Question

what is the enthalpy change during the process in @Abhisar which 100 grams H2O at 50C is cooled to ice at -30C? c(s) 2.09 J/g-C c(I) 4.18 J/gC c(g) 1.84 J/gC heat of vaporization= 40.67 kJ/mol heat of fusion= 6.01 kJ/mol

Answer 1

@Abhisar how u get started

Answer 2

@Abhisar hello

Answer 3

@jfraser @aaronq i need help

Answer 4

@4n1m0s1ty this question ugh

Answer 5

Ok, so we are looking at a phase transition of water. So here are a few things I want to ask you:

1.) What state are we starting in (gas/liquid/solid)? and what are we ending in (gas/liquid/solid)?

2.) Once you know which state the water is in we can figure out whether to use the heat of vaporization and/or heat of fusion. The heat of vaporization is when the phase transition is between gas<-->liquid (for water this occurs at 100 C), while the heat of fusion is between liquid<-->solid (for water this occurs at 0 C).

3.) So now that we know information from the two steps above, we can solve for the enthalpy. The enthalpy change is going to be equivalent to the energy required to change the temperature of the water from 50 C to 0 C (using the enthalpy change value for liquids), and then either heat of fusion or vaporization, and the finally going from 0 C to -30 C (what is phase does ice belong to).

So hopefully this helps. One thing I want to point out is that the heats are given in kJ/mol so you probably need to do some conversions to get to J/g. You'll need the molecular weight of water to do this.

Answer 6

@4n1m0s1ty work it out step by step pplz

Answer 7

Ok, lets go through the first step. What state is the water initially in at 50 C, and what does it turn into at -30 C?

Answer 8

@4n1m0s1ty a liquid

Answer 9

@4n1m0s1ty am I correct

Answer 10

@4n1m0s1ty hello u there

Answer 11

Sorry connection issues.

Yes that correct

Answer 12

Now what does it turn into at -30 C?

Answer 13

@4n1m0s1ty gas

Answer 14

@4n1m0s1ty hello

Answer 15

Yep

Answer 16

@4n1m0s1ty whats next

Answer 17

@4n1m0s1ty i need the work done, i need help hello

Answer 18

@4 its hw

Answer 19

So, no we know we are going from a liquid to a solid. If read my explanation above, we know that we need to use the heat of fusion.

However, its in kJ/mol. so we need to convert the sample that we have from grams into mols? do you know how to do this?

Answer 20

@4n1m0s1ty no, u there

Answer 21

@4n1m0s1ty this is a new chapter 4 for me

Answer 22

Yeah, sorry I tend to lurk around in different threads while waiting for a reply.

So yeah we need to convert from grams to mols. So basically you need to take the 100 g of water that you have and divide by the molecular weight (I think its like 18.02 g/mol, but I would do a quick google search to make sure.)

Answer 23

Yeah its 18.02 g/mol, so just do 100 divided by 18.02 to get the mols of water in the sample.

Answer 24

@4n1m0s1ty after that...

Answer 25

So once you have the mols of water you need to multiply that by the heat of fusion.

Answer 26

To get the energy change for the phase transition. Since, its in kJ just move the decimal place to the right 3 places to get Joules. We need to do this part since the other values are in Joules

Answer 27

Then after that we need to calculate the energy required to change the temperature of the water from 50 C to 0 C, and then going from 0 C to -30 C. Its two seperate calculations since its two different phases.

Answer 28

@4n1m0s1ty i got 33,351 Joules

Answer 29

So for the 50 to 0 one you need to use the 4.18 value, and for the 0 to -30 you need to use the 2.09 value

Answer 30

Hold on let me check

Answer 31

Yep got the same thing

Answer 32

@4n1m0s1ty for the fusion

Answer 33

@4n1m0s1ty what the equation from 50 to 0C

Answer 34

@4n1m0s1ty hello

Answer 35

So you need to use the sample size 100 g multiplied by the temperature change (50 C) multiplied by 4.18

Answer 36

Thats for 50 to 0

Answer 37

You do the same thing for 0 to -30, but instead you need to multiply by 2.09 instead of 4.18 since it is in solid phase

Answer 38

@4n1m0s1ty next step

Answer 39

@4n1m0s1ty u dont use 18.02

Answer 40

Nope

Answer 41

Its in grams now not mols

Answer 42

@4n1m0s1ty when we use mols

Answer 43

Once you find those two values, just add it to the 33351 joules to get the total enthalpy change

Answer 44

You need to use mols only when you wnat to cancel out mols in the problem.

Answer 45

Out heat of fusion was in kJ/mol, so we need the sample size in mols to cancel it out to just be left with energy.

Answer 46

@4n1m0s1ty is delta T 30

Answer 47

Yes

Answer 48

Oh just remembered delta T is -50 since we are going down in temp for the 50 to 0 one

Answer 49

@4n1m0s1ty 30*100*?

Answer 50

30*100*2.09

Answer 51

Thats the enthalpy change per temp per gram for ice

Answer 52

@4n1m0s1ty 100*50*4.18= 20.9 KJ

Answer 53

@4n1m0s1ty u add 20.9 + 6.27 + 33351

Answer 54

One sec checking

Answer 55

@4n1m0s1ty teacher answer: -60.6 Kj

Answer 56

@4n1m0s1ty how pplz

Answer 57

One sec

Answer 58

Trying to see where the math is wrong

Answer 59

Ok, found it. So we need to be consistent in our units.

