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Points Sava AngA4 500 sample ol methanoll ICHJOH; was combusled in Ine presenca o 0o0 103 Hexcesspxygen in @ bomb calonmeler containing 9 olwater The temperature of...

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Points Sava AngA4 500 sample ol methanoll ICHJOH; was combusled in Ine presenca o 0o0 103 Hexcesspxygen in @ bomb calonmeler containing 9 olwater The temperature of Ihe water increused Irom 24,80 "C la 29.76"C Thle hear capaclly 0l the calorlmeter was 2657 JIC, The specilic heal of water i5 4BA Jn"C How much heal In kJ, was produced by tha combustion of the melhanol sample? (Requirestat Ieast] Bleps ol calculatons "489 kJ"100 kJEiakJEff12 kJ

Points Sava Ang A4 500 sample ol methanoll ICHJOH; was combusled in Ine presenca o 0o0 103 Hexcesspxygen in @ bomb calonmeler containing 9 olwater The temperature of Ihe water increused Irom 24,80 "C la 29.76"C Thle hear capaclly 0l the calorlmeter was 2657 JIC, The specilic heal of water i5 4BA Jn"C How much heal In kJ, was produced by tha combustion of the melhanol sample? (Requirestat Ieast] Bleps ol calculatons "489 kJ "100 kJ EiakJ Eff12 kJ



Answers

One gram of liquid benzene is burned in a bomb calorimeter. The temperature before ignition was $20.826^{\circ} \mathrm{C}$ and the temperature after the combustion was $25.000^{\circ} \mathrm{C}$. This was an adiabatic calorimeter. The heat capacity of the bomb, the water around it, and the contents of the bomb before the combustion was $10000 \mathrm{JK}^{-1}$. Calculate $\Delta_{\mathrm{f}} H^{\circ}$ for $\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{l})$ at $298.15 \mathrm{K}$ from these data. Assume that the water produced in the combustion is in the liquid state and the carbon dioxide produced in the combustion is in the gas state.

To calculate the heat released program of the hydrocarbon. We need to know the amount of heat that was produced during the combustion of the hydrocarbon in the bomb calorie emitter. The heat of the reaction will be equal to negative the heat of the calorie emitter. The heat of the calorie emitter will be the heat capacity of the calorie emitter provided at 11.9 kila jewels per kelvin or kill jules per degree Celsius, multiplied by the change in temperature, which will be the final temperature, 23.55 degrees Celsius minus the initial 20.0 degrees Celsius. And we get negative 39 37 killer jewels of energy released. Because this occurred in a bomb calorie meter, it was a constant volume. So we can calculate Q. V. The heat released at constant volume per gram of the hydrocarbon by taking the heat that was released divided by the grams of the hydrocarbon that released that amount of heat. And we get negative 2590 killer jewels per gram hydrocarbon.

Hi there. In this problem, we are using a bomb calorie emitter to calculate how much energy is released from the burning of our sample. Um, so what we need to know here is the equation the amount of energy released is gonna be opposite in sign from the amount of energy absorbed and that amount of energy absorbed, he's gonna be the energy absorbed by the water that is in the bomb. Calera bitter, plus the energy absorbed by the bomb calorie emitter itself. So we're looking to calculate both of thes um, a little more detail in terms of the equation we want to use. We know that the equation for calculating the amount of energy absorbed by the water is going to be waters specific heat times the mass of the water times how much the temperature changes the map, the way we're going to calculate the amount of energy from the calorie emitter he is. We're going to have to take the heat capacity of the the bomb. And we're going to have to multiply that times the change in temperature E All right, So this is our plan now. We just need to start adding some values and solving for energy. Let's see, the specific heat of water is 4.1 84 Jules, for every gram degrees Celsius, we have 675 g of water, and the change in temperature is going to be 26.9 degrees, which was our final temperature and subtracting our initial temperature from that. All right, that takes care of the energy absorbed by the water for the energy absorbed by the bomb calorie emitter itself. We need to take the heat capacity, which is 534 Jules, for every degree Celsius that it changes. And again, we need our delta t. How much the temperature changed. Final temperature 26.9 minus initial temperature. 23.4. He's gonna be in degrees Celsius. And all of this I'm gonna put in brackets. So we remember, too, that it is gonna be negative when we're done going through the calculations. Now it is time for the calculator. Check out some units here. See that grams are gonna cancel degrees Celsius. They're gonna cancel the east degrees. Celsius will cancel. So we're left with a number of jewels of Let's see, when I calculate this I get 98 are 9885 jewels absorbed by the water. He and 1800 and 69 jewels of energy absorbed by the bomb calorie meter. Adding these together reversing the sign. I get negative. Okay, 11,754 jewels of energy. If I'd like to express this in killer jewels just to make the number a little more manageable I can do that by dividing by 1000 jewels in every kilo. Jewell, He had my answer. Rounding two significant figures would be 11.7 killer jewels of energy. Just the amount of energy released in this reaction. Thank you so much for watching.

