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(a) The tube of a mercury thermometer has an insi…

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Problem 71 Hard Difficulty

An iron cube floats in a bowl of liquid mercury at $0^{\circ} \mathrm{C}$ (a) If the temperature is raised to $25^{\circ} \mathrm{C}$ , will the cube float higher or lower in the mercury? (b) By what percent will the fraction of volume submerged change?

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Physics 101 Mechanics

Physics for Scientists and Engineers with Modern Physics

Chapter 17

Temperature, Thermal Expansion, and the Ideal Gas Law

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Video Transcript

Okay, So for this problem, I will draw the diagram first. So it says that there is a block off iron which is floating on mercury. So this is ironed and we have mercury. The initial temperature is given as zero degrees Celsius. So we assume that when this block of iron was dropped and the smug re, both of them attain thermal equilibrium and both of them are the same temperature, the temperature has increased to some D final as 25 degrees Celsius. And the question is that what happens do the block off iron that this block of iron sinks more or it pops up slightly. That is the question. So before we start the problem, as I tell my students, the first thing is that we need to highlight what are the concepts which are applicable to the scenario? I want other physical models which will be using in this problem. So the first physical model is obviously term Oh, expansion. Because as temperature rises, the volume off mercury and the volume of iron increases and thermal expansion formula is given by HVT is equals two. We not one class beater delta T, where beauty is the final volume V note is the initial volume be does the coefficient off volume expansion and Delta T is the rights of temperature. Okay, if the volume increases, then the density also will change. According to this formula, density is masked by volume, the master mating the same save the volume increases, the density is going to fall and the third is will use the principal off flotation. Why we're using this principle is because the block of iron floats on mercury, as the density of iron is years last, has combatted. Density of mercury is going to float on principle. A flotation also includes our committee's principal with state said. When an object is partially or completely summers in a fluid experiences a buoyant force which is equal to the weight of the fluid displaced by the object now dismissed portions have highlighted. This is submerged, so the weight of this fluid displaced is the buoyant force. The weight of the fluid displaced by this green portion off iron is the buoyant force, which is equals to the volume off the submarine's portion off iron times. The density off mercury Times Gene and I got into the principle of flotation, F G should be equal. Steffy. So we're going to use this. So these are the three key concepts which are used in the scenario. The next thing to think is how do we connect these three d arrive at a conclusion, So we need to see what is being asked in the problem. The problem asks that we need to find out whether, if we increase the temperature, what happens to the block, whether it goes up or sinks more. So obviously it's got something to do with the forces. Now. These forces are dependent on the density, which in turn is dependent on the volume, which in turn, you know, which can actually get it by this formula, Bt's we not one plus beater Delta T. So approach will be that we'll start with our equation and then we'll arrive at an equation for density comparison and then they will compare the density comparison, relax. We compared the densities to the Boyne forces and once we compared the buoyant force to the force of gravity, we can come to a conclusion whether it sinks or the crops up slightly. I hope the approach is clear, so I'll repeat. We'll start with the volume comparison will connect the density and then we'll connect the buoyant force to the force of gravity. All right, so another thing do not here is that the coefficient off volume expansion for mercury is greater than the coefficient of volume expansion for iron. Which means that for a rice off temperature, mercury expense more s compared to our and by using the formula, density is equals to mass by volume. And we know the density is proportional inversely to volume that if the volume rises, the density falls. We can conclude that the full in the density off Marguerite or the decrease in the density of mercury is greater then the decrees in the density off iron, because the increase off the volume off mercury is greater than the increase of volume of firing. Therefore, we can safely conclude this, and we're going to use this to solve our problem. Okay, so once you have gorgeous equation next I'll draw the diagram once again out here, and I'll draw the forces on this. We have the force, which is the buoyant force, and we have the force on gravity since the object is floating the net force zero in the via direction. So the F b, which is the buoyant force, equals the F G, which is the force of gravity. Now, as I have had it before, to the force or the buoyant force is equals to the weight off the floor displays by this green portion off. All right, which is equal to the volume off the sub Morris portion of farm times. The density off our times gravity Sorry, the density off McGary times gravity. So that is our boy in force. And the force of gravity is equals two the volume off the whole iron block times the density off the iron block times gravity. So in this equation, GMG cancels and bigger equation that the volume of the submerged portion off iron divided by the volume off the iron block is equals. Do the density off. I'm divided by the density off mercury. Okay, so but this equation, we can clearly see that the decrees in the density of mercury is more than the decrease in the density of iron. So which means that distant nominator here decreases more as compared to this new measure here. So we can safely assume that the volume of the submerged portion off iron increases slightly and therefore the block will sink. Or we can say that it will float, nor in mockery. You can also see that the F G is equals to the weight of the block. But F b changes according to this equation. So in this equation, we can clearly see that F B is dependent on the density off. Mercury and density of mercury decreases Maur as compared to the density of iron. So therefore, on the left, inside of the density of mercury decreases more than this volume off the sub Morris portion of Iran should increase likely to compensate for the change and hence the blocks would sink more. So that is our answer to part one that the block sinks are a florist lower in mercury. Okay, For the third part of the situation, we need to find out what is the fractional change. Okay, so the fraction of the volume submerged here, the fraction off the volumes submerged is given by the equation we mercury divided by we aren't I'll say it displaced, and we'll drop this subscript later on. So the fractional change is given by V McGary divided by we all right. Minus we not the initial We not more agreed to run it by me. Not all right. Derided by the knot Mercury divided by re not aren't just a fractional change, you know, in the volume which is changing here. So in this case, I will use the equation. V Mercury is equals two. We not mockery. One less beater. Mercury times Delta t All right, so I'll replace it on the next page. So, replacing the first time in the previous equation we get we not Mercury one place, Beto Mercury Delta t do added by we not if e one place beata f e Delta t. That's the first time of the previous equation minus we not more green. Directed by we not there e divided by v, not mockery divided by three, not f e. So, as you can see clearly in this case, there this term distraction in this fraction And this fraction can be taken common in the numerator. And if we cancel out with that in the denominator, So we're just left over with one bless beater, more great Delta t divided by one place. Beata. I'm Delta t minus one, and I'll just plug it into the equation. We get one place 1 80 times tend to the bar off Negative six. And Dr T is our 25 divided by one place 35 times turned to the par negative six, which happens to be the coefficient or volume expansion for iron and minus one. If you saw that, you will get 1.0, 45 divided by one point or 875 my calculator and you get 3.6 times 10 to the power. Negative three in the Cutler, which ends up with 0.0 36 which is 0.36 percent. Okay, so that's the answer. For the second part off the problem, which is the fractional change off, the volume is 0.36 percent.

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Physics for Scientists and Engineers with Modern Physics

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Problem

(a) The tube of a mercury thermometer has an insi…

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Physics for Scientists and Engineers with Modern Physics

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Video Transcript

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics

Andy Chen

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Physics for Scientists and Engineers with Modern Physics

Andy Chen

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

Douglas C. GiancoliPhysics for Scientists and Engineers with Modern Physics

Andy Chen

Marshall Styczinski

Farnaz Mohseni

Aspen Fenzl

Douglas C. Giancoli

Physics for Scientists and Engineers with Modern Physics