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A sophomore with nothing better to do adds heat to0.350 $\mathrm{kg}$ of ice at $0.00^{\circ} \mathrm{C}$ until it is all melted. (a) What is thechange in entropy of the water? (b) The source of the heat is a very massive body at a temperature of $25.0^{\circ} \mathrm{C}$ . What is thechange in entropy of this body? (c) What is the total change inentropy of the water and the heat source?

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a) 0.428b) $-0.392$c) 0.036 $\mathrm{k} \mathrm{J} / \mathrm{K}$

Physics 101 Mechanics

Chapter 16

The Second Law of Thermodynamics

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

Hope College

University of Winnipeg

McMaster University

Lectures

03:15

In physics, the second law of thermodynamics states that the total entropy of an isolated system can only increase over time. The total entropy of a system can never decrease, and the entropy of a system approaches a constant value as the temperature approaches zero.

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.

04:17

A sophomore with nothing b…

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20.25. A sophomore with no…

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03:29

20_23

08:06

20.58. A $0.0500-\mathrm{k…

09:00

In order to cool 1 ton of…

09:34

In order to cool 1 -ton of…

13:55

(a) After 6.00 $\mathrm{kg…

0:00

Find the entropy change wh…

05:03

Making lce You place 0.410…

01:52

Twenty grams of ice at pre…

00:43

03:18

Water and Ice A mixture of…

in this problem. We're going to use this formula to determine the changes of entropy in these processes, and we're also going to need a latent heat of freezing water here, which is this value. And so now we'll start off our day and we need to figure out the heat flow into the ice. And so the heat flow into the ice is equal to the master's laying here, which is able to 0.35 kilograms. These values were given in the problem and then lf is given here as 3.34 times, 10 to the fifth Jules per kilogram and the result is 1.17 times 10 to the fifth. Jules, this is what goes in for Q. And it's positive in this case since he'd is entering the system. Now Tea is also getting a problem, and it's 273 Kellen and it gave mine. We do have to use Kelvin anytime we use this formula. If you're given the temperature and Celsius, you must convert to Kellen first, and so dealt us is 1.17 times 10 to the fifth Jules over 273 come in and the result of this it is 429 jewels for cover. So if Q is positive, then Delta Asa's positive, and that's what we got here for Part B. We have the same heat except as offset sign, because now it's leaving the system that we're concerned with in part B. So it's negative 1.17 times 10 to the fifth, Jules, because this is the heat flowing out of the source and the tea is now 298. Kelvin, that's 25 degrees Celsius converted into government, which is what the large body of water is. That And so, Adele, tha s is equal to again Q over tea and employing in these values, its negative 393 Jules Berg, Alan. And so since you is negative, Delta s is negative. So this system actually lost entropy and the other system gained entropy. And Garcia is concerned with the overall change in entropy. And what we do is we take this change in entropy and we add it to this change in entropy. When you do that, you get total change in entropy of 36 Jules Per Kelvin and notice that this is positive as second Law of thermodynamics tells us it must be

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