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Ch.9 - Thermochemistry: Chemical Energy

Chapter 9, Problem 145

We said in Section 9.1 that the potential energy of water at the top of a dam or waterfall is converted into heat when the water dashes against rocks at the bottom. The potential energy of the water at the top is equal to EP = mgh, where m is the mass of the water, g is the acceleration of the falling water due to gravity 1g = 9.81 m>s22, and h is the height of the water. Assuming that all the energy is converted to heat, calculate the temperature rise of the water in degrees Celsius after falling over California's Yosemite Falls, a distance of 739 m. The specific heat of water is 4.18 J/(g·K).

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Hello everyone today. We have the following question. Gravitational potential energy is converted to thermal energy due to the vibrations at the point of impact with a surface. It is given by the formula PE or potential energy is equal to the mass Or where mass is equal to MG is equal to acceleration due to gravity which is 9.8 m/s squared. And height is our age. And an experiment a 2.5 kg aluminum sphere is dropped from a height of 654 m to the ground, assuming that all the potential energy is converted to thermal energy, calculate the change and temperature of the aluminum sphere and then we have the specific capacity here. So first we need to recall our formulas for potential energy as well as our specific heat. So potential energy we have in the question here. Mass times acceleration due to gravity times our height and for our specific heat it is our mass times are specific heat capacity times the change in our temperature. And we also need to be made known of one conversion factor that one jewel is equal to one kg meters squared per second squared. Now we're going to set these two equations equal to one another for what was given to us in the question. And so we have our mass here which is 2.5 kg. The acceleration due to gravity. This will be found in the reference text 9.8 m per second squared And our height was set to be 654 m on the right hand side, our masses once again 2.5 kg. However, we must convert this two g because when it comes to specific heat, our units are in grams. So we use the conversion factor that one kg is equal to 10 to the third grams. And then lastly we multiply by our specific heat capacity .897 jewels. Program Celsius and multiplied by our change in temperature which we do not have and what we are trying to find. Ultimately, when we simplify both of our sides, we get 16,023 kilogram meter squared over second squared is equal to Or that is going to be equal to 16,023 jewels. As we did use this conversion factor in step one. And so lastly We're going to take that value. Our 16,023 jewels. We're gonna convert that to C by using the conversion factor that 1°C is equal to 2,242.5 jewels. Our units of joules cancel out. And we were left with 7.15°C as our change in temperature. And we have now solved the problem overall. I hope that this helped. And until next time
Related Practice
Textbook Question

Methanol (CH3OH) is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane (CH4):

Step 1. CO(g) + 2 H2(g) S CH3OH(l) ΔS° = - 332 J/K

Step 2. CH3OH1l2 → CH4(g) + 1/2 O2(g) ΔS° = 162 J/K

(e) In what temperature range is step 1 spontaneous?

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Textbook Question

Methanol 1CH3OH2 is made industrially in two steps from CO and H2. It is so cheap to make that it is being considered for use as a precursor to hydrocarbon fuels, such as methane 1CH42: Step 1. CO1g2 + 2 H21g2 S CH3OH1l2 ΔS° = - 332 J>K Step 2. CH3OH1l2 S CH41g2 + 1>2 O21g2 ΔS° = 162 J>K (f) Calculate ΔH° for step 2.

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Textbook Question
Ethyl chloride 1C2H5Cl2, a substance used as a topical anes-thetic, is prepared by reaction of ethylene with hydrogen chloride: C2H41g2 + HCl1g2 ¡ C2H5Cl1g2 ΔH° = - 72.3 kJ How much PV work is done in kilojoules, and what is the value of ΔE in kilojoules if 89.5 g of ethylene and 125 g of HCl are allowed to react at atmospheric pressure and the volume change is - 71.5 L?
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Textbook Question
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Given 400.0 g of hot tea at 80.0 °C, what mass of ice at 0 °C must be added to obtain iced tea at 10.0 °C? The specific heat of the tea is 4.18 J>1g °C2 and ΔHfusion for ice is + 6.01 kJ>mol.
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Textbook Question
Imagine that you dissolve 10.0 g of a mixture of NaNO3 and KF in 100.0 g of water and find that the temperature rises by 2.22 °C. Using the following data, calculate the mass of each compound in the original mixture. Assume that the specific heat of the solution is 4.18 J>1 g °C2 NaNO31s2 S NaNO31aq2 ΔH = + 20.4 kJ>mol KF1s2 S KF1aq2 ΔH = - 17.7 kJ>mol
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