Question

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

Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation.

27.776 g-Measured mass of metal
26 mL- Distilled water measurement
25 C- Distilled water temp
100 C- Temp of metal
31.7 C- temp of mixture

Answer 1

can you figure out q_water ? using m= 26 g , c= 4.18, and ΔT=(31.7-25)

Answer 2

i'm sorry what do you mean by that?

Answer 3

how would I find that out?

Answer 4

your formula
qwater = m × c × ΔT.

says: multiply 3 things together: the mass, c (which is 4.18 because they told you) and the change in temperature. The water started at 25 C and went up to 31.7C

Answer 5

ok hold on right quick

Answer 6

it would be 728.156, that would be the qwater correct?

Answer 7

yes. now for the metal

they do not tell you c (that is what you have to find) but you know the rest of the info

Answer 8

@phi : hold on, I thought you just told me what c was?

Answer 9

@phi: which was 4.18?

Answer 10

c is the specific heat capacity of a material. It is different for different things.
they want you to find c for the metal

Answer 11

@phi: well they also give us a unknown metal that we had to calculate. This is the unknown metal table:

It goes in the same order.

25.605g
24 mL
25 C
100 C
0 C

Answer 12

@phi : they then tell us to do the same.

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 = qunknown metal

Using the formula qunknown metal = m × c × ΔT, calculate the specific heat of the metal.

Answer 13

ok, but first let's finish your first problem

Answer 14

Your first problems says (for the metal)
27.776 g-Measured mass of metal
100 C- Temp of metal
31.7 C- temp of mixture

It looks like the metal was hot, and they put it into water. the water heated up
(the water absorbed the heat 728.156 Joules according to your calculation)

the metal gave off that amount of heat and cooled down

Answer 15

the idea is that
qmetal = m × c × ΔT
but q_metal (the amount of heat the metal lost), equals the amount of heat that went into the water. so we can say q_metal= 728.156 Joules

Answer 16

now fill in as much as you know in the formula
qmetal = m × c × ΔT

(you know everything except "c" , which is the specific heat of this metal)

Answer 17

so then to fill in the formula it would be 728.156 Joules= 26 g (x) c (x) 6.7

Answer 18

@phi: Am I correct?

Answer 19

that is for the water (and c for water is 4.18 )

we want to do the same thing for the metal

Answer 20

ok now can you explain to me how you got that 4.18 so that I can try to see how I would get c for the metal?

Answer 21

They told us c for water is 4.18

***the specific heat capacity of water is 4.18 J / (g × °C) ***

we used the formula
q_water = m × c × ΔT
to find q_water= 728.156 Joules

The idea is that heat is energy. It took 728.156 J of energy to heat up the water.

where did that energy come from ? It came from the metal. The metal "lost" 728.156 J of energy

If we use the formula
q_metal = m × c × ΔT
replace q_metal with 728.156 J
also, fill in the mass of the metal, and how much the temperature of the metal changed.

Answer 22

Ok so now it is for metal: 728.156 J= 25.605 x c x (0-25) Correct?

Answer 23

I think we should stick to the first problem you posted.

Use the data from your experiment for the metal in your calculation.

27.776 g-Measured mass of metal
26 mL- Distilled water measurement
25 C- Distilled water temp
100 C- Temp of metal
31.7 C- temp of mixture

Answer 24

ok we are on the first problem, we are back to it, a clean slate.

We have just figured out the Joules, now what would be the next step?

Answer 25

q_metal = m × c × ΔT
replace q_metal with 728.156 J
also, fill in the mass of the metal, and how much the temperature of the metal changed.

Answer 26

ok wait so now we are on the 2nd question?

Answer 27

This is the question you posted. We have done the water part and got 728.156 J
now we are doing the metal

Answer 28

oh now I get it, we are doing the =qmetal correct?

Answer 29

yes

Answer 30

ok so now that I get it lol how would be find c?

Answer 31

The next step is

q_metal = m × c × ΔT
replace q_metal with 728.156 J
also, fill in the mass of the metal, and how much the temperature of the metal changed.

Answer 32

So it is now 728.156 J= 27.776 x 75?

Answer 33

how did you get 75 ? (and be sure to leave the c in the equation)

Answer 34

100 degrees C is the temp of the metal and 25 degrees C is the distilled water temp

Answer 35

I think the 25 C is the water before you put the metal into the water.
after a little while, the water heated up to 31.7 (and the metal cooled down to 31.7) Does that make sense ?

Answer 36

no it doesn't.... sorry :(

Answer 37

if you have some water and put a chunk of hot metal into the water, what happens to the temperature of the water?

Answer 38

the temp would get hot

Answer 39

and the metal would cool down until the water and the metal reached the same temperature.

That is what is going on here.

when they tell us

25 C- Distilled water temp
100 C- Temp of metal
31.7 C- temp of mixture

the 25 C is the water and the 100 C is the hot metal

31.7 C is the temperature of the metal in the water (the mixture) after the water heated up and the metal cooled down until both are at the same temperature. (notice that when you mix two things together they reach the *same* temperature... it would be weird if you put hot water in your cup and the cup got colder and the water got hotter... it doesn't do that... they water and the cup come to the same temperature)

Answer 40

oh ok I get that analogy. thanks so now what will happen next.

