Ch 21: Heat and TemperatureWorksheetSee all chapters
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Ch 01: Intro to Physics; Units
Ch 02: 1D Motion / Kinematics
Ch 03: Vectors
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Ch 21: Heat and Temperature
Ch 22: Kinetic Theory of Ideal Gasses
Ch 23: The First Law of Thermodynamics
Ch 24: The Second Law of Thermodynamics
Ch 25: Electric Force & Field; Gauss' Law
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Zeroth Law of Thermodynamics
Thermal Expansion
Introduction to Heat
Changes in Temperature & Specific Heat
Changes in Phase & Latent Heat
Temperature Change Across Phases
Phase Diagrams, Triple Points and Critical Points
Heat Transfer

Concept #1: Calorimetry


Hey guys in this video we're going to talk about calorimetry which is how you measure heat transfer during the processes that involve temperature changes or phase changes let's get to it. As I said calorimetry is the process of measuring how much heat certain processes take like if you want to convert ice into water or if you want to convert ice at -15 degrees Celsius into water at 30 degrees Celsius or if you then want to convert water into vapor any of these processes we want to figure out how much heat they take this is why we use calorimetry we use the M CAT equation and the latent heat equation to theoretically compute these values but calorimetry involves actually experimentally doing it. This can also be used to measure specific heats and latent heats for substances which we then use in our M CAT in our latent heat equation to make these theoretical predictions in calorimetry the fundamental idea is the heat gained by one equals the Heat lost by the other. So if I have two substances A and B that are in thermal contact with one another and A kicks over Q A it loses Q A and B gains Q B and then however much Q A lost has to equal however much Q B gained so If substance sorry however much B gained so if B gained 15 joules of energy A had to lose 15 joules or Q A had to be -15 joules that's why we have a negative sign in calorimetry problems that negative sign is crazy important so don't forget now the only way that calorimetry works is if your substances are completely isolated from the environment that way for instance substance A can't send heat out into the environment or substance B can't receive heat from the environment so you need very very good insulating, thermal insulating to completely isolate thermally A and B so that heat can only transfer between them there can be no outside influence on those systems let's do an example to see how this works in practice. 500 milliliters of cola has 50 grams of ice in it if the cola starts at 27 degrees Celsius how much ice melts for the cola to reach 2 degrees Celsius the density of cola is 1000 kilograms per cubic meter the fusion blah blah blah the rest are constants we don't need to read those assume no heat loss so what we are assuming is that the heat is only transferred between the cola and ice and the environment has no effect on that heat transfer. So the condition for calorimetry is that whatever heat the ice loses, the cola gains and since we're talking right now about heat transfers relating to changes in temperature we're talking about the cola going from 27 degrees Celsius to 2 degrees Celsius we need to use our M CAT equation for Q cola so Q cola is going to be the mass of the cola times the specific heat of the cola times the change in temperature for the cola we know the change in temperature for the cola because we're told the initial and the final temperature we know the specific heat of the cola but we don't know the mass of the cola, we are however told that there's 500 milliliters of cola and that the cola has a density of a 1000 kilograms per cubic meter, that's the same density as water and it's equivalent to 1 gram per milliliter so if the math of the volume of cola is 500 milliliters then the mass is 500 grams so 1 gram for every milliliter, so this is 0.5 kilograms which is 500 grams times the specific heat of cola which is 0.85 remember that this is in kilojoules so your final answer is going to be in kilojoules and the change in temperature is 2 degrees Celsius minus 27 degrees Celsius right the final minus the initial even though they are in degrees Celsius because this is a change in temperature it's equivalent to Kelvin plugging all this in we get 10.6 sorry -10.6 kilojoules right the cola is losing heat because it's decreasing its temperature and the ice is absorbing all of that heat what is the ice doing with that heat when it absorbs that ice is melting if we're talking about melting, melting is a phase change right from frozen water to liquid water so we need to talk about latent heat in this problem, so the heat for the ice is going to equal our latent heat equation which is actually going to equal a negative Q cola this is our condition for calorimetry that means that the mass of the ice equals negative Q cola over L. Q cola is already negative so this becomes positive which is good because mass can only be positive and this is in kilojoules so the latent heat of fusion for water better also have kilojoules in it and the latent heat of fusion of water is 334 and it is in kilojoules so we're good to go. Dividing that we get 30 sorry, we get 0.032 kilograms. So the mass of ice that melts is 32 grams. The number 50 grams never came into this problem it was irrelevant how much ice you have but it's good to know that you had enough ice for 32 grams of it to melt to give the cola that 2 degree Celsius temperature you wanted so it's really nice and cold to drink. Alright guys this wraps up our discussion on calorimetry. Thanks for watching.

Practice: 1g lead pellets are heated to 350 K. How many of these pellets must be added to 500 mL of water so that the equilibrium temperature is 25°C if the water is initially at 10°C? The specific heat for water is 4.186 kJ/kg K and 0.128 kJ/kg K for lead.

