Ch 18: Heat and TemperatureWorksheetSee all chapters
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Concept #1: Specific Heat

Transcript

Hey guys, in this video we want to talk about something called the specific heat. We spent a lot of time talking about qualitatively, how heat can change the temperature of a substance. Now we want to talk about quantitatively how much heat is required to change the temperature of a substance. Let's get to it. Remember guys that heat is an exchange of energy that affects the temperature of a substance and this is something I've been harping on since the beginning. The question we want to focus on now is how much heat exactly is required to change the temperature of a substance. The amount of heat to change the temperature of a substance should depend upon two things, how much substance you have or the mass of the substance and the particular material the substance is made out of. Some materials give up heat very sorry some materials change the temperatures very easily. Aluminum foil for instance you can bake something in the oven at a high temperature for a long time, a little while, maybe 20, 30 seconds after, you pull the aluminum foil out of the oven it's already back at room temperature. Very little heat needs to be transferred for it to change its temperature wildly. The specific heat which is given by a lower case C gives us the amount of heat required per unit mass to change the temperature and it's defined as Q over M delta T. Now it's important to understand specific heat is per unit mass. Remember that the amount of heat depends upon two things, the mass and the material. If you remove the mass from the equation by saying it's the amount of heat per unit mass, then the specific heat only depends on the material of a substance. So the specific heat is an inherent quality of a material. It does not change with the amount of material that you have because it's per unit mass and it's very very specifically defined that way. Rearranging this equation we would rewrite it as Q equals M, little C, delta T. This is sometimes referred to as the M CAT equation because if you pretend that the delta is a capital A, it spells M CAT and that's how some students remember it. Let's do a quick example. Water has a specific heat of 1 calorie per gram degree Celsius, remember specific heats are inherent to the material to the substance so water no matter how much we have of it will always have this specific heat. What is its specific heat in SI units? I'll call this question A. How much heat is required to increase the temperature of 0.05 kilograms of water by 15 Kelvin, I'll call this question B. Okay, so question A, we are doing a conversion here so we need to set up a conversion problem. 1 calorie and this is lower case calorie per gram degree Celsius Sorry if this is going to be times. We have three units that we need to convert calories need to become joules, grams need to become kilograms, degree Celsius need to become Kelvin. So we're going to need three conversion factors at least and this is going to equal some amount of joules per kilogram Kelvin. First let's do calories. We know from mechanical equivalence that one calorie lower case is 4.186 joules. So that takes us from calories to joules. Now a gram to a kilogram is very easy, 1000 grams is 1 kilogram and that takes us from a gram to a kilogram. Now the tricky one is degrees Celsius to Kelvin that's the last we need because that's going to cancel the degrees Celsius in the denominator and give us Kelvin. You have to think about what this number or what this unit signifies. Is this degree Celsius a unit of temperature or a unit of change in temperature? Because that's very very important. Remember, Celsius and Kelvin are not the same unit if you're talking about temperature but as I've harped on over and over and over, a change in temperature in degrees Celsius equals the change in temperature in Kelvin. So their units are the same for changes. If we look at this equation, we're talking about a change in temperature. So for a change in temperature it's one degree Celsius equals one Kelvin, they're one to one, they're the same for changes in temperature. So plugging this into our calculator, we get 4186 Joules per kilogram Kelvin which is sometimes simply written as 4.186 kilojoules per kilogram Kelvin. A lot of these specific heats are very very large. They're usually in the thousands of joules. So typically the unit that you see associated is a kilojoule per kilogram Kelvin not necessarily joule per kilogram Kelvin. Now Part B. that was to find how much heat is required to increase the temperature of 0.05 kilograms of water by 15 Kelvin. So delta T is 15 Kelvin, C we know for water is 4.186 kilojoules per kilogram Kelvin because we found that out in part A and we know the mass is 0.05 kilograms and all we need to do to find the heat is use the M CAT equation. So Q is M C delta T which is 0.05 kilograms times 4.186 times 15 Kelvin. Now don't forget what the units are of our specific heat. Our specific heat is in units of kilojoules per kilogram Kelvin so our output for the heat is also going to be in kilojoules. And this is 3.14 kilojoules. Like I said a lot of these problems are going to end up being in the thousands of joules so often kilojoules is the common unit you're going to find. Alright guys, that wraps up our discussion on specific heat. Thanks for watching.

Practice: A piece of copper has a 0.0005 m2 area and a 5 m length. What heat is required to increase the length by 0.5 cm? The density of copper is 8960 kg/m3 , the specific heat is 385 J/kg K and the expansion coefficient is 1.7 x 10-5 K-1 .