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An unknown mass of each substance, initially at 23.0 C, absorbs 1.95 * 103 J of heat. The final temperature is recorded as indicated. Find the mass of each substance.a. Pyrex glass (Tf = 55.4 C)b. sand (Tf = 62.1 C)c. ethanol (Tf = 44.2 C)d. water (Tf = 32.4 C)

d.$$496.9$$

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

University of Central Florida

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Lectures

00:42

In thermodynamics, the zer…

01:47

A spontaneous process is o…

01:07

An unknown mass of each su…

05:42

03:41

(II) A 215-g sample of a s…

06:19

(a) What is the specific h…

01:56

Two substances, A and B, i…

07:02

Suppose that 25 g of each …

01:22

Two substances, $\mathrm{A…

01:59

A 155-g sample of an unkno…

01:55

02:19

A 155 -g sample of an unkn…

03:13

Calculate the final temper…

06:35

The same heat transfer int…

00:54

A piece of unknown substan…

02:49

A piece of unknown solid s…

07:21

A 20.3 -g sample of an unk…

07:08

A certain substance has a …

06:11

Calculate the heat change …

02:45

Three portions of the same…

01:30

Suppose that 25 g of cach …

05:36

so in this problem were given a heat. Q equals one point nine five times ten to the third. Jules were given an initial temperature of twenty three point zero degrees Celsius, and we want to find the mass of various substances at various final temperature is using this information. And so the first thing we can do is take our equation for heat. Q equals m The mass times C s the specific heat times Delta T and rearrange this to solve for M. And so we get m equals Q over. See Ass Delta T. And so now that we have this, we can begin to solve this problems or first substance is Pyrex glass. And so we use this equation. M equals R. Q is the same for all these one point nine five times ten to the third Jules. And we divide that by these specific heat of Pyrex glass, which is zero point seven five and then our delta's. He will be our final temperature, which is given as fifty five point four, minus the twenty three point zero our initial temperature and solving. This gives us a mass of eighty point two five grams So that's your final answer for letter, eh? Now we'LL go to a ah, let her be. And this time we have sand So again we'LL use our same equation m equals Q Oversee s Delta T In this case, our cue is the same one point nine five times, ten to the third Jules, we divide that by the C s specific heat for sand, which is zero point eight four and then our delta T is just going to be our final temperature, given a sixty two point one minus r twenty three point zero, which is our initial temperature. And when we saw for this we get a mass of fifty nine point three seven rams. So four letter C. This time we have ethanol. And so again, we use the same equation. M equals Q Oversee ass Delta T. Our cue remains the same and now our specific heat is two point for two. That's what it is for ethanol. And our final temperature is forty four point to subtract our initial temperature to get our temperature change. And when we do this, we get a mass of thirty eight point zero one grams. So now finally to move on to letter D, which is water. So again we take our same equation. M equals Q Oversee ass Delta T Our cue is again the same one point nine five times ten to the third. Jules R. C s in this case is four point one eight and our change in temperature are final. Temperature is thirty two point four, our initial temperature is twenty three point zero, and so that gives a mass of forty nine point six three grams. And so that is our final answer for letter D.

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