So we have a problem that involves burning hard hydrocarbon and a bomb calorie meter. So first, what we need to think about is the act of combustion inside the calorie emitter. We know that the medium it's in his water. So as the hydrocarbon is combusted, it's going to heat up the bomb. Men heat up the water, so we know that the change in temperature here is going to be positive. As an exile. Thermic reaction occurs in a closed system, and then we also know at a thermal equilibrium that the heat given off by the combustion will be equal to the heat absorbed by the calorie Mader. So the problem is asking about the heat given off by combustion per gram of original substance, but it doesn't give us enough information to solve for the heat given. So we have to do since we know it's in thermal equilibrium, as we have to solve instead for the heat absorbed. And then what we'll do is use the mass value given for the hydrocarbon to find the heat given off by the combustion in terms of some energy value over grams. So here we have all the given data for the problem now. First, what we do need to do is set up in equation for the heat absorbed because this is what we can solve for. So we know this is gonna be equal to the heat of the water and the heat of the calorie maner. Now we need both of these values because it is a They're both part of the system that's absorbing the heat. So first, me the massive water in order to put a plug it into the emcee Delta T equation. So we have a volume of water of two point 550 leaders, but we don't have a mass. So what we can dio as weaken convert this to a mass using the density of water, which is one I am all right, one gram per middle leader. And then we have to do one more little conversion to get the correct units of volume cancelled where there's 1000 middle leaders in the leader. So all over volumes will cancel and we know we're left with 2550 grams of water. So now we have all the information we need to figure out the heat absorbed by the water and calorie emitter. So plugging all the values into the heat of water equation we know the mass, so the mass will cancel the grand bias from the specific heat were the specifically capacity of water is 4.184 grands and then we also have the change in temperature is 3.55 grams which you know from the problem, So we'll be left with a jewell value. And so the energy or the heat absorbed by the water is equal to 37,000 875 0.7 jewels and then the second part here we can solve for the heat of the calorie under. Now we don't take into the mass of the calorie motor into consideration. All we have here is the heat capacity and we have the change in temperature. Now we know the change in temperature is 3.55 degrees Celsius. But the heat capacity of the calorie meter is given in Calvin. So we need to do a little quick conversion where we know that to convert Celsius to Calvin, we simply just have to add 273.15 So this becomes 403 Jules per degree. Calvin multiplied by 200 270 6.7. Calvin Komen's cancel and we're left with a jewell value of 111,500 and 10.1 jewels. Now, since these heat values these energy values air so high, it is pretty easy just to convert to kill a jewels. And the problem doesn't specify whether has to be in jewels or killed, Jules. So just to save yourself some writing, we know that this heat of water is gonna be equal to 37 point 87 five. Kill a Jules, remember, You just divide jewels by 1000 to convert to kill it, Jules, and the heat of the calorie emitter is going to be equal to 111 0.51 zero killer jewels. You just move the decimal place over three times still left. So now we have all the heats we need and we can figure out the heat given off by the combustion program of hydrocarbons. So here we see that the heat given is gonna be equal to these two heat is absorbed, uh, added together, and this will give us a final energy of 149 0.385 killer jewels. Now, this is the heat given off by the combustion of the hard to carbon. But we're not done because the problem wants to know the heat given off per gram of substance. So we know that Q given going to be equal to 149 0.385 killer jewels. Then we just have to divide by the mass of hydrocarbon given in the problem, which is 1.5 20 grams. So finally we know from combustion doing this division that the hydrocarbon burning gives off 90 to 98. Sorry. Point to a killer jewels per gram burned

Okay. Another heat transfer question. So this time we are going to mirror TNT right direction of TNT. The energy released during the explosion. So it says right, you have .74g of TNT. and during the reaction the co kilometer right. The temperature increased from 23.4 to 26.9. And the heat capacity is given. Of the kilometer is 534 and it contains right 675 millimeter water. So it asked how much heat was produced by the combustion or exploring the TFT sample. So normally we say right, we regard this whole system, the whole all the things as a system. So the system. But he changed all the energy change of the whole system for the zero since its insulating system And the energy change of the whole system actually equals two. The energy change of the reaction, Right? That's what we want to calculate plus and to change it was a solution. Right? It leads to because the And you will be transferred to the solution. But here you say in the question also gives the heat capacity of the caliber cemetery, right? Which means you have to take the kilometre into account. So you have to add color barometer. Right? And we can do a arrangement right? Which means now, right? The energy transfer or energy change for the reaction equals two minus sign. Right? The change of the solution plus the energy change. Oh, calorie meter, right? And then we can expand it. Right? The energy change of the solution actually equals two. The specific heat of the water ray times the mass of the water and times the temperature chance. And plus right, the energy change of the Karam meter equals two here. Right? You say that capacity is already given, so equals to the capacity of the kilometer. Right? Times the temperature change. And then we can do a rearrangement right? Equals minus sign pause. Right here, capacity with a kilometer times Delta team. Now we can just plug in the number. Right? So Keep minus sign. So the specific heat of the water is 4.1 8 4. And the mass of water here, right is actually uh huh. 675. Okay. And plus the heat capacity is Of the cholera meter, which is 534. And then you have times right? The change of the temperature. So the final temperature is 26.9. The initial temperature is 23.4 and this Is equals two. Right? Use your calculator. You'll find actually roughly equals two. 11.75 kinkajou. Right? And this is the answer. And you will find right there has a minus sign. So which means distraction is the exact same correction.


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