August 08, 2022, 09:15:19 AM
Forum Rules: Read This Before Posting

Topic: phase change practice with other variables  (Read 21010 times)

0 Members and 1 Guest are viewing this topic.

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
phase change practice with other variables
« on: February 05, 2009, 02:08:25 PM »
I can tell this worksheet is going to be a nightmare because im stuck on number one....

What mass of ice was converted from -20.00*C to liquid water at 78.00*C by 7560 calories?

i have gotten this far..

(.5 cal/g)(M)(0*C - 20.00*C)  +  (80 cal/g)(M)  +  (1 cal/g)(M)(78.00-0)= 7560 calories
|                                        |                          |
v                                        v                          v
10g(M)              +                  80g(M)          +          78.00g(M) = 7560 calories

is this correct up to this point?

i have no clue what to do next, i know it involves algebra though

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #1 on: February 05, 2009, 02:31:08 PM »
Dear reezyfbaby;

Your first equation line is correct!
But units in your last equation are wrong!,  -  they must be:   cal/g * (Mg)  = cal !

On the left you have written:  a * (M g)  +   b *( Mg)   +  c * (Mg) = 7560 cal.
Can you do the little algebra on the left side and finally substitute correctly backward: a = 10 cal/g, and so on?

I hope to have been of help to you.
Good Luck!
ARGOS++

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #2 on: February 05, 2009, 02:40:05 PM »
so....

10 cal/g (M) + 80 cal/g (M) + 78.00 cal/g (M) = 7560   ?

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #3 on: February 05, 2009, 02:46:34 PM »

Dear reezyfbaby;

Not exactly!  -  But already much better!
Can you count the left side together and keep the units, also for your M?

Good Luck!
ARGOS++

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #4 on: February 05, 2009, 02:51:24 PM »

Dear reezyfbaby;

Not exactly!  -  But already much better!
Can you count the left side together and keep the units, also for your M?

Good Luck!
ARGOS++

are you asking me to distribute the 10 cal/g into the (M) ?

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #5 on: February 05, 2009, 02:56:44 PM »

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #6 on: February 05, 2009, 02:58:07 PM »
168.00 cal/g (M g)?

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #7 on: February 05, 2009, 03:07:21 PM »

Dear reezyfbaby;

Exact!
If you now complete the equation you get:   168.0 cal/grams * M grams =  7560 cal.
Now you can solve for M grams.

Good Luck!
ARGOS++

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #8 on: February 05, 2009, 03:21:29 PM »
would i take

168.0 cal/grams * M grams =  7560 cal

and divide it by 7560 cal to get rid of the cal and then be left with grams?

i got 45.00 g when i divided the two.

just going out on a limb here, algebra is not my forte either.

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #9 on: February 05, 2009, 03:23:34 PM »

Dear reezyfbaby;

Correct!

Good Luck!
ARGOS++

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #10 on: February 05, 2009, 03:42:38 PM »
i think i got this next one right so far....

A 150.00 g piece of -50.00*C ice is exposed to enough energy (85.00 kJ) to convert it to liquid water at what temperature?
my work....
(this first part is taking 85.00 kJ into calories...)

85.00 kJ * 100j / 1 kJ * 1 cal / 2.092 = 4063  (i used 2.092 because my chem teacher said 1 cal = 4.1845 J and since we start at ice i split it in half.)

(.5 cal/g)(150.00g)(0*C - -50.00*C) + (80 cal/g)(150.00g) + (1 cal/g)(150.00g)(F - 0*C) = 4063 calories... (the F is my place holder for the final temp)

3750 cal/g + 12000 cal/g + 150.00 cal/g(F - 0*C) = 4063 calories...

15750 cal/g + 150.00 cal/g * F = 4063 calories.

do i add 15750 to 150.00 and get 15900 and divide it by 4063?

macman104

• Retired Staff
• Sr. Member
• Posts: 1644
• Mole Snacks: +168/-26
• Gender:
Re: phase change practice with other variables
« Reply #11 on: February 05, 2009, 03:49:57 PM »
Why would you divide by 2 just because you are starting with ice?

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #12 on: February 05, 2009, 03:55:55 PM »

mr cool

• Regular Member
• Posts: 81
• Mole Snacks: +0/-1
Re: phase change practice with other variables
« Reply #13 on: February 05, 2009, 04:00:25 PM »

ARGOS++

• Sr. Member
• Posts: 1489
• Mole Snacks: +199/-56
• Gender:
Re: phase change practice with other variables
« Reply #14 on: February 05, 2009, 04:07:31 PM »