[mr cool]

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++]

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]

so....

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

[ARGOS++]

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]

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++]

[mr cool]

168.00 cal/g (M g)?

[ARGOS++]

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]

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++]

Dear reezyfbaby;

Correct!

Good Luck!
                    ARGOS⁺⁺

[mr cool]

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]

Why would you divide by 2 just because you are starting with ice?

[ARGOS++]

[mr cool]

[ARGOS++]