Copper calorimeter help

In summary, the problem involves finding the final temperature and amount of steam that condenses when 100 g of steam at 100 degrees Celsius is passed into a calorimeter containing 200 g of water and 20 g of ice at 0 degrees Celsius. The equation used is Q lost = Q gained and the final temperature can be between 0 and 100 degrees Celsius. The heat required to condense the steam, melt the ice, and heat the water is calculated. The final temperature is then determined using this information.
  • #1
lollypop
33
0
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. :confused:
 
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  • #2
lollypop said:
when they say that the can , the water and the ice are in thermal equilibrium at atmospheric pressure, what would be their temperatures??
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
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.
i used Qcan+(Qice+mL)+Qwater = Qlead
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
lollypop said:
i used Qcan+(Qice+mL)+Qwater = Qlead
mc(T-0)+mc(T-0)+mL+mc(T-0)= mc(T-255), using 0 for what was in the can, is this right?
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
doc al:
thank you for ur fast reply, yes i got it right this time, the final temp would be 21.4 :)
 
  • #6
Excellent!
 
  • #7
I have a similar problem

I'm working on a similar 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?

:confused:
 
  • #8
mithril said:
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?
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. :smile:
 
  • #9
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. exchange 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
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
Welcome to PF, Electro!
Electro said:
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.
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: [itex]Q_{steam} = m_{steam}L_v[/itex].
Then find out how much heat is required to (1) melt the ice [itex]Q_1 = m_{ice}L_f[/itex] and then (2) heat the water and calorimeter to 100 C [itex]Q_2 = m_{water+calorimeter}c (100)[/itex]. Crank out [itex]Q_1 + Q_2[/itex] and compare it to [itex]Q_{steam}[/itex].

Let me know what you discover.
 
  • #12
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.?
Thank you for your help
 
  • #13
Electro said:
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.
Careful! I said compare those energies, not set them equal. When you compare [itex]Q_1 + Q_2[/itex] with [itex]Q_{steam}[/itex] you of course find that they are not equal! [itex]Q_{steam}[/itex] 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 [itex]Q_1 + Q_2 = m_{steam-condensed}L_v[/itex] and solve for the amount of steam that condenses.
Maybe the ice is at 0 and the water + calorimeter are at another temp.?
Thermal equilibrium requires that the final temp of everything be the same.

Let me know if my explanation makes sense.
 
  • #14
Right Doc. :smile: 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).
Thank you for your help.
 

1. What is a copper calorimeter and how does it work?

A copper calorimeter is a device used to measure the heat released or absorbed during a chemical reaction. It works by insulating a reaction vessel with a copper container and measuring the change in temperature of the surrounding water.

2. What are the advantages of using a copper calorimeter?

Copper is a good conductor of heat and has a high specific heat capacity, making it ideal for accurately measuring heat changes. Additionally, copper is relatively inert and does not react with most substances, making it a suitable material for use in chemical reactions.

3. How do I set up and use a copper calorimeter?

To set up a copper calorimeter, fill the copper container with a known amount of water and record its initial temperature. Then, add the reactants to the reaction vessel and measure the change in temperature of the water over time. This change can be used to calculate the heat released or absorbed during the reaction.

4. What are some common sources of error when using a copper calorimeter?

Some common sources of error when using a copper calorimeter include heat loss to the surroundings, incomplete mixing of the reactants, and heat loss due to evaporation of the water. It is important to control these factors as much as possible to obtain accurate results.

5. Can a copper calorimeter be used for all types of chemical reactions?

Copper calorimeters are suitable for a wide range of chemical reactions, including exothermic and endothermic reactions. However, they may not be suitable for reactions that produce corrosive substances or high temperatures, as these can damage the copper container.

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