Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

Find H, E, q, and w for the freezing of water at -10.0 C. The specific heat capacity of ice is 2.04 J/g # C and its heat of fusion (the quantity of heat associated with melting) is -332 J/g.

-310.5692

Chemistry 102

Chemistry 101

Chapter 9

Thermochemistry

Thermodynamics

Chemical reactions and Stoichiometry

Carleton College

University of Central Florida

Drexel University

University of Toronto

Lectures

00:42

In thermodynamics, the zer…

01:47

A spontaneous process is o…

02:25

Calculate the quantity of …

03:16

A 20.0 -g sample of ice at…

07:35

A $20.0-\mathrm{g}$ sample…

03:31

02:45

Given that the heat of fus…

04:13

How much energy in kilojou…

02:09

The heat energy required t…

01:49

The energy required to mel…

0:00

04:10

An ice cube with a mass of…

01:14

You have a 1.00 -mole samp…

05:40

From the data below, calcu…

04:11

02:16

The freezing point of merc…

04:14

02:54

How much heat (in kJ) is r…

00:53

When water is supercooled,…

05:17

How much heat (in $\mathrm…

00:47

What quantity of heat is e…

01:10

Use standard enthalpies of…

So this problem wants us to find Delta H Delta E Q and W for water freezing at negative ten point zero degrees Celsius. And so first we'LL write down What we do know. So we know that the heat capacity of ice is two point zero four Jules Program degree Celsius. We know that the Delta H of fusion, which is associating with melting at zero degrees Celsius, is negative three thirty two jewels program, and we know that the heat capacity of water is four point one eight four Jules per gram degrees Celsius. And so now we can use this information in order to find the delta H of fusion at negative ten degrees Celsius. And so how we do that is we take the Delta H of fusion at zero degrees Celsius, and that is equal to Delta H of liquid water from zero to negative ten point zero degrees Celsius, plus the delta H of fusion for water at negative ten point zero degrees Celsius, plus the delta H of ice from negative ten point zero degree Celsius, two zero degrees Celsius. And so we can rearrange this to solve for the Delta H of fusion at negative ten point zero degrees Celsius, which is what we want on so that will be equal to Delta H effusion at zero degrees Celsius, minus delta H of the liquid water from zero to negative ten point zero degree Celsius. Ah, minus the delta H of ice from negative ten two zero degrees Celsius. And so we can plug in all these values. And so the Delta Age of Fusion, at zero degrees Celsius, is negative. Three three two jewels program. The Delta H of liquid will be the heat capacity of liquid water. Four point one eight four Jules program degree Celsius times negative ten point zero degrees Celsius is that is our delta T minus the delta h of ice. So we'LL use our he capacity for ice to point zero four Jules program degrees Celsius and multiply that by ten point zero degrees Celsius Is that is our delta T. And so doing all this gives us adult H effusion of negative three ten point five six Jules program. So this is our Delta H, and we also know that this equals Q because for a constant pressure, which we are assuming Delta H equals. Q. So we have our Delta h and we have our cue. Okay, now we can go about finding r. W. And so first we confined Delta V. So for one gram of water, which is this going to be the easiest to calculate? Delta V equals one, divided by zero point nine one six seven centimetres cubed program minus one point zero centimeters Cube program. And that gives us a zero point zero nine one centimeters Cube program, or nine point one times ten to the negative Fifth Leaders programme. So now we confined our p delta V so pressure times the change in volume are pressures one point zero atmospheres and are changing volume. We just found to be nine point one times ten to the minus fifth Leaders programme. And now this is the change in volume that the water is undergoing while it moves from liquid with ice. And so that gives us nine point one times ten to the negative Fifth Leader atmospheres per gram which is equal to zero point zero zero nine two jewels program. So now that we have that r w is negative. P. Delta v. R work and so that's just going to be negative. Zero point zero zero nine two jewels program. So we have our W. We have our Q and we have our Delta H. So now to find our Delta E. That is just equal to Delta H minus P. Delta V. And so it's just our Q minus. R w. And so we take our cue three ten point five six minus R W zero point zero zero nine two and therefore our Delta is equal to negative three ten point five six nine two jewels program, and that is our final answer. Since there is no mass given, we have to determine all of these energies on a per gram basis, and so that is our final answer.

View More Answers From This Book

Find Another Textbook

In thermodynamics, the zeroth law of thermodynamics states that if two syste…

A spontaneous process is one in which the total entropy of the universe incr…

Calculate the quantity of heat required to convert $60.1 \mathrm{g}$ of $\ma…

A 20.0 -g sample of ice at $-10.0^{\circ} \mathrm{C}$ is mixed with 100.0 g …

A $20.0-\mathrm{g}$ sample of ice at $-10.0^{\circ} \mathrm{C}$ is mixed wit…

Calculate the quantity of energy required to convert $60.1 \mathrm{g}$ of $\…

Given that the heat of fusion of water is $-6.02 \mathrm{kJ} / \mathrm{mol}$…

How much energy in kilojoules is needed to heat 5.00 $\mathrm{g}$ of ice fro…

The heat energy required to melt $1.00 \mathrm{g}$ of ice at $0^{\circ} \mat…

The energy required to melt $1.00 \mathrm{g}$ of ice at $0^{\circ} \mathrm{C…

Given that the heat of fusion of water is -6.02 kJ>mol, the heat capacity…

An ice cube with a mass of 20 $\mathrm{g}$ at $-20^{\circ} \mathrm{C}$ (typi…

You have a 1.00 -mole sample of water at $-30 .^{\circ} \mathrm{C},$ and you…

From the data below, calculate the total heat (in $\mathrm{J}$ needed to con…

You have a 1.00 -mole sample of water at $-30 .^{\circ} \mathrm{C}$ and you …

The freezing point of mercury is $-38.8^{\circ} \mathrm{C} .$ What quantity …

How much heat (in kJ) is required to warm 10.0 g of ice, initially at -10.0 …

When water is supercooled, it freezes at a temperature below $0.0^{\circ} \m…

How much heat (in $\mathrm{kJ} )$ is required to warm 10.0 $\mathrm{g}$ of i…

What quantity of heat is evolved when 1.0 L of water at $0^{\circ} \mathrm{C…

Use standard enthalpies of formation to calculate the standard change in ent…

03:37

Two compounds with general formulas AX and AX2 haveKsp = 1.5 * 10-5. Whi…

05:11

Bromine can form compounds or ions with any number of fluorine atoms from on…

02:27

When HNO2 dissolves in water, it partially dissociates accordingto the e…

03:14

Consider the titration curves (labeled a and b) for equal volumesof two …

03:13

An aqueous KNO3 solution is made using 72.5 g of KNO3 diluted toa total …

01:38

When 1 mol of a gas burns at constant pressure, it produces 2418 J of heat a…

06:45

A solution containing sodium fluoride is mixed with one containingcalciu…

00:57

How does hybridization of the atomic orbitals in the central atom of a molec…

01:50

A 0.95-m aqueous solution of an ionic compound with theformula MX has a …

01:33

Water softeners often replace calcium ions in hard water withsodium ions…