Question

Thermochemical equation.....

Calculate the calories required to convert 84.2g of H2O(s) at 0.0 degrees C to H2O(g) at 100 degrees C.

Answer 1

How many degrees do we want to raise the water?

Answer 2

100?

Answer 3

how many grams do we want to heat?

Answer 4

84.2g

Answer 5

How many calories does it take to raise 1 g of water 1 degree?

Answer 6

is it 4?

Answer 7

What is the definition of a calorie?

Answer 8

ummm, Im not sure, i no it equals 4.18 J if we need that for this?

Answer 9

Either of two units of heat energy.

Answer 10

84.2g x 100 x 4/ 18?

Answer 11

there are two things involved here
1) raise temperature of water from 0 to 100 degrees -> use the specific heat for water
2) convert liquid water at 100deg to gas at 100deg -> use latent heat of vaporization

Answer 12

total heat = sum of heat for both the steps

Answer 13

wait.. another one
step 0) convert SOLID water at 0 to liquid water at 0 -> latent heat of fusion

Answer 14

total heat = sum of all three heats

Answer 15

336.8/ 84.2g x 100 for step 1 = .04

Answer 16

start with step 0

Answer 17

ok

Answer 18

us the equation button below so I can understand.

Answer 19

o ok never saw that before, and im not sure how to do step 0

Answer 20

same exact thing like step 2 just a different value for Lf

Answer 21

can u give me the equation, and ill plug in the numbers, because i dont know the equation

Answer 22

\[Q_0=mL_f=84.2{\rm g}\times L_f\]
Lf is the latent heat of fusion.. should be in a table in your book

Answer 23

506.0?

Answer 24

84.2 x 6.01

Answer 25

you cannot solve without the unitz

Answer 26

84.2 g
6.01 what?

Answer 27

84.2g x 6.01 H2O

Answer 28

\[H _{2}O\]

Answer 29

H2O is the chemical formula for water.
It is not a unit for a physical quantity
units are like.. grams for mass, second for time, calori or joule for heat, etc..

Answer 30

you are looking for the latent heat of fusion.

Answer 31

i.e., the amount of heat required to convert 1g of ice to water.

Answer 32

delta H\[_{fus}\] (KJ/mol) = H2O 6.01

Answer 33

that is the latent heat of fution for ice or water.

Answer 34

ok. so, this value is in moles
and the amount we have is in grams
so, first convert 84.2g to moles for water

Answer 35

how mny moles of water do we have in 84.2g?

Answer 36

84.2g x 6.01 x \[10^{23}\]

Answer 37

= 5.06 x \[10^{25}\]

Answer 38

no

Answer 39

1mole of water = 2(1)+16=18g

Answer 40

4.6 moles?

Answer 41

so, how many moles do we have in 84.2g?

Answer 42

good
lets use 4.68

Answer 43

84.2/18= 4.7

Answer 44

o ok 4.68 works to

Answer 45

now ou can multiply apples with apples.

Answer 46

28.1

Answer 47

\[Q_0=4.68{\rm mol}\times6.01{\rm kJ/mol}\]
=?

Answer 48

28.1 what ?
joules,
kilo joules,
bananas,
calories,...?

Answer 49

28.1 KJ/mol

Answer 50

we have to be careful or evrything will go waste

Answer 51

notice that the mol gets cancelled.
so,
\[Q_0=28.1kJ\]

Answer 52

so, we apply this much heat and all that ice will turn to water

Answer 53

now we heat the water up to 100deg
\[Q_1=mc\Delta T\]

Answer 54

not sure how to do that... 28.1KJ x 100 deg?

Answer 55

m=84.2g
c=4.186J/g
delta T = 100-0=100

Answer 56

35246.1 J/ Degrees C

Answer 57

g degrees C

Answer 58

heat is just joules or KJ or cal or Kcal no more things with it.
so, \[Q_1=35246J=35.25kJ\]
since our first one was in KJ, we make all our heats to KJ

Answer 59

now, the liquid water is ready to be turned to steam and blow up:)

Answer 60

so, we do Latent Hat for Vaporization.
\[Q_3=mL_v\]

Answer 61

4.68 x 40.7 Delta H vap ( KJ/mol)

Answer 62

4.68 moles

Answer 63

190.5KJ

Answer 64

nice work

Answer 65

I cannot verify the values for the latent heats but thats the steps to do this problem.

Answer 66

so, total heat for the reaction = sum of all the heats
\[Q=Q_0+Q_1+Q_2\]

Answer 67

when you add, make sure they all have the same units. (in our case, it is kJ)

Answer 68

the final heat will also be in kJ

Answer 69

what is latent heat?

Answer 70

but I think the question asks for the value in calories... so do the necesary conversion

Answer 71

Latent heat is the amount of heat required to change the physical state of a unit mass of a subtance at a constant temperature.

Answer 72

so where would i find latent heat, in another table?

Answer 73

like from solid to liquid at its melting point
or from liquid to solid at melting point

from liquid to gas at boiling point and backwards

Answer 74

what elements or things do i need to find them for?

Answer 75

any element or compound that can change the states will have the corresponding latent heat

Answer 76

ok, im sorry but im really confused right now

Answer 77

what with?

Answer 78

i have no idea how to change latent heat or get it

Answer 79

it will be provided to you in the test.
In homeworks, they expect you to use the textbook (as they usually have the values for common materials)

Answer 80

you do not change latent heats. they are the properties of that material. you have to take it as given

Answer 81

4.18 for water?

Answer 82

4.18J/gC is the "specific heat for water"

Answer 83

Latent heat of melting - 334 kJ/kg

Answer 84

for water

Answer 85

its not negative the dash is like an equal sign

Answer 86

right.. units are different but representing the same quantity.

Answer 87

now what?

Answer 88

did you add up the three heats??

Answer 89

what other heats do i need to find? for what elements would it be 190.5 and 32.25?

Answer 90

???
are we still talking about this question or jumping out randomly?

Answer 91

no thats what we got before, im not sure what i need to find the heat for now?

Answer 92

ADD em up

Answer 93

the three heats that we found

Answer 94

wait, im just wondering are those the heats?

Answer 95

yes the "Q" is the symbol for heats

Answer 96

kJ is the unit of measurement for heat

Answer 97

o ok

Answer 98

so i got 32.25 + 190.5 + 334= 556.8KJ