hello bobcats today we're going to discuss heat transfer between substances and thermal contact so when two substances are in thermal contact in other words you have a high a substance with one a higher temperature than the other substance the hotter body or the higher temperature body or higher temperature substance let's say would be a better way to say it the higher temperature substance will transfer heat to the lower temperature substance until both substances reach the same temperature the same temperature so if you think about it if i take um a piece of metal that's really hot and i place it in um uh some water that's cooler what happens is the hot metal will transfer some of the heat to the water and as it does that the metal cools down and the water warms up and they both reach at a point where they have the same temperature which we call thermal equilibrium and the one with the higher uh specific heat will change temperature less than the one with the lower specific heat now in this case what we also need to understand is that the heat lost by the higher temperature substance and we'll call that q lost is equal in magnitude and opposite in sign we'll put it underline that q loss equal i'm in magnitude opposite in sign of the heat gained by the lower temperature substance and the one that is the one it gains will say a negative q gain okay substance and we'll say that's a negative q of gained energy or heat so what we can say then is that the amount of heat lost by one substance has to equal the amount of heat gained by the other substance and it's equal and opposite in sign so a positive heat loss is equal to a negative heat gained in a system now remember for q we can use the equation m c delta t that is it what q is so i can say the m c delta t of the substance that has lost heat is equal to a negative m c delta t of the substance that has gained the heat okay and this is a very important concept here so again let me refresh you with this if i put a piece of metal in water and the metal is a high temperature the metal will lose heat and go down in temperature and the water will gain the heat and go up in temperature but in this case this is a loss and this is a gain now so let's look at an example um this is the example that's on the powerpoint let me find my piece of paper here so this is the problem that you have on the powerpoint that we're going to go through right now using these these ideas up here so you can call this a calorimetry practice with heat transfer and the way i do problems like this as i read it i write down all my variables so this is a 25 gram sample of metal so i have my mass of my metal and that's going to be 25 grams so that's that then we have a temperature of the metal and that's the initial temperature so that's the initial temperature of the metal and that's going to be 87.7 degrees celsius is placed in 37.4 grams of water so the mass of the water is 37.4 grams and the water is at a temperature of 10.3 degrees initially so the initial temperature of the water is 10.3 degrees celsius and then the final temperature of the whole system so the temperature final is going to be 15.4 degrees celsius now because the final temperature is thermal equilibrium that means both the metal and the water are at that temperature the final temperature now the equation remember is that the heat of the metal because that's the higher temperature one is going to equal and the heat of the water but it's going to be an opposite value okay negative value instead so whatever is lost by a metal is gained by the water so remember the equation that we can use is going to be m c delta t and i'm just going to put a parenthesis around it so that i can identify as the metal is equal to the opposite value of an equal of the m c delta t of the water well i have the mass of the metal and i have i don't have specific heat of the metal in fact that's what we are looking for the specific heat so i know that i'm going to have to isolate for this value right there that c i don't have the change in temperature for the metal so i'm going to do that right now now remember the change in temperature is always temperature final minus temperature initial and so in this case for the metal the the change in temperature for my metal is going to be the final which is 15.4 degrees celsius minus the initial 87.7 degrees celsius so my change in temperature for the metal is going to be give me a second i got to calculate it out is going to be a negative 72.3 degrees celsius and the change in temperature for my water again final minus initial so the final is the equilibrium thermal equilibrium or 15.4 degrees celsius minus the initial temperature which is 10.3 degrees celsius so my change in temperature for my water is going to be 5.1 degrees celsius a positive okay so there is the change in temperature for the metal change in temperature for the water so now remember to isolate this all i have to do is divide both sides by the mass of the metal and the change in temperature of the metal so i'm going to rearrange this and write this as the specific heat of the metal because remember this is all about the metal is equal and opposite to the mass times the specific heat times the change in temperature of the water all over the mass and change in temperature of the metal mass times change in temperature of the metal now i'm just using this w and m just to identify which