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A copper calorimeter can with mass 0.100 $\mathrm{kg}$ contains 0.160 $\mathrm{kg}$ of water and 0.018 $\mathrm{kg}$ of ice in thermal equilibrium at atmospheric pressure. If 0.750 $\mathrm{kg}$ of lead at a temperature of $255^{\circ} \mathrm{C}$ is dropped into the can, what is the final temperature of the system if no heat is lost to the surroundings?

$21.264^{\circ} C$

Physics 101 Mechanics

Chapter 14

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Lectures

02:20

A solid, liquid, or gas is one of the three main states of matter. Solids have a definite shape and volume, and retain their shape when a force is applied to them. They are rigid, and do not flow to take on the shape of a container, but retain their own shape. Solids are held together by intermolecular forces, which are usually chemical bonds. Liquids have a free-flowing, continuous surface, and take the shape of a container. They flow to fill an available space. Their particles do not have a definite shape or volume, and they are not rigid. Gases have no definite shape or volume. They are not held together, and are not rigid. They flow to fill an available space. Gases are often described as being the state of matter with the lowest density.

03:25

The First Law of Thermodynamics is an expression of the principle of conservation of energy. The law states that the change in the internal energy of a closed system is equal to the amount of heat energy added to the system, minus the work done by the system on its surroundings. The total energy of a system can be subdivided and classified in various ways.

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Hey, everyone, this is question number 74 from chapter 14. In this problem, we're given a cut copper kalorama of mass. Your 0.1 kilograms were told that it contains their 0.16 kilograms of water and 0.18 kilograms of ice and then sat atmospheric pressure. We're told we're given that we drops your 0.75 kilograms of lead at 255 degrees Celsius into the counter ometer, and then we're asked to find the final temperature of the system. So we're back to Kalorama Tree. So are two equations for this Our cubicles emcee Delta T. And then if we have a phase change, it's Q equals L F er V, depending on if it's fusion or vaporization times m. So these air two questions will be using, and we're trying to find the final temperature of the system, which all of the elements in this system have in common, so we can go ahead and define or start this problem by defining Q for each element of the system. So we have the can, the water, the ice and the final temperature, and all four of them will have in common. The unknown tea, which will be the final temperature. So for the can Q. C C equals mass, your 0.1 kilograms see for copper is 390 Jules per kilogram, kay and then Delta T is our final temperature, which we don't know our high temperature minus our initial temperature. And because we have ice in the system and where atmospheric pressure or initial temperature is going to be zero degrees Celsius. And if you play the inter calculator, you get 39 Jules per okay times t t. Here being our final temperature. Okay, so now we can define water, so Q W No face changes here, so it's going to be the same thing. Emcee Delta t So am 0.16 kilograms zero for C for water is 4.19 times 10 to the third Jules, her kilogram kay and then tea for water again because we're dropping something much higher in temperature way and assume that tea is the final temperature that we don't know. It's going to be higher and then minus our initial zero degrees Celsius. And if you put that in your calculator, you get 670 0.4 Jules per k Time's team again are unknown. Now we can define Q for the ice. No, for the ice because our temperature is we're going to assume is increasing. The ice is going to melt, so we're gonna have to take into account the change in phase of the ice. So cue for the ice is going to be the the energy it takes to melt in, so m l f. So a massive ice is 0.18 kilograms and then heat effusion for ice is 3.34 times 10 to the fifth Jules per kilogram K and then we have the increase. Teo, whatever our final temperature is, which is going to be plus emcee Delta T. So am against your point No. 18 kilograms. See, for now, water now that it's melted for 0.19 time send of the third. I'm relieved the unit so I can try and squeeze this. And then tea is going to be Delta team's going T minus zero. And if you plug that into your calculator, you're going to get expression because there's two terms in this and you get 612 jewels plus 75.4. Jules. Okay, Times T. Okay, so now we define cue of the Len. That is equal, Tio. No face changes. So we're just going to say it's Elsie Delta T. So am 0.75 kilograms. See, for lead is 130 Jules for kilogram kay and then our delta T t minus 2 55 TV in the final temperature of the of the system. And if you plug that in, you get que ele is equal. Tio 97 0.5 Jules per K times T minus 2.486 time. Send the fourth, Jules. Okay, so we have 1/4 impressions now, Theo. Russian ship were told that we have no heat loss, So our Sigma Q summation of Q is equal to zero. So now we can just combine everything and solve for tea, which is what we have in common. So we have 39 Jules per k Times T plus plus 6 70.4 Jules per K times T 12 Jules plus 75.4 Jules per K Times t and then plus 97.5. Jules per K Times T minus 2.486 times 10 in the fourth. Jules. Okay, so big Long expression combined like terms and weaken solve for tea minusca the algebra. But you end up getting tea is equal to 1.85 times 10 to the fourth Jules divided by eight, 182.3 jewels. And that gives us a temperature of 21.4 degree. See? So that's the final temperature of the system. Make note that is greater than zero degrees Celsius. So our assumption that the ice completely melted was correct.

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