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How much heat (in $\mathrm{kJ}$ ) is needed to convert $866 \mathrm{g}$ of ice at $-10^{\circ} \mathrm{C}$ to steam at $126^{\circ} \mathrm{C} ?$ (The specific heats of ice and steam are $2.03 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ and $1.99 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}$ respectively.)

$2674.3 \mathrm{kJ}$ is needed

Chemistry 102

Chapter 11

Intermolecular Forces and Liquids and Solids

Liquids

Solids

Drexel University

University of Kentucky

Lectures

04:08

In physics, a solid is a state of matter characterized by rigidity and resistance to changes of shape or volume. Solid objects have a definite volume, they resist forces (such as pressure, tension and shear) in all directions, and they have a shape that does not change smoothly with time. The branch of physics that studies solids is called solid-state physics. The physical properties of solids are highly related to their chemical composition and structure. For example, the melting point of ice is significantly lowered if its crystal structure is disrupted.

03:07

A liquid is a nearly incompressible fluid that conforms to the shape of its container but retains a (nearly) constant volume independent of pressure. As such, a liquid is one of the four fundamental states of matter (the others being solid, gas and plasma). A liquid is made up of tiny vibrating particles of matter, such as atoms, held together by intermolecular bonds. Water is, by far, the most common liquid on Earth. Like a gas, a liquid is able to flow and take the shape of a container. Most liquids resist compression, although others can be compressed. Unlike a gas, a liquid does not disperse to fill every space of a container, and maintains a fairly constant density. A distinctive property of the liquid state is surface tension, leading to wetting phenomena.

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to approach this question. We're going. Consider each of the different cues that we're gonna have taking place as this transition occurs. So starting off we have solid water, which then we heat up from starting at negative 10 degrees Celsius, which is gonna dio all the way till we hit the transition point, which is solid water at zero degrees Celsius, which is then going to turn toe liquid water, which is also at zero degree Celsius. And then that is gonna heat all the way up until we have liquid water at 100 degrees Celsius, which is then going to turn into steam, or gaseous water, which is also at 100 degrees Celsius. And lastly, that is going to go two gaseous water. Sorry. This should be gaseous, not liquid, which is gonna be at 100 and 26 degrees Celsius. So we're going to denote this Q s for solid. This is gonna be Q f perfusion. This is gonna be cute. L for Liquid. This is gonna be Q v for vaporization. And lastly, we have qg for gaseous. And what we need to find is Q total, which is simply the son of all of these things. So let's go ahead and figure out what each of these are gonna be. Okay, starting with Q solid. We're gonna have, um, 866 grams times 2.3 Jules per gram. Degree Celsius times the temperature change, which we know for this is going to be zero minus 10 degrees Celsius. Yeah, and if we calculate that out, we're gonna find that that equals 17,579 0.8 jewels. Okay, Next up, Q f profusion were given a Moeller value. So we're gonna need to convert this 866 grams into moles. So we're going to multiply that by one mole, huh? 18.2 grams. You can get this for water just right off your periodic table, and then we're going to multiply this by 6000 and 10 jewels Terminal. This is also 6.1 kilo jewels per mole, as you will find in your data table. And that comes out to be 1,960,000. 274 274 jewels. And then you can move on now to the Q of our liquid. For this, we just have our 866 grams again. Multiply this by 4.19 Jules per gram. Degree Celsius times are temperature change, which is going to be 100 degrees Celsius minus zero to re Celsius. And this value here is going to equal 288,000. My apologies, not 288,000. This value is going to equal 362,000 854 jewels. Now we have our vaporization for vaporization Q. V. We have Q V equals that 866 grams times one more her 18.2 grams. Because again, this is a Moeller value that we're gonna be working with. And that is times 40 houses and 700 and 90 Jules Permal. The heat of vaporization of water is incredibly large. 40.79 kilo drills Permal, and if we calculate that out, we get a value off. 200 and 88,000 826.9. My apologies. It seems that I have these values flipped for Q fusion and Cube vaporization. So I'm gonna go ahead and swap these because I got the values in the wrong spot on my calculator. So that should be 1,960,000 274. Which does make sense given that we do expect this to be larger than Q F. And lastly, we have Q of our guests, which is 866 grams times our heat of our guests, which is 1.99 Jules per gram degree Celsius times are temperature change, which is 126 to 3 Celsius minus 100 degrees Celsius for our gas. And we go ahead and calculate that it and we will find that is, 44 1000 806 jewels. Okay, now, if we go ahead and some all these values up for Q total, we just simply are all these numbers up? We find that Q total is gonna equal 2674 0.3 Hyla jewels. We've gone ahead and done that conversion there because this value is gonna be so incredibly large that it's simply worth our time to convert it over tequila jewels and remembering We do need to round this off. Two significant figures were working with three significant figures, as that's what we're limited with with our 866 grams. So that's gonna equal to 0.67 times 10 to the three que jewels, which also equals 2.67 Mega Jules. Either one of these would be considered acceptable, and if you really want it would be 2.67 times 10 to the six jewels. So again, we've just gone ahead and systematically summed up each of these different Q values for those transitions that have taken place. And then we send them all together to find the total energy of this reaction.

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