Answer 41

The next step is

q_metal = m × c × ΔT
replace q_metal with 728.156 J
also, fill in the mass of the metal, and how much the temperature of the metal changed.

Answer 42

ok so now it would be the mixture correct?

Answer 43

for the temperature (both the water and the metal reach the same temperature)

Answer 44

ok so it would be 728.156= 27.776 x 31.7?

Answer 45

two things: (1) be sure to KEEP the "c"
(2) ΔT is the change in the temperature of the metal. How many degrees did it change ?

Answer 46

ok so it is 728.156= 27.776 x c x ?

I don't get it when you say how many degrees did it change.

Answer 47

read this carefully

25 C- Distilled water temp
100 C- Temp of metal
31.7 C- temp of mixture

the 25 C is the water and the 100 C is the hot metal

31.7 C is the temperature of the metal in the water (the mixture) after the water heated up and the metal cooled down

Answer 48

is the answer 31.7?

Answer 49

since the temp becomes the same?

Answer 50

31.7 C is the temperature of the metal after it cooled down.
but it cooled down from its starting temperature. How many degrees did it drop ?

Answer 51

wait can you rephrase that in another way?

Answer 52

what was the temperature of the metal before we put it in the water ?

Answer 53

the temp was 100?

Answer 54

yes, the metal starts at 100C and drops to 31.7 C
ΔT= 100-31.7

Answer 55

I feel like crying right now, you had to fight and claw your way into my brain so that I can understand thank you!

Answer 56

ok so now I have to subtract 100-31.7 to get the C?

Answer 57

almost. ΔT= 100-31.7

remember we are doing
728.156= 27.776 x c x ΔT

Answer 58

ok so the T= 68.3

Answer 59

728.156= 27.776 x c x 68.3

Answer 60

and the equation (so far) is
728.156= 27.776 x c x 68.3

we can simplify the right side by multiplying 27.776 x 68.3

Answer 61

you should get
728.156= 1897.1 c (we can leave out the x, we know we are multiplying)

to find c, we divide both sides by 1897.1 like this:
728.156/1897.1 = c * 1897.1/1897.1

which simplifies to
728.156/1897.1 = c
or
c = 728.156/1897.1

Answer 62

I got it, but don't go too fast on me like that or I could get lost. Um so C= 0.38?

Answer 63

yes, c = 0.38 J / (g × °C)

that means it might be copper or zinc

Answer 64

which part means it might be copper or zinc, the (g x C)?

Answer 65

the units are Joules (that is energy) per gram (how much metal) per degree
we can try to match 0.38 J / (g × °C) to a table of specific heat capacity of different metals, and the closest I can find are copper and zinc which both have 0.39 J / (g × °C)

Answer 66

oh ok that would make sense since this is a zinc table

Answer 67

now for you unknown metal

Answer 68

phew ok I think I understand now for the most part, let me show you I get it . Here's the table, I'll show you what to do.

25.605 g-Measured mass of metal
24 mL- Distilled water measurement
25 C- Distilled water temp
100 C- Temp of metal
0 C- temp of mixture

Answer 69

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 = qunknown metal

Using the formula qunknown metal = m × c × ΔT, calculate the specific heat of the metal.

Answer 70

I have a problem with your data. If we start with water at 25C and add metal at 100C the mixture CAN NOT go to 0º C. The water should get hotter than 25 and the metal should cool down (but not to 0)

Answer 71

when I did the lab that's what I got, do you want to re do the lab on that part?

Answer 72

I think we better.

Answer 73

ok hold on I'm doing that right now give me 2 min

Answer 74

@phi : i got 28.7 C this time around

Answer 75

OK, can you do the water part?
qwater = m × c × ΔT

Answer 76

um, first we find the joules right?

Answer 77

yes, but the first thing is write down m (mass in grams) c (4.18) and ΔT (28.7 C - starting temp of water)

Answer 78

quick question where is the 4.18 coming from again?

Answer 79

it is the known c for water.

Answer 80

25.605 x 4.18 x (100-28.7)

Answer 81

no. first we do the equation for the WATER.

how much water do you have? what temperature did it start at ?

Answer 82

24 mL?

Answer 83

I did not do the experiment. When you just did it, how much water did you use ?

Answer 84

hold on, this is a virtual lab

Answer 85

yes, I assume this is not real or it would take a lot longer to do.

Answer 86

yea, btw thanks for having a lot of patience with me, this is not my thing esp when it involves math. Reading and English is my thing so when you need help it could be like a collab

Answer 87

but hold on

Answer 88

i used 25 g of water when I first start.

Answer 89

can you list all the data right now?

Answer 90

ok. now we use

qwater = m × c × ΔT

m stands for the mass of the water. They give the water in ml, but water is 1 g per ml
so that means we can just use the number they give us.

c is 4.18 (because we are told that)
ΔT is the difference between the temp of the water and the temp of the mixture.

can you fill in the numbers?

Answer 91

what do you mean for the number they give us?

Answer 92

how much water is used in the experiment ?

Answer 93

rephrase that

Answer 94

cause I'm confused when you say that

Answer 95

how much distilled water is used in the experiment?

Answer 96

measurement or temperature?