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A pure gold ring and pure silver ring have a total mass of 14.9 g. We heat the two rings to 62.0 C and drop them into 15.0 mL of water at 23.5 C. When equilibrium is reached, the temperature of the water is 25.0 C. What is the mass of each ring? (Assume a density of 0.998 g>mL for water.)

6.185

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Carleton College

University of Maryland - University College

University of Kentucky

Lectures

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So this problem gives us a combined mass of gold and silver, and we are heating up this gold and silver and putting it into water. And then the water will absorb all of the heat that the golden silver is losing. And so the first thing we'LL do is, since we don't know the individual masses of gold and silver, we'LL call the massive gold X and the mass of silver fourteen point nine minus x Since the total list fourteen point nine. If the massive gold is X, the massive silver has to be fourteen point nine minus x. So now we confined the heat. Released by the water are the heat absorbed by the water. Rather, Q E equals M C Delta T. The mass is going to be our fifteen point zero millilitres times zero point nine nine eight grams. Permit a leader to convert that into grams, then our specific heat of water. See, it's four point one eight for Jules per gram degrees Celsius, and then our change in temperature is going to be twenty five degrees Celcius minus twenty three point five degrees Celsius, and so that gives us a heat of water of ninety three point nine five jewels. So this is the heat the water absorbs, and it's equivalent to the heat that the gold and silver lose. So now we can evaluate the gold and silver separately. So we'LL start with the gold. And so for the gold again, Q E equals M C Delta T R M is axe. That's our massive gold, the specific heat of gold, a zero point one two six Jules per gram degree Celsius and then the delta T is sixty two minus twenty five, both our degrees Celsius, and that gives us four point six six two. So now we can do the same thing for silver. Q equals M C. Delta T equals fourteen point nine minus X times. The specific heat of silver is zero point two three three Jules program Degrees Celsius, and then we have our same delta t as with gold. And so that gives us one twenty eight point four five to nine minus eight point six to one x. So this is our mass or our heat rather off silver. And so now if we take our heat of gold and heat of silver and add them together. So we take four point six six two acts the heat of our gold, plus the heat of our silver one. Twenty eight point four five to nine, minus eight point six to one x. This will equal the heat of our water ninety three point nine five Julius. And so now we can just solve for acts. So we first combined like terms we get negative three point nine five nine X equals negative thirty four point five zero two nine and therefore X equals eight point seven one five. So therefore Ah, that is our mass ofthe gold is eight point seven one five grams in our mass of silver is fourteen point nine grams minus eight point seven one five grams or six point one eight five grams. So that's our massive silver and this is our massive gold

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