Finding Final Equilibrium and the like

[nycginger]

Hi. I'm having incredible difficulty on my homework (2 problems left that are due tomorrow)for my physical science class. My professor is kind of making us rely on ourselves to learn this stuff. This is the first collegiate class I've taken that has actually made me feel stupid. The syllabus says this class only requires college algebra, but another professor mentioned we needed physics to succeed in this course.

Please help me. I am totally lost. I have the formulas here for one problem we did in class, but now that I look at my notes for finding the final equilibrium temperature, it quite frankly looks like a different language.

Here are the two problems that are giving me trouble:

1) A 2500-g block of lead at 95C is dropped into an insulated cup holding 1200 g of water at 22C. Determine the final equilibrium temperature.




2) A 350-g metal cube at 85C is placed in a Styrofoam beaker holding 150 g of water at 22C. The final equilibrium temperature is 36C. Determine the specific heat of the metal. From this value, identify the possible metal using the Internet containing tables of specific heat




I've tried using Q=MC[tex]\Delta[/tex]T and the answer should be in celsius. I've tried converting to Kelvin (not even sure if that's necessary) to see if it would help in the slightest. I'm really feeling discouraged. Can you please help?



M1*C*DT1=M2*C*DT2=0

The example he gave us in our notes reads:

1 calorie/kg celscius

(Tf-2s)=-0.1kg (0.3 cal/kg celsius) * (Tf-100)
tf-25=-0.03 (tf-100)
Tf-25=-0.03Tf+3 (Distribute the -0.03 to both sides)
1.03Tf-25=3 (Distribute 25 to both sides)

1.03Tf=28
Tf= 27.18 degrees celsius.

This example would be a lot easier to understand if he didn't pull numbers out of thin air instead of using the actual numbers. :-(

Please help!

 

[gneill]

Let's take one problem at a time. For the first problem, do you have the specific heats of lead and water?

The idea is to set up and equation that relates the heat lost by the hotter substance to the heat gained by the cooler substance (the lead and the water in this case). The specific heats tell you how much heat moves into or out of a given mass of the substance for a given change in temperature. Or, stated in reverse, how much the temperature of a given mass of a substance will change if you put in or take out a certain amount of heat.

Thus, for example, suppose you have mass M of a substance with specific heat C. Then if the change in temperature is ∆T, the amount of heat that's moved in or out (depending upon the sign of ∆T) is given by MC∆T.

If you want to find an equilibrium temperature, then you don't have the actual ∆T. What you have is the starting temperatures of the samples, say T1 and T2. If you suppose that the mutual final temperature will be some value Tf, then you write the ∆T's as

∆T1 = (T1 - Tf)

∆T2 = (T2 - Tf)

Assuming that the final temperature is going to lie somewhere in between T1 and T2, one of those ∆T's is going to be negative. Thus the sum M1*C1*∆T1 + M2*C2*∆T2 should be zero. Why don't you try setting up the equation for the first problem?

 

[nycginger]

Let's take one problem at a time. For the first problem, do you have the specific heats of lead and water?

The idea is to set up and equation that relates the heat lost by the hotter substance to the heat gained by the cooler substance (the lead and the water in this case). The specific heats tell you how much heat moves into or out of a given mass of the substance for a given change in temperature. Or, stated in reverse, how much the temperature of a given mass of a substance will change if you put in or take out a certain amount of heat.

Thus, for example, suppose you have mass M of a substance with specific heat C. Then if the change in temperature is ∆T, the amount of heat that's moved in or out (depending upon the sign of ∆T) is given by MC∆T.

If you want to find an equilibrium temperature, then you don't have the actual ∆T. What you have is the starting temperatures of the samples, say T1 and T2. If you suppose that the mutual final temperature will be some value Tf, then you write the ∆T's as

∆T1 = (T1 - Tf)

∆T2 = (T2 - Tf)

Assuming that the final temperature is going to lie somewhere in between T1 and T2, one of those ∆T's is going to be negative. Thus the sum M1*C1*∆T1 + M2*C2*∆T2 should be zero. Why don't you try setting up the equation for the first problem?

I only have the information listed. He gave us no other information for #1.

