# Copper calorimeter help

1. Apr 21, 2004

### lollypop

hi everybody:
my problem says the following:
A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure.
If 0.750 kg of lead at a temperature of 255 degree celcius is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

when they say that the can , the water and the ice are in thermal equilibrium at atmospheric pressure, what would be their temperatures??
I am not sure how to set up the equations. should i use Qwater+Qice+Qlead+...= 0 where the Q=m*c*delta T and solve for T.
Delta T should be T-255???? , what T do i use for what was in the can before adding the lead?
thanks.

Last edited: Apr 21, 2004
2. Apr 21, 2004

### Staff: Mentor

If ice and water are in thermal equilibrium, what temperature must they have?

Here's how to set up this problem: find initial temp of can, water, and ice. (That's easy! See above.) You know the initial temperature of the lead. Everything will reach the same final temp, call it T.

The heat that will flow from the lead (mcΔT) will equal the heat absorbed by can (mcΔT), water (mcΔT), and ice (mL + mcΔT). Don't forget that the ice must first be melted (mL) before you can raise its temperature.

3. Apr 22, 2004

### lollypop

I tried doing the calculations but i still get it wrong. I'm still not sure as to which delta T to use for what was in the can before adding the lead.
mc(T-0)+mc(T-0)+mL+mc(T-0)= mc(T-255), using 0 for what was in the can, is this right???

4. Apr 22, 2004

### Staff: Mentor

Almost. Since 0 < T < 255, the heat released from the lead will be mc(255-T) ... you had the sign wrong. Other than that, if you have the right values for m, c, and L, it looks good. If you still get the wrong answer, post the numbers that you are using.

5. Apr 22, 2004

### lollypop

doc al:
thank you for ur fast reply, yes i got it right this time, the final temp would be 21.4 :)

6. Apr 22, 2004

### Staff: Mentor

Excellent!

7. Jun 23, 2004

### mithril

I have a similiar problem

I'm working on a similiar problem and while I have the solution i'm not sure why.

You said

"Almost. Since 0 < T < 255, the heat released from the lead will be mc(255-T) ... you had the sign wrong. Other than that, if you have the right values for m, c, and L, it looks good. If you still get the wrong answer, post the numbers that you are using."

isn't deltaT always = Tfinal - Tinitial? And in this case the initial value was 255 so why mc(255-T)? Besides that the temperature of the lead decreased so wouldn't you expect the change in temperature to be negative?

8. Jun 23, 2004

### Staff: Mentor

Welcome to PF!

It's just a question of how you like to set up your equations. Some folks like to use: Qgained = Qlost. In that case your values had better both be positive.

Others like to use: Q1 + Q2 = 0 (the net heat flow is zero). In that case Q1 = - Q2, and yes you'd have Δt = Tfinal - Tinitial.

As long as you know what you're doing, it's a matter of taste.

If you still have a problem, post it and you'll get plenty of help. Start a new thread.

9. Jun 26, 2004

### Electro

Hello,

Cruising the internet, I saw this wonderful website. I am a high school senior (in fall college freshman). As I have a lot free time now in summer, I preparing for college. The problem I have has to do with temp. exhange but still I can't solve it right. It says:

Determine the result (final temp. and how much steam condensed) when 100 g steam at 100 Celcius is passed into 200 g of water and 20 g of ice at 0 Celcius in a calorimeter whose water equivalent is 30 g.

I tried to solve it: Q lost= Q gained
or:
Q (ice to melt)+ Q (water ice to warm) + Q (water to warm) + Q (calorimeter to gain) + Q (steam to condense) + Q (water steam to change temp.)=0

I can't go to the correct answer. The signs are ok but still there's sth missing.
Thank you.

10. Jul 3, 2004

### maverick280857

Sometimes in calorimetry you should draw a rough line diagram to represent the temperatures of the things given to you, for the initial and thermodynamic equilibrium states, so that you will never make a mistake in setting up the equations. (Somewhat like energy lines). But that again isn't a golden rule...

11. Jul 3, 2004

### Staff: Mentor

Welcome to PF, Electro!
The final temperature can only be: (a) 100 C (if only a portion of steam condenses) or (b) some T between 0 and 100 C.

Don't be afraid to play around with it. Let's just guess that the answer is (a). (If we are wrong, then we move to b.)

First find out how much heat would need to be removed from the steam to condense it all: $Q_{steam} = m_{steam}L_v$.
Then find out how much heat is required to (1) melt the ice $Q_1 = m_{ice}L_f$ and then (2) heat the water and calorimeter to 100 C $Q_2 = m_{water+calorimeter}c (100)$. Crank out $Q_1 + Q_2$ and compare it to $Q_{steam}$.

Let me know what you discover.

12. Jul 4, 2004

### Electro

Thank You for your answer Doc. Fortunately I have only the answers of this exercise (final temp.= 100 d. Celcius and 49 g of steam condensed). The way I solved the exercise is the same as you did. I find the energy given when steam condenses + the energy when this water cools and equal that with the energy when ice melts+energy of water ice to warm + water to warm+calorimeter energy. But the result is contradictory. Maybe there is something I don't get in this exercise? Maybe the ice is at 0 and the water + calorimeter are at another temp.?

13. Jul 5, 2004

### Staff: Mentor

Careful! I said compare those energies, not set them equal. When you compare $Q_1 + Q_2$ with $Q_{steam}$ you of course find that they are not equal! $Q_{steam}$ is the energy released if all the steam changes to water; by comparing this to the energy needed to heat the the ice + water to 100 C, you'll find that only a portion of the steam needs to condense to heat up that water. To find the amount of steam condensed, you can either take ratios of the energies or set $Q_1 + Q_2 = m_{steam-condensed}L_v$ and solve for the amount of steam that condenses.
Thermal equilibrium requires that the final temp of everything be the same.

Let me know if my explanation makes sense.

14. Jul 6, 2004

### Electro

Right Doc. I found first the mass of the steam condensed and it was 49 g. After this I began setting the equalities and as a mass of steam I used 49 g instead of 100. The result was exactly 100Celcius (final temp).