one i'm looking at and so now i'm going to substitute in for those and so that's going to be a negative value and the mass of the water remember we're looking at mass of the water and that's this one right here is 37.4 grams the specific heat of water is oh i don't have it written down but it's something we should remember memorize it so c of the water is equal to 4.184 joules over grams times degrees celsius okay and it's a specific heat of the metal that we're looking for we don't know what that is so this specific heat is 4.184 joules over grams degree celsius times the change in temperature of the water so the change in temperature of the water is 5.1 degrees celsius all over the mass of the metal which is way up here 25 grams times the change in temperature of the metal so the change in temperature the metal is this one a negative 72.3 degrees celsius and so when i look at my units my grams cancel my that cancels and so i get the units for specific heat still and the negative divided by a negative gives a positive and so my specific heat of this metal is going to be 0.44 joules over grams times degrees celsius and that is iron okay so we'll practice one more and then i will do another video so let's look at the next one that i have for practice i've already got it printed out so again we should call these calorimetry practice problems for heat transfer i forgot to call to give it a proper name so that's the calorimetry practice problems for heat transfer because we're transferring heat from a high temperature object to a low temperature object okay so another practice problem let me get a different color pow pen just because i don't want to keep doing the same color i want to be different and then this one again i'm going to write out my variables this one we have that's 10 grams of silver so the mass of my silver which is ag is equal to 10.0 grams and it is heated to 100 degrees celsius so that's my initial temperature so the initial temperature of the silver is equal to 100 degrees celsius then it is added to 20 grams of water so my mass of the water is 20 grams and the water starts out at 23 degrees celsius so the initial temperature of the water is 23.0 degrees celsius now when they hit thermal equilibrium the temperature was measured at 25 degrees celsius so that is the t final for both of them the water and the silver and that's going to be 25.0 degrees celsius now the specific heat so we got to talk about specific heat here so the specific heat of water is the c for the water again is 4.184 joules over grams degrees celsius and then we've got to find the specific heat of silver so the specific heat of silver is what we don't know and then we already know that it's all about change in temperature so i'm going to go ahead and do my change in temperature for the silver is remember its final which is 25 .0 minus initial which is a hundred point zero so the temperature change is going to be a negative 75.00 degrees celsius that's for the water i mean the silver sorry and the change in temperature for the water is the final 25 minus the initial 23 and so that comes out to be two or nine that's is it two or three two degrees celsius okay so now i have all my information and again we're going to go back to that equation where the q of silver is equal and to the opposite value of the q of the water because the energy loss has got to be equal to energy gain they just have to have different or opposite signs so q again is equal to m c delta t that's what q is so i'm going to substitute that in since that's for the silver and then the negative q is negative m c delta t for the water now remember the one that i need to isolate for is the specific heat for the silver so i just divide both sides by the mass and change in temperature of silver so the specific heat of the silver is going to equal to the negative value of the mass times specific heat times change in temperature that's mcat of the water divided by remember it's just the mass and change in temperature of silver so that's going to be the mass times the change in temperature of the silver now i'm just going to substitute all my values in so the specific heat of silver is equal to a negative or opposite sign of the mass of the water remember this is water on top mass of water was 20 grams the specific heat of water is 4.184 joules over grams degrees celsius and then the change in temperature of the water which we figured out down here is 2.0 degrees celsius and that's all over the mass of the silver which was 10 grams times the change in temperature of the silver and the change in temperature of the silver we figured out down here is a negative 75 degrees celsius and then i make sure my units work out so grams cancel grams celsius crams celsius the specific units are joules per gram degree celsius and then of course the negative divided by negative gives me positive and i haven't figured out this value so i got to get a calculator real we're going to quick my calculator right here so we take since the negative divided by negative is positive i'm not going to worry about the negative so it's going to be 20 times 4.184 times 2 divided by 10 divided by 75 enter and so the specific heat of silver is 0.223 [Music] joules over grams degrees celsius okay and so that's how we solve the problems for using calorimetry for heat transfer between two substances in thermal contact and i'll see you next video