M1*C1*∆T1 + M2*C2*∆T2=0

I've read this over a couple of times and still feel pretty lost.

Is this anywhere close to right?

I googled the specific heat of lead and it says it's 0.031.

Tf-22=-0.12kg(0.031cal/kg degrees celsius) * (Tf-0.25kg)

then maybe

Tf-22=-0.00372(Tf-0.25)
Tf-22=0.00372TF-0.00093
+0.00372Tf +0.00372Tf

0.00372Tf-21.9963=-0.00093?

I'm really confused. How am I doing?

P.S. This is still not really making sense, but thank you for helping me with this. I really appreciate it.

 

[gneill]

I only have the information listed. He gave us no other information for #1.

M1*C1*∆T1 + M2*C2*∆T2=0

I've read this over a couple of times and still feel pretty lost.

Is this anywhere close to right?

I googled the specific heat of lead and it says it's 0.031.

Tf-22=-0.12kg(0.031cal/kg degrees celsius) * (Tf-0.25kg)

then maybe

Tf-22=-0.00372(Tf-0.25)
Tf-22=0.00372TF-0.00093
+0.00372Tf +0.00372Tf

0.00372Tf-21.9963=-0.00093?

I'm really confused. How am I doing?

P.S. Thank you for helping me with this. I really appreciate it.

The value that you found for the specific heat of lead is okay, but beware that it seems to be specified in units of kcal/(kg K). That's fine if you also specify the specific heat of the water in the same units.

The equation you wrote first is the form you want to begin with. But when you started entering actual numbers, it seems to have gone somewhat astray. For example, note that you are trying to take the difference between a temperature and a mass in kilograms (in red). So something went off the track with your algebra. I don't see where the 0.12kg came from.

You might have a go at expanding your formula:

M1*C1*∆T1 + M2*C2*∆T2=0

with the symbolic values for the ∆T's plugged in and solving for Tf. Do it all symbolically first, then plug in the appropriate values at the end.

 

[nycginger]

The value that you found for the specific heat of lead is okay, but beware that it seems to be specified in units of kcal/(kg K). That's fine if you also specify the specific heat of the water in the same units.

The equation you wrote first is the form you want to begin with. But when you started entering actual numbers, it seems to have gone somewhat astray. For example, note that you are trying to take the difference between a temperature and a mass in kilograms (in red). So something went off the track with your algebra. I don't see where the 0.12kg came from.

You might have a go at expanding your formula:

M1*C1*∆T1 + M2*C2*∆T2=0

with the symbolic values for the ∆T's plugged in and solving for Tf. Do it all symbolically first, then plug in the appropriate values at the end.

I changed grams to kilograms.

I'm still confused by the ∆T. I'm not sure where it comes from or what I'm supposed to put there.

I know that ∆T stands for the temperature change (Tf-Ti). I'm just utterly confused as I'm supposed to get there.

The 0.25kg and the 0.12kg came from converting grams to kilograms.

I really appreciate you helping me walk through this with my level of ignorance. I've been sitting here contemplating shoving a pen in my temple, so you're really helping me to try to calm down.

Okay

So...

If Heat = Q which is a joule or Newton meter...that is equal to mass (kg) which is why I converted.

Q=mass times C (specific heat) times ∆T or the temp change.

1200 * 95C * (95-22) + 2500 * 22* (22-95)=0

Am I anywhere close?

Symbolically?

1200* 95* ∆T1 + 2500 * 22 * ∆T2 = 0?

God, I wish this made sense to me. This kind of thing wasn't ever my forte. :-( I feel so darn stupid and I make the Dean's List!

 

[gneill]

You know that you have two different substances, each with its own starting temperature. When they're placed together, one will warm up and the other will cool down until they reach some compromise temperature. That's the Tf. The ∆T's are the individual temperature changes that the substances undergo to reach that final temperature, Tf.

Joules are energy, kilograms are mass. They're not the same thing. In fact, a Joule can be expressed in units of kg*m²/s².

The only free variable in the heat balance equation that we're looking at is Tf. All the rest are known values that are specified in the problem statement. Each one of them should only appear once in the equation.

