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      02-15-2011, 09:54 PM   #1
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Chemistry help

I need to get these by midnight tonight but I have no idea how to do them.

Problem 1-


Problem 2-

Last edited by aretardedorange; 02-15-2011 at 10:59 PM..
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      02-15-2011, 10:14 PM   #2
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are you working with Q=mcAT?
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      02-15-2011, 10:17 PM   #3
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Quote:
Originally Posted by skinnycow View Post
are you working with Q=mcAT?
sounds like it. Just remember to do your phase changes. Thats a diff formula.
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      02-15-2011, 10:17 PM   #4
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1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol
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      02-15-2011, 10:18 PM   #5
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Quote:
Originally Posted by skinnycow View Post
1) 63360 degrees C

2) 114153.62 degrees C
The answers would be in Joules.
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      02-15-2011, 10:21 PM   #6
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Quote:
Originally Posted by BrokenVert View Post
The answers would be in Joules.
oops sorry i meant J
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      02-15-2011, 10:23 PM   #7
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Quote:
Originally Posted by skinnycow View Post
1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol
geez. i must be a dumbass. i don't remember a thing in high school. i'm sure i took chemistry somewhere in those 4 years. hahaha

i did end up being a biochemistry major and graduating with honors out of college though... but i still would have to do some mental digging to do those problems, even tho they look simple as pie
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      02-15-2011, 10:24 PM   #8
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Yes that formula. But I reboot my computer and it gave me new problems. If you could do these two it would be much appreciated.

Last edited by aretardedorange; 02-15-2011 at 10:59 PM..
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      02-15-2011, 10:30 PM   #9
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-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)

Last edited by Inspired; 02-15-2011 at 10:36 PM..
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      02-15-2011, 10:30 PM   #10
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give me a sec lol
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      02-15-2011, 10:39 PM   #11
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Quote:
Originally Posted by twinturbo335 View Post
-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)


THANKS!
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      02-15-2011, 10:41 PM   #12
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1) 0.1222
2) 142718.4

the first one's tricky

on the first one, since the metal is in solution, you'll end up having the same temperature (between water and the metal) which means T-final for the water is also 61.1, so you throw that into the Q = mc(Tf-Ti) equation and you get the energy change by the water, Q = 329.28 since you have Q, which will fit into the Q=mcT on the metal equation, you can put everything in and find c 329.28 = (55.89)(c)(61.1 - 12.9) rearrange the equation and solve for c
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      02-15-2011, 10:47 PM   #13
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Quote:
Originally Posted by skinnycow View Post
1) 63360 degrees C

2) 114153.62 degrees C

I think that's right, last time i took chem was in hs lol
Isn't it -63360? It is t final minus t initial. 21-186
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      02-15-2011, 10:49 PM   #14
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Quote:
Originally Posted by twinturbo335 View Post
-1.2x10^5 for problem 2?
(1200g)(0.901J/gC)(21-133C)
This one is wrong. What you had before was correct.
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      02-15-2011, 10:49 PM   #15
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Quote:
Originally Posted by skinnycow View Post
1) 0.1222
2) 142718.4

the first one's tricky

on the first one, since the metal is in solution, you'll end up having the same temperature (between water and the metal) which means T-final for the water is also 61.1, so you throw that into the Q = mc(Tf-Ti) equation and you get the energy change by the water, Q = 329.28 since you have Q, which will fit into the Q=mcT on the metal equation, you can put everything in and find c 329.28 = (55.89)(c)(61.1 - 12.9) rearrange the equation and solve for c
1. Is incorrect . Thanks anyway though. Much appreciated!
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      02-15-2011, 10:49 PM   #16
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Quote:
Originally Posted by twinturbo335 View Post
Isn't it -63360? It is t final minus t initial. 21-186
yeah it is i didn't put the negative cause the question was worded funny 'how much is removed' but yeah it's technically negative
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      02-15-2011, 10:51 PM   #17
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Quote:
Originally Posted by aretardedorange View Post
1. Is incorrect . Thanks anyway though. Much appreciated!
awww i'm sorry i was trying to do it as quickly as i could and couldn't really look it through
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      02-15-2011, 10:51 PM   #18
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Quote:
Originally Posted by aretardedorange View Post
This one is wrong. What you had before was correct.
The first one I didn't put to 2 significant figures. thought it would be better than having a whole bunch of #s
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      02-15-2011, 10:53 PM   #19
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Quote:
Originally Posted by skinnycow View Post
awww i'm sorry i was trying to do it as quickly as i could and couldn't really look it through
No problem . I kind of got the hang of these problems (kind of) being the key word lol. Will attempt some of these tomorrow.

Thanks everyone that helped! Once again much appreciated!
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      02-15-2011, 11:04 PM   #20
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Quote:
Originally Posted by aretardedorange View Post
Yes that formula. But I reboot my computer and it gave me new problems. If you could do these two it would be much appreciated.

A 55.89 g sample of metal changes its temperature from 12.9 °C to 61.1 °C when it absorbed energy from a 78.7 g sample of water whose initial temperature was 66.2 °C. What is the specific heat of the metal in J/g·°C?

A skillet made of Aluminum (Al) weighing 1.2 kg is heated on a stove to a temperature of 133 oC. If the skillet is removed from the stove and allowed to cool to room temperature (21oC), how much heat must be removed from the skillet. The specific heat of Aluminum is 0.901 J/(g.oC)
is Q1 .0316?

C=Q/M(deltaT)

First I tried to solve for TF for the 2nd part.
(78.7g)(4.184)(Tf-66.2)

=85J

(55.89)(C)(61.1-12.9)

C=2693.9

85J/2693.9

.0316J/G [C]
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      02-15-2011, 11:09 PM   #21
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Quote:
Originally Posted by twinturbo335 View Post
is Q1 .0316?

C=Q/M(deltaT)

First I tried to solve for TF for the 2nd part.
(78.7g)(4.184)(Tf-66.2)

=85J

(55.89)(C)(61.1-12.9)

C=2693.9

85J/2693.9

.0316J/G [C]


booo it bugs me when i get something wrong!!!!
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      02-15-2011, 11:10 PM   #22
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Quote:
Originally Posted by skinnycow View Post
booo it bugs me when i get something wrong!!!!
I'm not sure BTW. Kind of rusty.....
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