Example #1: Equilibrium Temperature of Coffee in a Cup


Hey guys, let's do an example for 350 grams of coffee at 75 degrees Celsius is in a 0.15 kilogram glass at 27 degrees Celsius if the specific heat of coffee is blah blah blah, what is the temperature of the coffee at equilibrium? Assume no heat loss to the environment. Since we're talking about the temperature of the coffee at equilibrium we know the temperature of the coffee has to equal the temperature of the cup since the coffee and the cup are in thermal contact with each other but not in thermal contact with anything else right assume no heat loss we know that this involves calorimetry. So our calorimetry condition is that what ever heat the coffee loses, is heat gained by the cup and in both of these scenarios we're talking about changes in temperature due to the heat exchange between the coffee and the cup so we need to use our M CAT equations so Q for the coffee is going to be the mass of the coffee times the specific heat of the coffee times the change in temperature right so what's the final temperature T that equilibrium temperature, minus the initial temperature of the coffee. Now Q of the cup is going to be the mass of the cup specific heat of the cup times once again the delta T. Which is that final equilibrium temperature minus the initial temperature of the cup. But what I need in order for the calorimetry equation to apply the calorimetry condition to apply is I need -Q of the cup. So I'm going to make this -Q cup by simply reversing the order of the subtraction this is still going to be the mass of the cup the specific heat of the cup but now it's going to be the initial temperature of the cup first minus that final equilibrium temperature. Now our calorimetry condition says that this value has to equal this value, what we are told in the problem is that the coffee and the cup reach equilibrium what that means is that the final temperature T has to be the same for both the coffee and the cup, so I'm going to set the two things highlighted in yellow those two heat together. Right Q coffee equals - Q cup that means that the mass of the coffee.C heat of the coffee there's going to be a lot of writing guys just bear with me. There's going to be a lot of writing for you to.That's simply setting those two equations equal to each other don't forget though that these temperatures these equilibrium temperatures are the same because the coffee and the cup in equilibrium and it's what we're looking for, we're looking for the equilibrium temperature so we want to solve for those temperatures that I highlighted in green so we want to expand these coefficients. Right so that we can get the T's isolated on both sides of the equation sorry this is coffee on the left not cup, and still we're looking for those T's which are the equilibrium temperatures so I want to move all the equilibrium just to one side let me start giving myself some space here. I'm going to move this term over to the left side, and I'm going to move this term over to the right side that way I can have all my constants on the right and I can have my unknown temperature on the left.This term was negative on the right side so it's positive on the left side.This term was negative on the left side so it's going to be positive on the right side, and now I have those equilibrium temperatures both together on the left side but they each have a coefficient so I need to factor that coefficient that was a terrible brace I need to factor that coefficient out, when I factor it I get M, C times T where T is that equilibrium temperature that we're looking for and this equals all that stuff on the right. So what I want to do to get T by itself is I simply want to divide this coefficient which I have written in purple I want to divide it over to the right hand side. Now I have T by itself and that looks like all of this junk so much stuff to write, divided by the coefficient that I highlighted in purple. This is the point where we can actually plug in numbers because we know the mass of the coffee the mass of the cup the specific heat of the coffee the specific heat of the cup and the initial temperature for both the coffee and the cup so we can start plugging all of this stuff in finally. The mass of the cup is 0.15 kilograms right. The specific heat of the cup is 0.84 and this is in units of kilograms so make sure that all of your specific heat sorry kilojoules make use all of your specific heat have kilojoules the initial temperature of the cup was really quickly 27 degrees Celsius right room temperature because we're talking about absolute temperatures here we've got an isolated temperature over there, we've got isolated temperatures here and here there are no differences in temperature to be seen anywhere we cannot just use Celsius we have to convert from Celsius to Kelvin you can only use Celsius and Kelvin interchangeably when it's a difference in temperature but now that we're talking about an absolute temperature you cannot use them interchangeably so it was initially 27 Celsius so plus 23 takes us to Kelvin Now there was 350 grams of coffee which is 0.35 kilograms the specific heat of coffee was 8.9 and this is in kilojoules and initial temperature of the coffee was 75 degrees Celsius which I need to convert into Kelvin. The coffee 0.35 kilograms 8.9 is the specific heat the cup 0.15 kilograms 0.84 is the specific heat. Plugging all of that in we finally get a temperature of 346 Kelvin now you could leave it like that but the odds are since the problem gave all the initial temperatures in degrees Celsius you should find the final equilibrium temperature in degrees Celsius. So that temperature in degrees Celsius is going to be 346 minus 273 which Is just 73 degrees Celsius. So the coffee started at 75 degrees Celsius right once it reached thermal equilibrium with that cup it only lost 2 degrees Celsius to reach thermal equilibrium which is barely anything and that's because the cup required very very little heat to change its temperature dramatically. That's it for this problem guys. Thanks for watching.