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You decide to take a nice hot bath but discover that yourthoughtless roommate has used up most of the hot water. Youfill the tub with 270 kg of $30.0^{\circ} \mathrm{C}$ water and attempt to warm itfurther by pouring in 5.00 $\mathrm{kg}$ of boiling water from the stove. (a) Is this a reversible or an irreversible process? Use physicalreasoning to explain. (b) Calculate the final temperature of thebath water. (c) Calculate the net change in entropy of the system (bath water + boiling water), assuming no heat exchangewith the air or the tub itself.

a) The entropy of an isolated system can change but it can never decrease. since, the mixed temperature is in between $T$ 'and $T,$ so we can't decrease its entropy to its original states. Thus, the process is irreversible.b) 304 $\mathrm{K}$c) $-4267 \mathrm{J} / \mathrm{K}$, The net change in entropy of the system is, 475 $\mathrm{JK}$

Physics 101 Mechanics

Chapter 16

The Second Law of Thermodynamics

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

University of Washington

Hope College

University of Sheffield

Lectures

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we're told we wanted to take a nice hot bath, but discovered that our roommate has used up most of the hot water. Um, uh, we feel the tub with 270 kilograms of 30 degrees Celsius water and help the warm it further by pouring in five kilograms of boiling water from the stove. So here is our bathtub. We have our existing water, and here at 30 degrees C, we pour some hot water in to try to warm this up, and we are given all of these things. And then we also know that the total mass of water in the end after we poured in, it's just the sum of the mass of the hot and cold water. So who asked if this is our era is a reversible or irreversible possible well, boring hot water into the cold. Water is an irreversible process, and you could think of that is there's no way of extracting the hot water from the cold water without doing so additional work. So you could do what I suppose, if you had some thermal process heat pump that would transfer, take heat from the water warm some other water, but there's basically no way off. Once you've mixed the waters, there's no way off going back and un mixing. So then it asks us to calculate the final temperature of the bath water. Well, we can do that by knowing that the internal the change in internal energy, um, changing internal energy of the hot water must equal the change in internal energy of the cold water. And so the change in internal energy of the hot water equals the mass of the hot water times the capacity times of change in temperature of that hot water. So it's initially 100 degrees C on it, the final temperatures. What we're trying to figure the common. So we have the change in internal energy of the cold. Water is the mass of the cold water times the heat capacity, times the temperature and the cold water minus the final temperature that we're trying to determine. So we quit those two things, and we get that the the equation here and we get that the he could pass these cancel out, and if we rearrange some things, we find that you have the final temperature. This is the total mass equals the mass of the cold water times the temperature of the cold water plus the mass of the hot water times the temperature of the hot water. Now, if we saw for the final temperature, we see that actually, it's a way that a mass weighted average of the temperature. So we get the mask of the cold times the temperature of the coal divided by the mass of the total, plus the mass of the hot kind of temperature. They're hot, divided by the massive total and plugging those numbers. And we get 304 points for three. Calvin um, which is not a whole lot warmer than 30 degrees Celsius. So we didn't warm the water a whole lot. Now define the change in entropy, which is what we're asked for next, assuming that there's no heat exchange with the air of the cup itself. Well, to do that, we need to take a look at how the change in internal energy how that changes with temperature. So the change in internal energy of the hot water is the mass of the hot water times the heat capacity, the hot water, times the temperature that it's at minus temperature. It was initially and we can then right Are are entropy statement in differential form. So that is the problem is here is that the heat is not being dumped at a certain temperature because the water and the both water, the hot and the cold, are changing temperature as they as they exchange heat. So we need to do an integral and that integral we confined because we know the differential of the of the entropy because the differential of the internal energy divided by capture. And so if we integrate this, we get the change of entropy equals the integral of the differential of the internal energy divided by t. And we can write that as one the girl of one over the temperature times the derivative of the internal energy with respect to the temperature of times t t. So we get an integral in terms of temperature in terms of the tea And if we put that stuff in OK, so how we need to figure out what the u. D t is. Okay. So for the hot water we have this U S a function of key so we can take the rebel expect to tea, and we get DVT because the mass of the hot water heat capacity. So that means that the change in entropy of the hot water is the integral from the from the initial state, the natural temperature to the final temperature of one over tea kinds of mass of the hot water cups of heat capacity that actually comes out of the integral. So we get an integral of one over TV T and remember from calculus that is the natural law. So we get that the mass of the hot water turns to keep capacity of hot water, kinds of natural log of the final temperature, minus the natural law of the initial temperature, and using our properties of the log rhythm, we can get that that is the mass of the hot water, trying to heat capacity times the natural law of the temperature ratio of the final temperature divided by the initial temperature. And if we plug in numbers, we have everything. We have everything here so we can plug your numbers and we get minus 4265 jewels per Calvin. So that's the change of entropy of the hot water, It's entropy went down. Now we need the change. Figure out the change of entropy of the cold water, and we go through the same process and get exchanging. Entropy is its mass of the cold water times the capacity water times the natural log of the final temperature, divided by the initial temperature and calculating fucking in numbers. There we get that that is 4739.5 jewels for Calvin. So the entropy of the cold water went up, which we'd expect because we better get a net a positive number for the changing in tribute system. And if we take 4739 0.5, minus 4265 we get that. The total change in entropy of the system is 704 175 jewels for Calvin

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