Calorimetry Problems, Thermochemistry Practice, Specific Heat Capacity, Enthalpy Fusion, Chemistry

so how can we solve this problem? How much energy is required to

heat

80 grams of water from 26 degrees Celsius to 48 degrees Celsius and gives us a

specific

heat

capacity

of the water? What equation do we need if we want to calculate the amount of thermal energy? required whenever you have a temperature change use this equation q is equal to m cap m c delta t we have the mass in grams which is 80 grams and we have the

specific

heat

capacity

which is 4.184 joules per gram per centigrade and the final temperature change minus 48 initial degrees centigrade minus 26 degrees centigrade that is 22 degrees centigrade now let's pay attention to the units the unit centigrade cancel the unit grams cancel and then we are left with the unit in joules so q is equal to 80 times four point one eight four multiplied by twenty-two, Therefore, seven thousand three hundred and sixty-three point eight joules are required to heat 80 grams of water from 26 degrees Celsius to 48 degrees Celsius.

Specific heat capacity indicates how much energy is required to heat one gram of water by one degree. degrees Celsius and that's 4.184 you might need to know that, but that's it for this question, that's all you have to do now. here is the second question: how much energy is required to melt 75 grams of ice and we are given a heat of

fusion

, so in the last example we had a temperature change problem we had to use q equals mcat but in this problem it is a phase change, we have ice that is a solid and is melting into a liquid when you have a phase change, you can use this equation q is equal to m times delta h in this case delta h of melting, but for phase change

problems

you have to pay attention to the units because sometimes q could be equal to n times delta h, which is the moles multiplied by the

enthalpy

value, but here this is joules per gram, so we want to use the mass instead of moles just keep in mind it has to be in joules it could also mean calories but you want it to be in a unit of energy usually joules so we'll use this equation so we have 75 grams of ice and the heat of melting of ice is 334 joules per gram, so what this means is that if you have one gram of ice, it takes 334 joules of thermal energy to melt that gram of ice, so if you want to melt 75 grams of ice, you have to multiply 75 times 334 to get that answer and it is equal to 25 and 50 joules, so that is the amount of thermal energy required to melt 75 grams of ice into a 70 gram piece of metal at 180 degrees.

Centigrade was dropped into a bucket containing 400 grams of water at 25 degrees Celsius, the final temperature of the mixture was 28 degrees Celsius, what is the specific heat capacity of the metal? Well, let's understand what is happening in this problem, so we have a bucket of water and a temperature is 25 degrees Celsius, we are going to place a piece of metal inside this bucket of water and initially the temperature of the metal is 180 degrees Celsius. , now heat will flow from hot to cold, so heat will be transferred from the metal into the surrounding water molecules now, as the metal loses thermal energy, the temperature will decrease and as the water molecules absorb thermal energy, the temperature will increase at the end of this heat transfer process, the final temperature will be the same as the temperature of the metal will decrease to a value and the temperature of the water will increase to a certain value until those two values ​​meet and then thermal equilibrium will be established and that is the final temperature and the mixture will remain at that temperature in this case. that temperature is 28 degrees Celsius using this information we can calculate the specific heat capacity of the metal the heat released by the metal is equal to the heat absorbed by the water now for this process to work it is necessary to add a negative sign to each side of the equation because the temperature of the water increases and the temperature of the metal decreases, so one side is negative and the other side is positive, so to balance it you need to introduce a negative sign since this is a temperature change problem. we can use the equation q is equal to m cap so we have negative m cat on the left and positive m c delta c on the right so the mass of the metal is 70 grams we are looking for the specific heat capacity of the metal the final temperature of the metal is 28 minus the initial temperature which is 180. the mass of water is 400 grams the specific heat capacity of water is 4.184 and the change in temperature of the water the final is 28 the initial is 25. so 28 minus 25 is three and if we multiply three by four point one eight four and four hundred on the right side this is equal to five thousand twenty point eight on the left side twenty-eight minus one eighty is negative one fifty-two times negative seven that's going to be positive ten thousand six hundred and forty times c now, to solve c we need to divide both sides by this number, so c is equal to five thousand twenty point eight divided by ten thousand six hundred forty, so that's point four seven two joules per gram per centigrade, so that's how you can find the heat capacity specific to the metal in this particular problem. 100 grams of metallic iron at 95 degrees Celsius were placed in 200 grams of water at 25 degrees Celsius.

