Question

Need help with chemistry work please help me understand? Calorimetry -

qwater= M x C x DeltaT
I finished that equation: 26.0 x 4.18 x 6.3 = 684.684.
I'm stuck on
qmetal = 27.776 x C x DeltaT?
IDK what to do.

full question:
Calculate the energy change (q) of the surroundings (Water) using the enthalpy equation qwater= M x C x deltaT

we can assume that the specific heat capacity of water is 4.18 J / (g x degrees C) and the density of water is 1.00 g/mL
the water has absorbed the heat of the metal. So, qwater=qmetal

Answer 1

@scitoteipsum

Answer 2

you mean to calculate the change in temperature of the surroundings, right?

Answer 3

yes

Answer 4

I'm confused because i don't know what the last part is with the qmetal part @electrokid

Answer 5

how much heat did the metal loose?

Answer 6

hold on, i will give you the chart, i got 6.3 for the first equation for delta t

Answer 7

Metal: Aluminum
Mass of metal: 27.776 g
Volume of water in the calorimeter: 26.0 mL
Initial temperature of water in calorimeter: 25.3 °C
Temperature of hot water and metal in hot water bath: 100.5 °C
Final temperature reached in the calorimeter: 31.6 °C

Answer 8

then,
\[\Delta Q_{\rm water}=m_{\rm water}\times c\times \Delta T\]

Answer 9

from your question, you are asked for the energy change of water. which is the heat LOST by water from 100 to 31.6 degrees

Answer 10

i had that, but didn't know if it was right, 68.9

So equation so far is
27.776 x C x 68.9

Answer 11

I'm just confused on C, i think C is .902 or somethin right cause i googled Specific heat capacity for aluminum and got a few diff answers from .897 to .91 is that right though?

Answer 12

we are dealing with the equation for water..
how is alluminium going to affect? Water and Al both are loosing heat

Answer 13

you said the temperature "inside" the calorimeter is 36 so, the temperature of both water and Al decreased

Answer 14

wait..
I did the qwater equation already and ended up with 684.684.. i need the qmetal unless im doing something totally wrong?

Answer 15

if they are asking for Q water, it should be negative and that is it!

Answer 16

so am i wrong?

Answer 17

I'm sorry.. i just don't catch on fast in chemistry

Answer 18

no. you are correct, with the approach. I did not verify the values. but otherwise, that is the heat lost...
you should get Q negative because
\(\Delta T\) = T final - T initial = negative

Answer 19

kapeesh?

Answer 20

so you take the final temp - intial temp and you wind up with..

Answer 21

31.6-100.5=

Answer 22

-6.3

Answer 23

-68.9

Answer 24

wait so what am i doing with my work if thats the answer?

Answer 25

that is just the \(\Delta T\)
we are finding Q.. for water
Q for water = mass of water * "c" of water * Delta T

Answer 26

so it'd be... 26.0 x C x -68.9?

Answer 27

c=4.186 J gm /K

Answer 28

can you tell me how you get C though?

Answer 29

it is a standard value. should be in the book with a small table giving you the "c" values for some substances.

Answer 30

i looked up specific heat capacity of aluminum and got .897 or is that something different?

Answer 31

the important things are
1) signs -> to tell if heat is lost or gained
2) units -> all units must be consistent -> add apples to a basket of apples and not bananas
3) use only the information provided. Check the question twice.

Answer 32

yep.,
c for Al -> 0.897 or approx 0.9 J gm /K

Answer 33

so what is c=4.186 J gm /K for then? what does that stand for?

Answer 34

that is the "c" for water

Answer 35

oh! yeah now im understanding, i have that in the equation at the top already, hm

Answer 36

yep

Answer 37

so the equation is 26.0 x 4.18 x 6.3 for qwater and
27.776 x .897 x -68.9 for qmetal?

Answer 38

and that is qwater=qmetal?

Answer 39

yes but I think you are asked just for the heat change in water.

Answer 40

just for sanity, could you enter the question exactly as it is

Answer 41

yes ill copy paste

Answer 42

it seems to me that you wrote the "yuour interpretation" of the question

Answer 43

Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qmetal

Answer 44

that's how its written, and i apologize for not grasping the concept of this with your help, I'm very appreciative though.

Answer 45

no problemo.. a good problem before sleep :)

Answer 46

how can this be the question whe it says nothing about the temperatures, masses, etc

Answer 47

well it was a lab

Answer 48

there you go
tell me more :)

Answer 49

and this is the questions concluding, all other stuff it gives is the data chart i put up about the aluminum

Answer 50

got ya

Answer 51

so the temperatures you have are for Al .. not the water...

Answer 52

Metal: Aluminum
Mass of metal: 27.776 g
Volume of water in the calorimeter: 26.0 mL
Initial temperature of water in calorimeter: 25.3 °C
Temperature of hot water and metal in hot water bath: 100.5 °C
Final temperature reached in the calorimeter: 31.6 °C
this is for everything

Answer 53

Temperature of hot water and metal in hot water bath: 100.5 °C

this is the misleading line... 100.5 is the temperature of the hot Al plate that you put in the cold water.

Answer 54

then you notice that the water temperature rose from 25.3 to 31.6 degrees

Answer 55

:)
and Al cooled from 100.5 to 31.6 degrees

Answer 56

so what does this

Answer 57

mean?

Answer 58

this means that as the Al plate cooled, it lost heat .. (Q is negative)
that heat is gained by the water (Q is positive)

The aim of the lab is to show hat both the ques will have the same magnitude

Answer 59

the calorimeter makes sure that the heat from Al does not go anywhere and stays inside.

Answer 60

yes, so.. the equations would be negative? is my teacher like mind f'ing my head? excuse my language but..

Answer 61

only one is negative. I am not aware of your teacher.
Maybe you did not understand him.

Answer 62

and but what?

Answer 63

wait, so the water is negative while the Al is positive? or otherway around

Answer 64

read 3lines above

Answer 65

yes only

Answer 66

is nega

Answer 67

loss = negative
gain = positive

Answer 68

even in real life, it is valid

Answer 69

OH THE AL LOST IS AND THE WATER GAINED

Answer 70

PING

Answer 71

and the world become gright again

Answer 72

SO THE EQUATIONS HOW WE WORKED THEM OUT ARE CORRECT

Answer 73

ITS JUST NOT ASKING FOR them like that right

Answer 74

yes. I did not check the numbers..
AGAIN: you should get the Q from Al and water to be equal.

Answer 75

thats the whole point!@

Answer 76

wait.. how do you do that..