti, The initial temperature is 20 degrees Celsius. And I make this expression zero. Our space has narrowed. I'm sorry. It looks a little scary now, you have a great formula. The unknown is hiding here. Can we solve it? Yes, we can. Look, you just have an unknown. The unknown T is the end. These are the same variables. This temperature is water and copper is the temperature to be reached. The whole expression here, the orange expression will give a negative assessment, because copper loses a certain amount of energy and the expression of water is positive, because it gains heat. Both will be corrected. And it gives you zero.

This is what we need. We just need to find the end of T, we hit them for that, T combines the last variables We find the end of t. First I will knock everything open. I will combine the last terms of t and T will calculate the non-final expression. So I will separate the expressions. I try to find the end of T and get 21.58 degrees Celsius. When you look at it, you can say that we did something wrong. 21.58? Water started with 20. The temperature almost did not rise. Yes, that's what we said, the specific heat capacity of water is very high, you can add a lot of heat and the temperature will not change much. Consider that we could add this dish as well. This pot also absorbs some heat, that is, we may have another expression here, Q of the vessel and if we take this into account, or we could throw another cube here and one here.

If you collect all the incoming Qs the heat you gain or lose if you added you would get zero, because if the heat is not lost, that heat will be transferred here. No heat will be generated or lost. Simply contact materials will be transferred inside. This was also the key to resolving the issues, for special heat capacity issues of this type. That's how you set it up and then you found the unknown. In this case, T is the end. Sometimes a variable that is not given to you one of them may have a mass, or the specific heat capacity of one, find the answer without making a difference. Let me ask you another question. Suppose we take the same amount of water, At 2 pounds and 20 degrees Celsius, but this time i want to know that to boil and evaporate all the water here how much heat should I add? Well, the first thing we have to do is to find the boiling point.

And the boiling point of water It is 10 degrees Celsius. With the Q MCAT formula, MC is written with delta T. The mass is 2. The specific heat capacity of the water is 4186. Temperature change, good boiling point is 100, that is, I increase the water temperature from 20 to 100. T is equal to the last 100. T initially to 20. And I know the boiling point must add heat to reach, 669760. I do not pay attention to the main figures, but this is a calculation. But it is not enough to boil. It's just the temperature of the water It was to raise it to 100 degrees Celsius. Not enough. Bring the water to 100 degrees Celsius if you want, he just stands here. But it does not boil.

We must continue to add heat. How much heat should we add? When the water reaches 100 to evaporate the juice you boil and evaporate you need to know the temperatures. Because we boiled in this case because evaporation heat is needed we evaporate the liquid. If we turned a liquid into a solid we would call the heat of melting. Formula for melting and evaporation temperatures looks like it. Q is needed to change the state of the aggregate which is the amount of heat. This happens when the state of the unit changes, melting and evaporation temperatures. Specific heating temperature occurs when changing.

This is the calculated temperature To increase to 80 degrees Celsius indicates how much heat is needed. This calculation will tell us that When you are at 100 how much heat we need to add Let's turn the aggregate state of water from liquid to vapor. Melting and boiling temperatures The formula is similar. Q is equal to mL. m is the mass. The more mass, the more heat you need to add. L is the specific heat of melting and boiling. This term is similar to the specific heat capacity, but you have to change the temperature instead of showing how much heat is needed how much you need to change the state of the aggregate indicates that heat is needed. And it turns out that the boiling point of water the specific heat capacity is very large. 2,260,000 coul kilograms per kilogram. This means that one kilogram of water boiling point To convert 1 kilogram of steam 2 260 000 coul energy needs. So if this juice at 20 degrees if I want to steam first with mc delta T I have to bring it to 100 degrees Celsius.

Then I add a little heat, that is, m multiply by L. It weighs 2 kilograms. L for water is equal to 2 260 000. I get boiling water to bring the temperature Need 669 760 coul and then him to turn the liquid into a seal 4,520,000 coul, this in general 5 189 760 coul gives, that is, 2 kilograms of water at 20 degrees the amount needed to convert to steam. Now I want to show you something. Let's delete it. Instead of starting with water, let's say we start with ice.

Let's say you started with a big piece of ice and 3 pounds of ice. He is very cold. Just not zero. Let's say the initial temperature minus 40 degrees Celsius. Our question is that how much heat should we add 3 pounds of ice 3 pounds turn to steam? But we don't just bring it to the point of evaporation, I want to make it warmer than 100 degrees Celsius. I want the T to end Let it be 160 degrees Celsius.

How much heat to do it should i add To better watch it and bring it to life in our eyes I am typing the temperature on the vertical axis and how much heat by creating a function I will calculate what we need to add. Let me tell you briefly how to do it. In simple terms, Q is equal to mc delta T. m is equal to 3 kilograms. c, c also has a value. We will talk about this for a minute. Delta T. Well, delta T, my final temperature It is 160 degrees Celsius. The initial temperature is minus 40. Remember the negative sign. And I included a special heat capacity I solve the problem, this is wrong. You can't do that. What specific heat capacity should you include? Water? Ice? Steam? Each has a different price. And there are two aggregate changes. First, ice turns into water and then after a while the water evaporates. You can't ignore these aggregate changes. So we do not. That's what you have to do. We start with minus 40 degrees Celsius. I know it's above the arrow, but i don't care the zero point of the vertical line.

And we will add heat to it will reach zero degrees Celsius and we must stop here. We have to stop at zero degrees Celsius, because the state of the aggregate changes every time we must stop. How much heat is this now? We can find mc delta T. m is 3 kilograms. The specific heat capacity of ice is 2090 and the final temperature is zero, our initial temperature, T last exit initial negative 40. Remember the negative sign. We calculate this and find the amount of heat. 250 800 coul. But this was just to bring the ice to the melting point. Now we have to melt it. What will the graph look like when it melts? Ice during melting the temperature will remain constant. When the ice cube melts the temperature does not change. All energy is spent on breaking communications and turn ice into water. How much heat is this? When the state of the aggregate changes, Q is equal to mL.

m is 3 kilograms. Special heat capacity for melting, evaporation we can not use. The solid turns into a liquid. The melt needs a special heat capacity and this heat capacity for water Divide 333,000 couls by a kilogram, This also means that 999,000 coul heat The amount of ice is zero degrees Celsius we can turn sapphire into water. Now you see how it works. Where should we take water from scratch? Not to 160. Here, because 100 degrees Celsius it turns to steam. And we have to stop when the state of the unit changes, because the specific heat capacity will change. To find Q here We can use mc delta T, The specific heat capacity of water is 4186. Delta T, last 100, is the initial zero. We buy 1 255 800 coul. Now we have to evaporate the water. How much heat do you need? The state of the unit is changing. That is, we will do it in mL, the mass is 3 kilograms. Specific heat capacity of evaporation It is 2,260,000, ie the water evaporates 6 780 000 coul to convert need energy and we must look at one step.

Steam at 100 degrees Turn to steam at 160 degrees. We must also do a mc delta T calculation, The mass of steam is 3 kilograms. The specific heat capacity of steam is 2010. The final temperature is 160. The initial temperature was 100, which also gives us 361,800 coul. It needs so much heat that minus 40 degrees ice cube Let's turn to steam at 160 degrees..

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