What is the final temperature of the mixture? This is like in the previous problem. This is another heat transfer problem, the heat released by the metallic iron is equal to the heat absorbed by the water sample, so q of affine is equal to q of h2o and let's add a negative sign to make this work now the temperature The final temperature will be somewhere between 95 and 25 degrees Celsius. If we average the two numbers it equals 60. If you add 95 and 25 and then divide it by 2 you should get 60. Do you think the final temperature will be between 60 and 95 or 25 and 60 what?

Would you say now notice that the specific heat capacity of water is much greater than that of metallic iron? It is 4.184 compared to 0.45. The substance that has a higher specific heat capacity can absorb much more energy without significantly changing its temperature, so a substance that has a very high specific heat capacity will generally experience a small change in temperature, so the final temperature will be close to 25ºC, it will be lower than 60ºC and we also have twice as much water compared to iron in terms of mass, so now let's figure it out. so negative m cap is equal to mcat so we have 100 grams of metallic iron the specific heat capacity is 0.45 delta t is the final temperature minus the initial temperature of the metallic iron which is 95 and that is equal to the mass of water which is 200 times the specific heat capacity of water is 4.184 times the final temperature minus the initial temperature of the water, which is 25.

So 100 times 0.45 is 45. 200 times 4.184 is 836.8, so now what we have to do is do is distribute 45 times negative tf, that is 45 negative tf. and negative 45 times negative 95 is equal to positive 4275 now if you multiply those two, it's going to be 836.8 times tf and then 836.8 times negative 25. it's negative twenty thousand nine twenty, let's move this forward, so what should we do to continuation? get tf by itself, so let's add 45 tf to both sides, so 836.8 plus 45 will be 881.8 tf. Next, let's add twenty thousand nine twenty to both sides, so twenty five thousand 195 is equal to 881.8 tf, so now let's divide both sides by 881.8.

So 25195 divided by 881.8 gives you a final temperature of 28.57 degrees Celsius. This is how you can find the final temperature when you mix two substances at different temperatures, you need to know the specific heat capacity of each of them and then you can solve it. Using the process we just performed in a coffee cup calorimeter, 2.6 grams of calcium chloride were dissolved in 260 grams of water at a combined initial temperature of 23 degrees Celsius, the final temperature was 26.4 degrees centigrade and we calculate the

enthalpy

change of the reaction, so we are. Given this reaction, how can we find the enthalpy change?

Normally, the enthalpy change is usually in units of kilojoules per mole, so we can estimate it by taking q of the reaction, which is usually in joules, but we will have to convert it to kilojoules. and we need to divide it by the moles, we can easily find n if we convert grams to moles, we can do that, so go ahead and pause the video and work on this problem and then unpause when you're ready to see the solution. Now how can we find the q of the reaction? The energy released by the reaction or absorbed by it is equal to the energy absorbed or released by the water, but it will be negative somewhere and once we have q we could divide it by n and we will get delta h, so let's find out how much thermal energy was absorbed or released by the water sample and let's use the equation q is equal to mcat, so we have 260 grams of water, that is the mass, the specific heat capacity is 4.184 the final temperature is 26.4 minus the initial temperature of 23 , so this is equal to 3698.7 joules.

By the way, this particular reaction would you say it is endothermic or exothermic knowing that the temperature of the solution increased from 23 to 26.4, what would you say now? that the q value for water is positive for h2o the temperature rose which means the water absorbed heat so for water it is an endothermic process but for the reaction it is exothermic because the reaction had to release energy at the surrounding water molecule so that the temperature of the water can rise, so whenever the temperature of the solution rises, that tells you that the reaction was endothermic, if the temperature of the solution fell, the reaction would be endothermic, let's say that if the temperature dropped, that means that the water lost thermal energy that it released. thermal energy to the reaction meaning the reaction absorbed or gained thermal energy so it is endothermic for the reaction exothermic for water if the temperature of the solution dropped but in this case it is the other way around so keep in mind Note that q of the reaction is equal to negative q of water, so it will be negative 3698.7 joules, so we have an exothermic reaction in this particular example.

