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$\bullet$ The icecaps of Greenland and Antarctica contain about 1.75$\%$ of the total water (by mass) on the earth's surface; the oceans contain about $97.5 \%,$ and the other 0.75$\%$ is mainly groundwater. Suppose the icecaps, currently at an average temperature of about $-30^{\circ} \mathrm{C},$ somehow slid into the ocean and melted. What would be the resulting temperature decrease of the ocean? Assume that the average temperature of ocean water is currently $5.00^{\circ} \mathrm{C}$

1.76 $\mathrm{C}^{\circ}$

Physics 101 Mechanics

Chapter 14

Temperature and Heat

Thermal Properties of Matter

The First Law of Thermodynamics

University of Michigan - Ann Arbor

University of Sheffield

McMaster University

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everyone, This is question number 84 from Chapter 14 in this problem were given that 1.75% by mass of water on the earth is an ice caps 97.5% by masses in the oceans. We're told that the temperature of the ice caps is mine started degree Celsius. They slip into the ocean and melt. And the were told that the current ocean temperature is equal to five degrees Celsius, so without this given information were asked to find the temperature decrease in the ocean. So this is one of those temperature change problems with a system that we have two different parts that we have to take into account, and eventually they're going to reach equilibrium and get to this final temperature that we don't know. So we're going to assume that no heat loss so cue of the ice plus que of the ocean is going to equal zero. So this will be the equation that we base the rest of this problem off of No. So we can right expressions for Q of the ice and cue of the water said everything equal to zero, and then the only thing that we're not going to know is the T final, which we can then use to answer the actual question. So let's define a couple things first Mass of the ice. We're not given a mass for the total water on Earth, but if we keep that is a variable, it will end up canceling so we can say mass in the ice equals point. A 175 just em, and massive ocean water equals 0.975 AM And that's just converting those two decimals from percentages. So now we have to right oppression for Q of ice and queue of water. So cue of ice. We have three stages to this. We have the ice increasing in temperature from minus 32 0 degrees. Then we have the temperature remain the same, but the ice melting. And then we have the temperature increase from zero degrees to the final temperature. So there's three parts to that. So the first part, like I said, increasing from minus 30 0 that's just going to be our equation. Emcee Delta T. And our Delta team. This case is going to be zero minus negative 30. So that's just going to be 30 degrees Celsius. And then we add the cue from the face change, which is M I l fusion for ice. And then we add the temperature increase to the final temperature, and that is going to be M I. C. I am psi Delta T Delta team's final minus initial, but our initial zero. So this is just gonna end up being t f, which is what we're looking for. So now we can do the same thing to find for ocean water, So q o W This is a simple one. It's just going to be emcee Delta T. Because there's no faint face change. So m o seawater, and this is going to be TF final minus initial, which were given five degrees C. So now we can essentially, you're just going to add both of these together in Summit to zero. So this is just going to be a plus, this stuff at the bottom Hey, and from there you said it equal to zero, and we can rearrange and solve for T F. And I am going to skip the algebra process. But if we rearrange and solve for t F without plugging numbers and yet to make it easier. We get minus Master the ice See the ice 30 degrees Celsius, minus mass of the ice lf the ice plus em of the ocean water, sea of water, five degrees Celsius you just end up factoring out a T f and solving for it. The middle on the bottom will be master the ice plus massive the ocean water time See a war. So after you do your algebra, make sure you get that equation and from here we can plug in our numbers. I'm probably going to end up going onto another line something to go to the next page. So T f is equal to minus mass 0.175 mm Sea of ice were given the problem 21 100 Jules per kilogram, kay. And then our delta weaken just say is 30 K because we know change in Celsius is the same as changing Kay and then minus 0.175 em multiplying our heat of vaporization 334 times 10 to the third Jules per kilogram fusion. Excuse me. Not vaporization plus massive waters. Your 0.975 mm times for 190 See for water, Jules per kilogram, Kay. Times five again change. And then all of that is over your point. 0175 m plus zero point 975 em times for 190 Jules per kilogram. OK, All right. And if you carefully multiply all of that, the the M's are going Tio cancel and you get T F is equal to 1.348 times 10 to the fourth Jules per kilogram over 4.159 times 10 to the third Jules per kilogram. Kay. So our jewels per kilogram cancel. Well, we have this k left, and we're left with R T f of 3.24 degrees Celsius. But we're asked for temperature decrease. So five minus our initial temperature of the ocean 3.24 gives us a temperature decrease of 1.76 degrees Celsius, and this doesn't seem to know much. And it's not that much because the mass of water in the ice caps, compared to the mass of water in the ocean is very small. So the temperature trained the temperature of the ice caps is going to have a very minimal change on the ocean

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