# Copper calorimeter final temperature

• Vuldoraq
In summary, the copper calorimeter 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^{o}C is dropped into the calorimeter can, the final temperature will be 28.3^{o}C.
Vuldoraq

## Homework Statement

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$$^{o}$$C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

c(copper)=390 (J/k*kg)
c(water)=4190 (J/k*kg)
$$L_{f(water)}$$=334*10^3(J/Kg)

## Homework Equations

Q=m$$L_{f(water)}$$

Q=mc$$\Delta$$T

0=$$Q_{1}$$+$$Q_{2}$$+$$Q_{3}$$

## The Attempt at a Solution

I said that the heat lost by the lead is equal to the heat gained by the calorometer. So,

$$-mc(lead)\Delta T(lead)=mL_{f(water)}+mc(copper)\Delta T(cal)+mc(water)\Delta T(cal)$$Than I noted that $$\Delta T(lead)\ and \Delta T(cal)$$ both share the same final temperature, say Tf. Letting $$\Delta T=Tf-Ti$$ and subbing in the Ti (initial temp) values and then rearranging to get Tf on it's own,

$$-mc(lead)*(Tf-255)=mL_{f(water}+mc(copper)*(Tf-0)+mc(water)*(Tf-0)$$

Therefore,

$$Tf=24790/877=28.3^{o}C$$

Which is apparently wrong. Can anyone please tell me why?

Your equation is correct. Try crunching the numbers again. I haven't got my calculator, but if the numbers given are correct you should get the correct answer.

Did you put the correct masses in the correct places? I noticed you wrote "mLf(water)" where I would have written "ice."

I used the following numbers,

$$-97.5*T_{f}+24862.5J=72.45J+39T_{f}+745.82T_{f}$$

Where the 72.45J is the latent heat of fusion of the ice, which I now can see is wrong. It should be 6012J. I foolishly used the waters specific heat incorrectly.

Redoing the above gives,

$$-97.5*T_{f}+24862.5J=6012J+39T_{f}+745.82T_{f}$$

$$Therefore T_{f}=18850/882.32=21.4^{o}C$$

Thanks for pointing that out, does this look right to you?

Hi everyone,

Please note i accidentally posted the same question twice. Sorry.

Vuldoraq

Hi,

21 degrees C is the right answer. Thanks for all the help.

Vuldoraq

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

A copper calorimeter is a device used to measure changes in temperature during a chemical reaction or physical process. It works by trapping the reaction or process inside a copper container, which allows for accurate measurement of the temperature change.

## 2. How is the final temperature of a reaction or process determined using a copper calorimeter?

The final temperature is determined by measuring the temperature of the reactants or substances before they are combined in the calorimeter, and then again after the reaction or process is complete. The difference between the initial and final temperatures is the change in temperature, which can be used to calculate the final temperature.

## 3. Why is copper a common material used in calorimeters?

Copper is a good conductor of heat, meaning it allows heat to flow through it easily. This makes it an ideal material for use in calorimeters, as it allows for accurate temperature measurements. Copper is also relatively inexpensive and readily available, making it a practical choice for scientific experiments.

## 4. Is it necessary to use a copper calorimeter for all experiments?

No, a copper calorimeter is not necessary for all experiments. Other materials, such as aluminum or steel, can also be used as calorimeter containers. The choice of material may depend on the specific experiment and the desired level of accuracy.

## 5. What factors can affect the final temperature measurement in a copper calorimeter?

Several factors can affect the final temperature measurement, including the accuracy of the temperature probe, the amount and type of substance used, and the insulation of the calorimeter. It is important to control these factors to ensure accurate and reliable results.

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