Question

Part I:
1. Calculate the energy change (q) of the surroundings (water). 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.fh
2. Calculate the specific heat of the metal.

Answer 1

@JoannaBlackwelder could you please help me?

Answer 2

so there is a lab and the photo i attached is the table. i have to answer those questions above and i really don't understand them... do you get it?

Answer 3

Sure. So, did you take the hot metal and drop it into the room temp water, wait for a bit, then measure the temp of the water again?

Answer 4

@JoannaBlackwelder yes

Answer 5

Ok. The idea is that all of the energy from the hot metal was transferred into the water. So we need to find how much energy the water took in.

Answer 6

q=m*C*ΔT
where q is energy, m is mass, C is specific heat, and ΔT is change in temp

Answer 7

Using this formula, can you calculate the energy absorbed by the water?

Answer 8

okay got the formula! so what do i have to plug in to m?

Answer 9

Use the volume of water and density of water to calculate mass.

Answer 10

i know that the measured mass from the data table is 27.776. that all

Answer 11

The table also states that the volume of water is 26 mL and density is given in the problem as 1g/mL

Answer 12

oh yes. do i plug in the measurements in this formula q=m*C*ΔT ?

Answer 13

Right, but be careful to use the measurements of water to find the heat increase of water.

Answer 14

q=m*C*ΔT
=27.776*
i am kinda stuck finding the c for specific heat and the temperature change.

Answer 15

m is the mass of water, remember? Try that one again.

Answer 16

Specific heat in given in the problem statement.

Answer 17

Change in temp is final temp - initial temp

Answer 18

specific heat 4.18
mass of water: 18.015
change in temperature:

Answer 19

How did you get the mass of water?

Answer 20

i was thinking of the molar mass of water which is 18.015 g/mol.

Answer 21

We could use that if we knew we were using 1 mole of water. But we don 't.

Answer 22

We are given the volume of water to be 26 mL.

Answer 23

then how would i find the mass of the water?

Answer 24

26 mL*(1 g/mL)=26g

Answer 25

oh so is the mass 26

Answer 26

Yes.

Answer 27

Can you find the change in temp?

Answer 28

q=27.776*26* change in temperature

Answer 29

You switched out the C not the m. Try that again.

Answer 30

q= 26* 27.776*change in temp

Answer 31

No. C is 4.18, m is 26.

Answer 32

oh yes sorry

Answer 33

No worries. Just try to be careful.

Answer 34

ok now know those two how do i determine the change in temperature

Answer 35

Final temperature of water - initial temp of water.

Answer 36

thanks but according to the data table there is many temperatures how do i know which on is the final temperature of water and the initial temp of water

Answer 37

The distilled water temp should be the initial temp. And the temp of the mixture should be the final temp.

Answer 38

31.6-25.5= 6.1

Answer 39

:)

Answer 40

q= 26*4.18*6.1= 662.948 :)

Answer 41

:) Awesome!

Answer 42

So now, we are assuming that all of the heat absorbed by the water came from the metal. So, the heat loss of the metal is 662.948 J.

Answer 43

Or -662.948J to show heat loss.

Answer 44

So now use the formula again, but use the measurements of the metal.

Answer 45

q=m*C*ΔT
q= 27.776*4.18*6.1
??

Answer 46

q = -663.949J, m is good, C is what we are solving for, and change in temp is temp of solution-temp of metal

Answer 47

31.6-101
??

Answer 48

Right. The temp goes down, thus the change in temp is negative.

Answer 49

yes so the answer for that is -69.4 ?

Answer 50

Right. :)

Answer 51

yay okay and the answer to the first part is 662.948 J ?

Answer 52

Yup!

Answer 53

Omg! Thank for all of your help and being so patient with me! ≧◉◡◉≦

Answer 54

Using sig figs, 663J

Answer 55

yep! okay!

Answer 56

You're welcome. :)

Answer 57

Wanna finish calculating the specific heat of the metal?

Answer 58

absolutely

Answer 59

Use the equation again, but with the numbers for the metal.

Answer 60

okay
q= 27.776* C* 31.6

Answer 61

I though you found the change in temp to be -69.4

Answer 62

And the heat added to the water came from the metal, so q=-663

Answer 63

that is what i meant opps
-663= 27.776*C*-69.4

Answer 64

:) No solve for C.

Answer 65

*Now

Answer 66

.3439

Answer 67

Looks good. Any ideas on the units?

Answer 68

hmm... what units?

Answer 69

The units of the specific heat.

Answer 70

umm i don't think so

Answer 71

*Hint. It is the same as the units for the heat capacity of water.

Answer 72

Joules / mole K
??

Answer 73

Are you sure those are the units of the heat capacity of water given in the problem?

Answer 74

celsius
??

Answer 75

Right. And mole should be g.

Answer 76

yes! Thanks again!

Answer 77

No worries :)