Let's take the general equation again, and rearrange it a bit for convenience

[tex] M_1 C_1 \Delta T_1 + M_2 C_2 \Delta T_2 = 0 [/tex]

[tex] M_1 C_1 (T_1 - T_f) + M_2 C_2 (T_2 - T_f) = 0 [/tex]

[tex] (T_1 - T_f) = \frac{M_1 C_1}{M_2 C_2} (T_f - T_2) [/tex]

Each of the variables can be assigned values from the problem statement:

M₁ = mass of water
M₂ = mass of lead
T₁ = initial temperature of water
T₂ = initial temperature of lead
C₁ = specific heat of water
C₂ = specific heat of lead

If you plug in the given values, making sure that consistent units are always used (don't mix grams with kg, or cals with kcals without conversion for examples), then you should end up with a straightforward equation to solve for the final temperature T_f. With this arrangement of the equation, all the masses and specific heats conspire to produce a single unit-less value that is the proportionality constant for the temperature changes that occur in the two substances. Does this help?

 

[nycginger]

You know that you have two different substances, each with its own starting temperature. When they're placed together, one will warm up and the other will cool down until they reach some compromise temperature. That's the Tf. The ∆T's are the individual temperature changes that the substances undergo to reach that final temperature, Tf.

Joules are energy, kilograms are mass. They're not the same thing. In fact, a Joule can be expressed in units of kg*m²/s².

The only free variable in the heat balance equation that we're looking at is Tf. All the rest are known values that are specified in the problem statement. Each one of them should only appear once in the equation.

Let's take the general equation again, and rearrange it a bit for convenience

[tex] M_1 C_1 \Delta T_1 + M_2 C_2 \Delta T_2 = 0 [/tex]

[tex] M_1 C_1 (T_1 - T_f) + M_2 C_2 (T_2 - T_f) = 0 [/tex]

[tex] (T_1 - T_f) = \frac{M_1 C_1}{M_2 C_2} (T_f - T_2) [/tex]

Each of the variables can be assigned values from the problem statement:

M₁ = mass of water
M₂ = mass of lead
T₁ = initial temperature of water
T₂ = initial temperature of lead
C₁ = specific heat of water
C₂ = specific heat of lead

If you plug in the given values, making sure that consistent units are always used (don't mix grams with kg, or cals with kcals without conversion for examples), then you should end up with a straightforward equation to solve for the final temperature T_f. With this arrangement of the equation, all the masses and specific heats conspire to produce a single unit-less value that is the proportionality constant for the temperature changes that occur in the two substances. Does this help?

I wish it did. *Sigh* I can honestly say I'd rather jump in front of a bus. I'm losing hope quickly. The school closes soon and I have to print this off tonight. :-(

Specific Heat of water: 4.186 joule/grams
Mass of water: 1200 grams
Mass of lead: 2500grams


1200 * 4.186 (22-Tf) + 2500 * 0.03 (95-Tf)=0

110510.4 -4.186Tf + 7125 -0.03Tf=0

Add -0.03Tf to both sides

?

110510.4-4.156Tf + 7125 = 0

-4.156Tf = 103385.4

Tf= 24876.179?

Even that looks wrong.

I don't get where I'm messing up!

Wait a second..


I did it again changing the grams to kilograms and ended up with 24.8231...

is that right? :D

 

[gneill]

I wish it did. *Sigh* I can honestly say I'd rather jump in front of a bus. I'm losing hope quickly. The school closes soon and I have to print this off tonight. :-(

Specific Heat of water: 4.186 joule/grams/K
Mass of water: 1200 grams
Mass of lead: 2500grams


1200 * 4.186 (22-Tf) + 2500 * 0.03 (95-Tf)=0

That 4.186 is in joules/g/K, whereas the 0.03 (which should be 0.031 for accuracy) is in cal/g/K. It should be one set of units or the other. Calories makes sense because then the value for water is 1cal/g/K. So:

1200 * 1 (22-Tf) + 2500 * 0.031 (95-Tf)=0

Expanding:

26400 - 1200 Tf + 7362.5 - 77.50 Tf = 0

Collecting terms:

33762.5 - 1277.5 Tf = 0

Tf = 26.4 °C