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When a 0.740 - g sample of trinitrotoluene (TNT), $\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{N}_{2} \mathrm{O}_{6},$ is burned in a bomb calorimeter, the temperature increases from $23.4^{\circ} \mathrm{C}$ to $26.9^{\circ} \mathrm{C}$ . The heat capacity of the calorimeter is 534 $\mathrm{J}^{\prime} \mathrm{C}$ , and it contains 675 $\mathrm{mL}$ of water.How much heat was produced by the combustion of the TNT sample?

$q_{\text {reaction }}=-1869 \mathrm{J}$

Chemistry 101

Chapter 5

Thermochemistry

University of Central Florida

University of Maryland - University College

Brown University

Lectures

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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.

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