# Calorimeter with ice and water

• Vuldoraq
In summary, the copper calorimeter can hold 0.160 kg of water and 0.018 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead is dropped into the calorimeter, the final temperature will be Tf=28.3 degrees Celsius.
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?

Hi Vuldoraq,

What numbers did you use in your final equation to get 28.3 degrees?

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,

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

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

Hi everyone,

Please note I accidentally posted the same question twice. Sorry about that. I would delete one, but it seems I am unable.

Vuldoraq

Hey,

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

Vuldoraq

## 1. How does a calorimeter with ice and water work?

A calorimeter with ice and water works by measuring the temperature change that occurs when a substance is placed in it. The ice-water mixture acts as a heat sink, absorbing the heat from the substance being tested. By measuring the temperature change of the ice-water mixture, the heat released by the substance can be calculated.

## 2. What is the purpose of using ice and water in a calorimeter?

The purpose of using ice and water in a calorimeter is to create a constant and known heat sink. This allows for more accurate measurements of the heat released by the substance being tested. The ice and water also help maintain a constant temperature throughout the experiment.

## 3. How is the heat capacity of ice and water taken into account in a calorimeter?

The heat capacity of ice and water is taken into account by using the specific heat capacity values for both substances in the calculations. This accounts for the energy required to change the temperature of the ice and water, and allows for more accurate calculations of the heat released by the substance being tested.

## 4. Can a calorimeter with ice and water be used to measure the heat of reaction for any substance?

Yes, a calorimeter with ice and water can be used to measure the heat of reaction for any substance as long as the substance and the ice-water mixture are in a closed system. This ensures that all the heat released by the substance is absorbed by the ice-water mixture and can be accurately measured.

## 5. How can the results from a calorimeter with ice and water be used to calculate the heat of fusion or vaporization of a substance?

The results from a calorimeter with ice and water can be used to calculate the heat of fusion or vaporization of a substance by using the equation Q = m * ΔH, where Q is the heat released or absorbed, m is the mass of the substance, and ΔH is the heat of fusion or vaporization. By measuring the temperature change and knowing the mass of the substance, the heat of fusion or vaporization can be calculated.

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