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We submerge a 32.5-g iron rod, initially at 22.7 C, into an unknown mass of water at 63.2 C, in an insulated container. The final temperature of the mixture upon reaching thermal equilibrium is 59.5 C. What is the mass of the water?

$34.8, \mathrm{H}_{2} \mathrm{O}$

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Carleton College

University of Central Florida

University of Kentucky

University of Toronto

Lectures

00:42

In thermodynamics, the zer…

01:47

A spontaneous process is o…

03:20

A 32.5 -g iron rod, initia…

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We submerge a 2.85-g lead …

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A 1.50 -kg iron horseshoe …

01:41

A $1.50-\mathrm{kg}$ iron …

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10:33

An 825 -g iron block is he…

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(II) A hot iron horseshoe …

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A 1.22 kg piece of iron at…

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A 25.5-g aluminum block is…

02:02

A $31.1-\mathrm{g}$ wafer …

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A $15.7-\mathrm{g}$ alumin…

05:12

A $50.0 \mathrm{g}$ piece …

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A 25.0 -g sample of pure i…

So in this problem, we're submerging on iron rod of known mass inside oven inside of water of unknown mass, and the iron rod goes from twenty two to twenty two point seven degrees Celsius to the equilibrium temperature of fifty nine point five. And the water goes from sixty three point two to re Celsius to that same equilibrium, tension, temperature. And so what we can use here is the formula Q. Gained equals Q lost. And so the iron is gaining heat and the water is losing the exact same amount of heat. And so we can rearrange this into Q gained minus Q lost equal zero. And now we can plug in. Ah, the formula for Q gain in the formula for que los. And so it's M one C one delta T one minus em to see two Delta T tube equals zero, and so are one. In this case is the iron and are to is the water. And so we know the mass of our iron is thirty two point five grams. Well, plug in these specific heat of iron, which is zero point four five, and then the temperature change the iron changes by fifty nine point five minus twenty two point seven. It actually goes from twenty two point seven two, fifty nine point five. And so we do final minus initial, and then our mass to is unknown. The specific heat of water is four point one eight and the water goes from sixty three point two two, fifty nine point five. And the reason we write it this way is because we want this Delta T to be a positive number so that the formula works out and this equal zero. And so if we do a little bit of math, we end up with five. Thirty eight point two as the multiplication of these terms minus R m too. Times fifteen point four six six, which is the multiplication of these terms, and that equals zero. And so we can rearrange this to solve for mass to, and that gives us a mass to, ah, thirty four point eight grams and that'LL be the mass of our water. And so thirty four point eight grams of water is our final answer

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