[student8607]

Consider metal chunks A, B & C of three different elements at 100 degrees Celsius. All have the same mass. Each is dunked into its own container of water at 20 degrees Celsius with the mass of water being the same as that of the metal chunk. If A, B & C produce final temperatures of 60 - 40 - 30 degrees Celsius, calculate their specific heats relative to the specific heat of water.


I don't even know where to start.

[ARGOS++]

Dear student8607;

Use:   q = c * m *  ∆T      and:   Q_gain = Q_loss.
As in your case the relative spec. Heat capacity is asked, you can simplify in the way that you set c for water = 1.0 J g^-1 °K^-1.

Good Luck!
                    ARGOS⁺⁺

[student8607]

does the question mean each metal was initially 100C or combined they were 100C...i'll assume they each were 100C

the mass of the metals is the same as the mass of water (1g)?


For A I got 313J

For B I got 333J

For C I got 343J

[ARGOS++]

Dear student8607;

Both of your first assumptions are correct!

If I assume that you used for c_Water something around 334 1.0 J g^-1 °K^-1, then I have you to tell, that the results at least for B & C are wrong.
(Please watch the units for your results!)

Hint:
     -  Insert  q = c * m *  ∆T      into:   Q_gain = Q_loss.
     -  Then conform to your suggestion you will find out that the masses m will cancel out.

Remember:
As in your case the relative spec. Heat capacity to Water is asked, your result for A should be spelled: "Equal" or "Same as for water".

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

[DrCMS]

Try again.

For the first 1 you have a mass of A (doesn't matter how much) at 100°C,
you add that to the same mass of water at 20°C and both end up at 60°C.  

What is the temperature change for the water?
What is the temperature change for metal A?

So what is the relationship between the specific heat of A and water?

[student8607]

Try again.

For the first 1 you have a mass of A (doesn't matter how much) at 100°C,
you add that to the same mass of water at 20°C and both end up at 60°C.  

What is the temperature change for the water?
What is the temperature change for metal A?

So what is the relationship between the specific heat of A and water?

They both change by 40°C?
1:1

[ARGOS++]

Dear student8607;

Correct!

And what you now get for B & C?

Good Luck!
                    ARGOS⁺⁺

[student8607]

3:1 and 7:1

[ARGOS++]

Dear student8607;

Finally you have to tell what is 3  and what 1, and what 7 and what 1, because that's also left to you.

Good Luck!
                    ARGOS⁺⁺

[student8607]

J/g C

--Now suppose that the same hot metals are dunked into a liquid other than water that has a smaller specific heat than water. How will the final temperatures are thermal equilibrium compare to those found in water? The mess of the liquid is the same as that of each of the metals.

Hmm.. A lower specific heat would mean less temperature change - so the temps would be higher?

[ARGOS++]

Dear student8607;

Sorry!, I meant it different:
3:1 means 1/3 of that of water, and 7:1 means 1/7 of that of water, because think about that the bigger temperature change must correspond to the smaller c!

Otherwise you are correct!

Good Luck!
                    ARGOS⁺⁺

[student8607]

Oh. Ok. I see.

For my second question then since the liquid has a smaller c....it would have a bigger temperature change. So the metals would be at lower temps? Correct>

[ARGOS++]

Dear student8607;

Exactly!,  -  The metal with the smaller c ends with the lower final temperature.

Good Luck!
                    ARGOS⁺⁺