Our next step is to take the grams of calcium chloride and convert them to moles, so what is the molar mass of calcium chloride? of calcium is approximately 40.08 and the molar mass of chlorine is 35.45 35.45 times 2 is 70.9 plus 40.08, that's approximately 110.98, so the molar mass tells you the relationship between grams and moles, for calcium chloride, if you have one mole of the substance, then you have a mass of 110.98 if we divide these two numbers 2.6 divided by 110.98 is equal to the zero point two three four three moles so now we can calculate the enthalpy of the reaction which is q divided by n so q is negative 3 698.7 n is this number and there is something else we need to do, we actually need to convert joules to kilojoules, so let's not forget to do that at the top, a kilojoule is equal to a thousand joules, so you have to divide it by a thousand, so it's negative three point six nine eight. seven kilojoules and let's divide it by 0.02343 moles so the final answer is negative 157 0.9 kilojoules per mole this is how you can find the enthalpy change of the reaction given the temperature change of the water so here it is Today's last question, calculate the energy required to heat 24 grams of ice to negative 20 degrees Celsius to vaporize at 250 degrees Celsius and we have everything we need we have the heat of

fusion

the heat of vaporization the specific heat capacity for ice and water and also the specific heat capacity for steam like Okay, so how can we solve this problem?

Well, first let's draw the water heating curve. Ice melts at zero degrees Celsius and water boils at one hundred. We start at a subzero temperature which is minus twenty as we add heat in the x. The axis will be the amount of heat added on the y-axis. This will be the temperature, so let's start at -20. As we add heat the temperature will increase towards zero once we get to zero the ice will start to melt but when that happens. the temperature will remain constant and then once all the ice has melted the temperature will start to increase once we have liquid water and once we reach 100 the water will start to boil into steam and then the temperature will increase again in this problem, you need to realize that there are five different q values ​​that we need to calculate. q1 is a temperature change problem.

We need to find out how much heat is required to raise the temperature of the ice from -20 to zero. You can go directly from -20 to 250 because the changes in specific heat capacity are different for ice, liquid water and steam, so you have to divide them into five parts. q2 is the energy needed to melt 24 grams of ice. q3 is the thermal energy necessary to raise the temperature of the water from zero to one hundred. q4 is. the energy required to vaporize liquid water into steam and q5 is the energy required to heat steam from 100 to 250.

So once we find these five different values ​​of q, we can add them together and get the total heat required to go from ice to negative . 20 degreescentigrade for steam at 250. so let's start by calculating q1, which is a temperature change problem, it is equal to mcat, the mass is 24. the specific heat capacity of ice is 2.03 and the temperature change we go from negative 20 to zero because ice melts at zero, so that's an increase of positive 20 degrees Celsius 24 times 2.03 times 20. is 974.4 joules now, what about q2 q2 is the phase change? will be the mass multiplied by the heat of fusion the mass is 24 grams The heat of fusion is 334 joules per gram and this is equal to 8,000 16 joules q3 is a temperature change problem so we need to use m c delta cm still is 24. the specific heat capacity of water is 4.184 and the temperature change is We are going from zero to 100, so the temperature change is 100 degrees Celsius, so if we multiply these values, you will get 1041.6 joules, like this which now we need to find q4, which is a phase change problem, is the mass multiplied by the enthalpy of vaporization, so that's going to be 24 times, this is in kilojoules and so far everything else is in joules, so so instead of using 40.7 kilojoules per mole we can use forty seven hundred, but actually Note that this is moles, so we have to convert grams to moles, so let's start with 24 grams of h2o, the molar mass of the water is about 18.

That's two plus 16. and the enthalpy of vaporization is 40.7 kilojoules per mole, so right now the unit grams of water cancel out and the moles cancel out, so the last thing we need to do is convert kilojoules to joules and that's one thousand joules per kilojoule. 24 divided by 18 is approximately 1.33 if we multiply it by 40.7 and then by a thousand, this is equal to 54,266.7 joules, so that is q4. Now our next step is to find q5, which is a temperature change problem, so m is 24, the specific heat capacity of the vapor is 2.01 and the temperature change we go from 100 to 150, that is, 100. to 250, that is a change of 150 and this is equal to 7236 joules, so to find the total heat absorbed we need to add the five values ​​of q, so this is equal to eighty thousand five hundred thirty-four point seven joules , as you can see most of the energy went into vaporizing the water.

It takes 54,000 joules to vaporize liquid water into steam, so that's where most of the energy went, but that's how you find the total, so that's it for this video, thanks for watching and have a great day.