Question

I have the problem...
"A 20g chunk of aluminum at 95 degrees C is placed into 20 gallons of water at 86 degrees C.

What is the final temperature of the mixture?
What mass of water is vaporized?
How much water is left over?

Hint... The density of water is 8.33 lbs/gallon... 1lb = 453.6g"
If I did it correctly the final temperature is 139.29 degrees C, but I'm unsure how to find the other two. I used the formula mc delta t = mc delta t + mHv. The only problem is that the answer I get doesn't come close to what my teacher gave me.

Answer 1

so you used: \(q=m*C*\Delta T\)

then: \(q_{Al}=-q_{water}\)

\(q_{Al}=m_{Al}*C_{Al}*(T_f-T_i)=m_{water}*C_{water}*(T_f-T_i)=-q_{water}\)

Answer 2

Yes.

Answer 3

and you got the wrong temperature?

Answer 4

Well, I had to convert the gallons to grams and I may have done that incorrectly.

Answer 5

\(\rho=\dfrac{m}{V}\)

\((m=8.33 lbs/gallon*{20 gallons})*453.6g/lb\)

is that what you did?

Answer 6

Yes.

Answer 7

oh, i didn't see you wrote that you did get the temperature.

okay, so find the q transferred by the Al chunk.

from there you have to subdivide it, the heat to get the water to 100 celsius, then the heat required to vaporize some of the water.

which is what you wrote "q= mc delta t + mHv" pretty much.

maybe you're just doing it wrong, post your work and I, or someone else, will correct it if it's incorrect.

Answer 8

Well, I did \[20,000\times0.9\times(950-x)=75569.76\times4.184\times(x-86)\] and then I got \[180\times(950-x)=316183.87584\times(x-86)\] which I followed with the distributive property and got \[17100000-1800x=316183.8754x-27191813.3222\] next I added 27191813.3222 to both sides and 1800x to both sides to get \[44291813.3222=317983.8754x\] and after dividing both sides by 317983.8754 I ended with 139.29.

Answer 9

hmm you made a mistake

20 g *0.9*(x-95)=75569.76 g*4.184*(x−86)

x=85.9994876107373397 = 86 degrees


If you think about it, the temperature of the metal shouldn't have gone up

Answer 10

though, this way, it doesn't make sense for them to ask you how much water evaporated...

are you sure \(C_p\) for Al, is 0.9 ?

also, was it, 20 g or 20 kg?

Answer 11

Its 20,000 g and 20 kg. I just checked my periodic table and it says that the specific heat of Aluminum is 0.90

Answer 12

ohh okay, you wrote 20 g in your first post. And the temp is 950 i guessing? you wrote 95 lol

Answer 13

Oh. Yes, I'm sorry. I was rushing. It's 1am. I finally decided this was going to take a while and slowed myself down, but yes, that is what I meant.

Answer 14

Yeah, i know. It's 1 am here too. Would you mind if i came back to this tmrw, or are you in a rush to finish it?

Answer 15

I has a Chemistry test first thing tomorrow morning, but I can't ask you to stay.

Answer 16

okay no worries, this shouldn't take too long. Another quick question, is the final temperature of the Al 950, or is that the initial?

Answer 17

Initial I do believe.

Answer 18

sorry, i forgot a negative sign.

it should read: \(-[20000*0.9*(T_f-950)]=75569.76 *4.184*(T_f-86)\)

\(T_f\)=132.5372542613215746= 132.5 celsius

now you can find \(q_{Al}=20000*0.9(132.5-950)\)= -14715000 J

so it gave off that much heat.

now you need to do this part in a few steps, the water needs to get to 100 celsius before it evaporates

Answer 19

I have a question. Why is the first part negative?

Answer 20

Because heat was given off by the aluminium towards the water.
When energy is given off it's given a negative sign, when it's absorbed it's positive.

Answer 21

That makes sense.

Answer 22

okay, cool.

Answer 23

Heating to 100: \(q_{water}=75569.76∗4.184∗(100−86)\)= 4426574.26176

so we minus that from:

i'm gonna change the signs to make this simpler:

\(q_{left over}=q_{Al}-q_{water}\)=14715000 J-4426574.26 J= 10288425.74 J


now we know that water can't be heated higher than 100 celsius without being evaporated, so the heat that is left, MUST go to those 2 processes together:

for water vapour C=1.996, i don't know \(H_{vap}\) off the top off my heat, but you plug that in and solve for m

\(10288425.74 J= m*1.996 *(132.5-100) + m*H_{vap}\)

Answer 24

Hv=334

Answer 25

is that in joules or kJ?

Answer 26

it's more than likely, kJ, so you need to convert it to J, multiply by 1000
so Hvap= 334000

\(10288425.74J=m∗1.996∗(132.5−100)+m*(334000)\)

solve for m to get the mass of water that evaporated.

Answer 27

I think its Joules because we've never had to convert it before.

Answer 28

Same concept thought. I understand.

Answer 29

okay, if it's in J, then theres not need to convert it. just make sure your units are the same everywhere.

Answer 30

Do you know how to do the rest?

Answer 31

You mean he water left over? If so, no.

Answer 32

so okay, once you find the mass of water that evaporated, you simply subtract that mass from the original (total) mass.

\(mass_{leftover}=mass_{total}-mass_{evaporated}\)

Answer 33

The original water mass?

Answer 34

yeah, the 20 gallons which were 75569.76 grams


so: \(mass_{leftover}=mass_{total}-mass_{evaporated}\)

\(mass_{leftover}=75569.76 \;g-mass_{evaporated}\)

Answer 35

Okay. That's what I thought, but I figured I should make sure before I assumed wrong or something. Thank you! I really appreciate you staying up and helping me.

Answer 36

no problem, i hope it helped and you do well on that test tomorrow !