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### Topic: Lab today (specific heat calculations) Help please  (Read 10984 times)

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#### mr cool

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##### Lab today (specific heat calculations) Help please
« on: May 04, 2009, 06:02:18 PM »
Lab data:
Aluminum---Copper---Alloy (pennies)
Temp of water---------------- 90.0 *C---91.0*C----91.0*C
Final calorimeter temp-------29.3. *C---29.8*C----29.7*C
Initial calorimeter temp-------24.6 *C---24.8*C----24.5*C

In this lab we setup a system wbere a testube full of aluminum was heated up with water in a 250 ml beaker surrounding it.  But thats irrelevent...

Onto the questions....

Pure samples....
1: calculate the change in temp of both metals....
29.3 -24.6= 4.7 *C  Change-----Aluminum
29.8 - 24.8 = 5.0 *C Change----Copper

2: calculate the heat gained by the water in each trial ...

How would i do this?

3:  calculate the specific heat of both metals (cant do it without help on number 2?)
4:the accepted theoretical value of copper is 0.389 J/g C. And zinc is 0.388 J/g C.  Determine your perceny error for your copper trials.  (i can do this easily if i had help with number 2)

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #1 on: May 04, 2009, 06:15:49 PM »
Dear mr cool;

What is the mass of water of the calorimeter??   It is required!

For 2.:
Use:  q = m * c * ∆T    For cwater see on:
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

For 3.:
Use:   q = c * m *  ∆T      and:   Qgain = Qloss.

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

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #2 on: May 04, 2009, 07:17:00 PM »
Dear mr cool;

What is the mass of water of the calorimeter??   It is required!

For 2.:
Use:  q = m * c * ∆T    For cwater see on:
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html

For 3.:
Use:   q = c * m *  ∆T      and:   Qgain = Qloss.

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

Good Luck!
ARGOS++

Water used = 50.0 ml so 50.0 g.

So m * C * temp change

50.0 g * C * 29.3-24.6 = heat gained?  What do i put as C?

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #3 on: May 04, 2009, 07:25:29 PM »

Dear mr cool;

From the link I gave you:  cWater(l) =  4.187 J °K-1 g-1

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #4 on: May 04, 2009, 07:28:41 PM »
So i did 50.0 * 4.187 * 4.7 = 983.945 ... That doesnt seem right.

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #5 on: May 04, 2009, 07:36:07 PM »

Dear mr cool;

But it is right for 4.7°K, - if you add the unit Joule.    Check for the units.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #6 on: May 04, 2009, 07:42:04 PM »
Oh lol, silly me, i forgot heat is not equal to temperature gained
Yea, 900 J is more logical now, 983.94 J / g * C... Correct?

Then, 50.0g * 4.187 (C?, not sure on units). * 5.0 = 1046.75 J/ g *C

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #7 on: May 04, 2009, 07:47:55 PM »
Dear mr cool;

No!  –  The unit is only Joule!    Do the calculation with all units and cancel them out!

But the values are correct.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #8 on: May 04, 2009, 07:53:30 PM »

Dear mr cool;

No!  –  The unit is only Joule!    Do the calculation with all units and cancel them out!

Good Luck!
ARGOS++

50.0 g * 4.187  *4.7 *C. I dont see how thise are canceling

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #9 on: May 04, 2009, 07:58:08 PM »

Dear mr cool;

What did I give you for units for Cwater?:
From the link I gave you:  cWater(l) =  4.187 J °K-1 g-1

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #10 on: May 04, 2009, 08:07:09 PM »
Ok well now #3 is purplexing me.

In my book it says change in heat energy = specific heat of sample * mass of sample * change in temp of sample.

But i have a feeling he wants us to switch up this equation a bit because we dont have specific heat.

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #11 on: May 04, 2009, 08:15:08 PM »

Dear mr cool;

For 3.) you have to solve the formula  q = c * m *  ∆T   for  cmetal and use the correct values for mass and ∆T.

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #12 on: May 04, 2009, 08:21:40 PM »

Dear mr cool;

For 3.) you have to solve the formula  q = c * m *  ∆T   for  cmetal and use the correct values for mass and ∆T.

Good Luck!
ARGOS++

What is cmetal

C * 7.8229 * 29.3 - 24.6. ?

#### ARGOS++

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##### Re: Lab today (specific heat calculations) Help please
« Reply #13 on: May 04, 2009, 08:39:58 PM »

Dear mr cool;

Do you know what means to solve an equation for a given parameter?
If yes,  -  then please repeat:
For 3.) you have to solve the formula  q = c * m *  ∆T   for  cmetal and use the correct values for mass and ∆T.
c = cmetal is the value you have to find/calculate!

Am I correct if I assume that  7.8229 mean the mass of aluminium?
And that (4.7°K) is not the temperature difference of aluminium!

Good Luck!
ARGOS++

#### mr cool

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##### Re: Lab today (specific heat calculations) Help please
« Reply #14 on: May 04, 2009, 08:44:10 PM »
29.3 - 24.6 = 4.7 = temp change of aluminum.

And i have no clue about the rest ofthe post, this is impossible.