# Solving Final Temp of Ice & Water Mixture

• Arooj
In summary, the final temperature of the mixture of 100g of ice at 0°C and 500g of water at 100°C can be approximated to be around 70°C after accounting for the heat of fusion and using the equation Qw=-Qx to solve for the final temperature. This is because the ice undergoes a phase change and the hot water remains water as it cools to the final temperature.
Arooj

## Homework Statement

A 100g piece of ice at 0.0°C is dropped into 500g of water at 100°C. What is the final temperature of mixture.

## Homework Equations

Q=mL (latent heat)
Q=mcΔt
Qw=-Qx (energy absorbed by water is equal but opposite to that of substance inserted into water)
specific heat ice: 2.09 * 10^3 J/kg * °C
specific heat water: 4.186 * 10^3 J/kg * °C
specific heat steam: 2.01 * 10^3 J/kg * °C
latent heat of fusion: 3.33 * 10^5 J/kg
latent heat of vaporization: 2.26 * 10^6 J/kg

## The Attempt at a Solution

Do you find the total heat including phase change heat for both substances, using final temperature as an unknown variable? If so:
Q = mcΔt + mL
Qw = (0.5 *4.186 * 10^3 *(Tf-100)) + 0.5 * 2.26 * 10^6
Qx= -((0.5 *4.186 * 10^3 *(Tf-0)) + (0.5 * 3.33 * 10^5))
I set both equal after simplifying, and obtained a final temperature of 264.14°C, which is too bizarre of an answer.

The water at 100C is still liquid, so the heat of vaporization is not needed.
However, the ice at 0C is solid, and the heat of fusion must be added to the ice to melt it.

Only the piece of ice undergoes a phase change. The hot water remains water as it cools to the final temperature. So, think about how you wrote Qw. Also, I think you used the wrong mass for the ice in your expression for Qx.

Thanks for the help. Not including the heat of vaporization, I got 28.08 C. I also did have a mistake in calculating the energy of the piece of ice where I put it the mass of the water instead. Is 28.08 a reasonable answer, or should it be closer to 100 C because the more massive object would have more of an influence on the temperature?

I get the final temperature to be quite a bit higher, in agreement with your intuition.

To get 28.08 I did
Qw=-Qx
(0.5 *4.186 * 10^3 *(Tf-100)) = -((0.1 *4.186 * 10^3 *(Tf-0)) + (0.1 * 3.33 * 10^5))
2093Tf - 209300 = -418.6Tf - 33300
2511.6Tf = 176000
Tf = 70.07 C

I think I have it, I might have made some original mistake in calculation.

That looks good.

As a rough approximation, ice at 0°C corresponds to (hypothetical) water at -80°C. So you heat 100g by 150° and cool five times this amount by 1/5 of the temperature difference. The .07 are the deviation from this approximation ;).

## 1. What is the process for solving the final temperature of an ice and water mixture?

The process involves using the principle of energy conservation, where the energy lost by the ice as it melts is equal to the energy gained by the water as it warms up. This can be represented by the equation micecice(Tf - Ti) = mwatercwater(Tf - Ti), where m is the mass and c is the specific heat capacity of the substance, Tf is the final temperature and Ti is the initial temperature.

## 2. What values do I need to know to solve for the final temperature?

You will need to know the mass and specific heat capacity of both the ice and water, as well as the initial temperatures of the two substances. These values can be found in a reference table or by measuring them directly.

## 3. Can this process be used for any ice and water mixture?

Yes, this process can be used for any mixture of ice and water as long as the ice is completely melted and mixed with the water. It is important to note that this process assumes that there is no heat loss to the surroundings.

## 4. What if there are other substances present in the mixture?

If there are other substances present in the mixture, their mass and specific heat capacity should also be taken into account in the equation. This will result in a more complex equation, but the principle of energy conservation still applies.

## 5. Is there a shortcut for solving the final temperature of an ice and water mixture?

Yes, there is a shortcut method called the "rule of mixtures" which can be used when the initial temperatures of the ice and water are the same. It involves finding the average specific heat capacity of the mixture and using it in the energy conservation equation. However, this method is only accurate for small temperature differences between the initial and final temperatures.

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