[BossSoss]

A 20 gram block of ice at -10 degrees celsius is dropped into 180 grams of water at 80 degrees celsius. What is the final temperature inside the cup?

I've already found the joules required to warm the ice from -10C to 0 which is (10C)(20g)(2.06) = 412j. Then, I found the joules needed to change the solid into liquid, which was (20g)(334j) = 6680j. I don't know what I'm supposed to do next. Anyone have an idea?

[ARGOS++]

Dear BossSoss;

  Use:   q = c * m *  ∆T      and:   Q_gain = Q_loss.
  Do you know c for water?

(Search the forums; there are several similar/identical questions.)

Good Luck!
                    ARGOS⁺⁺

[BossSoss]

Dear BossSoss;

  Use:   q = c * m *  ∆T      and:   Q_gain = Q_loss.
  Do you know c for water?

(Search the forums; there are several similar/identical questions.)

Good Luck!
                    ARGOS⁺⁺

Yes, the c is 4.184, correct? What's the and:   Q_gain = Q_loss for?
I'm not sure which steps to take for this problem..

edit: and yes, i've used the search function. i've only found topics that have problems that consists more than just water.

[ARGOS++]

Dear BossSoss;

Q_gain = Q_loss means that you should set the sum of all heat that the ice is gaining will be exact of the same amount of all heat the "hot" water is losing.
Additionally you know, that the final temperature ( = x°C) you are calculating for must be the same for ice and water!

With your pre-calculated results and this equation you are able to determine x°C.

Good Luck!
                    ARGOS⁺⁺

[BossSoss]

Dear BossSoss;

Q_gain = Q_loss means that you should set the sum of all heat that the ice is gaining will be exact of the same amount of all heat the "hot" water is losing.
Additionally you know, that the final temperature ( = x°C) you are calculating for must be the same for ice and water!

With your pre-calculated results and this equation you are able to determine x°C.

Good Luck!
                    ARGOS⁺⁺

Ah, I see where you're getting at now. So the heat that the ice is gaining should be 6680 + 412 = 7092j and.. I should subtract that from how much energy the hot water contains? And.. the hot water has 66944 joules, correct? Here's my work (200g)(80C)(4.184) = 66944. I'm not sure if thats entirely correct though.

[ARGOS++]

Dear BossSoss;

Not exactly!,  you have for the start only 180g hot water (80°C), but you end with 200g middle-hot water of temperature x°C!
But the basic idea seems correct.

Good Luck!
                    ARGOS⁺⁺

[BossSoss]

Dear BossSoss;

Not exactly!,  you have for the start only 180g hot water (80°C), but you end with 200g middle-hot water of temperature x°C!
But the basic idea seems correct.

Good Luck!
                    ARGOS⁺⁺

Mm.. so then its (180g)(80C)(4.184j) = 60249.6 - 7092 which equals to 53157.6. Where does the Qgain = Qloss come into play?

[ARGOS++]

Dear BossSoss;

You do it better step-wise:
7092 J + 20g * 4.184 Jg^-1°K^-1 * (x°C – 0°C)  =  180g * 4.184 Jg^-1°K^-1 * (80°C – x°C)

That’s where the Q_gain = Q_loss comes into play!

Now solve the equation for x°C!

Good Luck!
                    ARGOS⁺⁺

[BossSoss]

Dear BossSoss;

You do it better step-wise:
7092 J + 20g * 4.184 Jg^-1°K^-1 * (x°C – 0°C)  =  180g * 4.184 Jg^-1°K^-1 * (80°C – x°C)

That’s where the Q_gain = Q_loss comes into play!

Now solve the equation for x°C!

Good Luck!
                    ARGOS⁺⁺

Alright, but does the 7092 J + 20g = 7112?

[ARGOS++]

Dear BossSoss;

How are you able to add:  7092 Joule and 20 grams ?

Please study the equation very carefully!!

Good Luck!
                    ARGOS⁺⁺

[BossSoss]

Dear BossSoss;

How are you able to add:  7092 Joule and 20 grams ?

Please study the equation very carefully!!

Good Luck!
                    ARGOS⁺⁺

Subtract 20g from both sides?

[macman104]

Boss, please look at the picture below.  I recreated ARGOS's equation for you. 

First, your 7092 number, you should understand is your energy required to melt your water to 0C.

Please notice the part in red.  This is all one term of your equation (a q = c*m* :delta: t calculation).  Do you see, this term tells us the energy related to heating the water from 0 degrees to some degree x? 

On the right, you have the term (another q = c*m* :delta: t calculation) that determines the energy for cooling 180g of water at 80 degrees to some degree x.

Now follow ARGOS's instruction, and solve for x.

[BossSoss]

Hm, I see. I ended up with 63.5248566 as the final temperature. Is this correct?

[Borek]

Not checking anything else I can telll you you have shown to many significant numbers.

[ARGOS++]

Dear BossSoss;

You solved the equation correct!,  -  And your 63.5°C as final temperature is also correct!

Good Luck!
                    ARGOS⁺⁺