Answer 60

Since your teacher left it in kJ. We should just leave 33351 J as 33.351 kJ

Answer 61

When you add them together you get 60.6 kJ

Answer 62

@4n1m0s1ty show all the steps again. But how u get -60.6 kj

Answer 63

@4n1m0s1ty what are the values again

Answer 64

Ok so you found these correct?

20.9 kJ for 50 to 0
6.27 kJ for 0 to -30
33.351 kJ for heat of fusion

Answer 65

We just add them together, and get the enthalpy change. The negative just denotes the flow of energy. Its negative, so its going out of out system. This makes sense since the you are going from liquid water to ice, so you are losing heat to cool the water.

Answer 66

@4n1m0s1ty yes, i have another question: a substance has the following properties: heat of vaporization: 31.8 kj/mol heat of fusion: 2.51 Kj/mol Boiling point: 76C melting point: -23C c(s): 3.0 J/gC c(l) 2.5 J/gC c(g) 1.0 J /gC calculate the energy needed to convert 0.500 mol from -50C to 130C. Molar Mass=154 grams.

Answer 67

@4n1m0s1ty can u email as well the step by step process. klloganinfo@cox.net

Im kinda of guessing.

Answer 68

I kind of just told you how to solve the problem in this thread. Its the same problem just with different numbers.

Answer 69

@4n1m0s1ty do you use the heat of vaporization

Answer 70

Yes, you just account for both fusion and vaporization because you go from solid to gas

Answer 71

so just find enthalpy from fusion and enthalpy from vaporization

Answer 72

@4n1m0s1ty and why? I have another question: How much energy does it take to convert 130 grams of ice at -40C to steam at 160C i dont get this question

Answer 73

and add it to the temperature changes:
-30 to 0
0 to 100
100 to 130

Answer 74

Because at phase transitions (meaning water at 0 and 100 C) there is a certain amount of heat that needs to be released/captured in order for a phase transition to occur.

Answer 75

@4n1m0s1ty what values are those 4? the latter question or or former?

Answer 76

Thats for the second question you asked

Answer 77

@4n1m0s1ty and add it to the temperature changes:
-30 to 0
0 to 100
100 to 130

Answer 78

@4n1m0s1ty u mean converting to 130 grams question?

Answer 79

No for this one:

yes, i have another question: a substance has the following properties: heat of vaporization: 31.8 kj/mol heat of fusion: 2.51 Kj/mol Boiling point: 76C melting point: -23C c(s): 3.0 J/gC c(l) 2.5 J/gC c(g) 1.0 J /gC calculate the energy needed to convert 0.500 mol from -50C to 130C. Molar Mass=154 grams.

Answer 80

oh mistake its -50 to 0 not -30 to 0

Answer 81

@4n1m0s1ty i caught it

Answer 82

but yeah just do what we did for this first problem, and the second and third should be the same way to solve

Answer 83

@4n1m0s1ty 4 that question do I use 77 for m

Answer 84

@4n1m0s1ty have another question: How much energy does it take to convert 130 grams of ice at -40C to steam at 160C i dont get this question

Answer 85

Actually, I messed a bit, lets go back to this problem.

yes, i have another question: a substance has the following properties: heat of vaporization: 31.8 kj/mol heat of fusion: 2.51 Kj/mol Boiling point: 76C melting point: -23C c(s): 3.0 J/gC c(l) 2.5 J/gC c(g) 1.0 J /gC calculate the energy needed to convert 0.500 mol from -50C to 130C. Molar Mass=154 grams.

Like I said before you calculate heat of vaporization and fusion. But the substance here isnt water. So the boiling/freezing points are different the temperature changes are different, it should be

-50 to -23 (solid)
-23 to 76 (liquid)
76 to 130 (gas)

Answer 86

Also for this question:

How much energy does it take to convert 130 grams of ice at -40C to steam at 160C i dont get this question

What do you not understand about it?

Answer 87

@4n1m0s1ty u subtract -50+23, -23+76 and 76-130? idek

Answer 88

Ok so I think you might be misunderstanding some stuff so let me clarify.

Answer 89

@4n1m0s1ty those 3 values are what numeric value?

Answer 90

@4n1m0s1ty How much energy does it take to convert 130 grams of ice at -40C to steam at 160C i dont get this question: email me klloganinfo@cox.net

Answer 91

When we want to find the total enthalpy change we need to find the enthapy change from the phase transition and from the temperature change. I think you understand the phase transition stuff, so lets talk about the temperature change.

When we calculate the enthalpy change from temp. We need to calculate the full range, the equation is this:

\[Q=mc \Delta T\]

were Q is the heat, m is mass, c is the specific heat (those values they give you for each phase in the problem) and delta T is the temperature change. So when we calculate this, we need to do the final - the initial temperature for the delta T.

Answer 92

Also, I would really like to email you, but unfortunately, I'm not currently in a situation where I can at the moment. and its easier to just help from this site.

Answer 93

@4n1m0s1ty but 4 future reference whats ur email?