# Thermal equilibrium for ice being added to liquid water

• Mangoes
In summary, the conversation discusses a problem where 0.05625 m^3 of H2O(s) is added to a thermally isolated system of 20 L of H2O (l) with initial temperatures of 0 C for the ice and 95 C for the liquid water. The question is to determine the final state and temperature of the system. The conversation includes calculations for the heat of fusion and specific heat of water, but the final result is not consistent with the initial temperatures. The conversation also explores the concept of thermal equilibrium and the maximum amount of energy that can be extracted from the liquid water.
Mangoes

## Homework Statement

0.05625 m^3 of H2O(s) is added to a thermally isolated system of 20 L of H2O (l). If the ice was initially at 0 C and the liquid water was initially at 95 C, what is the final state and temperature?

## The Attempt at a Solution

I missed class for this so I'm going by what I remember as my book doesn't exactly cover this from what I can tell.

I looked up the density for solid water at 0 C and converted the volume to mass, there are 51.4 kg H2O (s)

I looked up the density for liquid water at 95 C and converted the volume to mass, there are 19.2 kg H2O (l)

Ice is going to gain energy as it goes from a phase change of solid to liquid. The liquid mass that was once ice is then going to gain energy as it reaches thermal equilibrium with the liquid water originally at 95 C. The source of the heat is going to be the liquid water. Since heat in = heat out,

q (phase change) + q (temp. raise) = q (temp. lower)

I calculated q (phase change) by looking up the heat of fusion of H2O which is 334 J/g.
q (phase change) = (334 J/g)(51,400g)

The specific heat of water is (4.184 J/g),
q (temp. raise) = (4.184 J/g)(51,400g) Δt

Similarly,
q (temp. lower) = (4.184 J/g)(19,200g) Δt

Putting it all together:

(334 J/g)(51,400g) + (4.184 J/g)(51,400g) Δt = (4.184 J/g)(19,200g) Δt

But this gives a Δt of 127.4 K which doesn't make sense considering the initial temperatures. It's been a while since I've covered this topic and can't really tell what I'm doing wrong. Would anyone be able to point me in the right direction?

Not all of the ice (solid water) can melt; there's not enough heat which can be extracted from the liquid water to do so. You basically have a big glass of ice water here.

That's a first for any of these problems I've ever seen.

How would I go about finding out the maximum amount of energy that can be extracted from the liquid water?

I'd assume that if not all of the ice can be converted to liquid water, then only enough can be converted so that which has been converted to liquid reaches thermal equilibrium with the liquid water... but then the remaining ice remains there without any change in temperature? Seems unintuitive...

Remember, once the liquid water and the ice have reached the same temperature, heat transfer stops. There's nothing that says ice and water can't coexist at 0 degrees C.

Since the container is 'thermally isolated', no additional heat can enter from the surroundings to melt the ice. It's one reason why a thermos bottle or a dewar's flask is a handy device to have on a warm day, or even a tall glass with a lot of ice cubes.

From the information given, it appears that the final state and temperature will depend on the specific heat capacities and densities of both ice and liquid water. It is also important to consider the total mass of the system, as the addition of 0.05625 m^3 of ice will change the overall mass and therefore the specific heat capacity of the system. Additionally, the thermally isolated system will prevent any heat exchange with the surroundings, meaning the energy for the phase change and temperature changes must come from within the system itself. To accurately solve for the final state and temperature, it may be helpful to set up an equation with all of the relevant variables and solve for the unknowns. It may also be useful to double check the values used for the specific heat capacities and heat of fusion, as these can vary slightly depending on the source.

## 1. What is thermal equilibrium?

Thermal equilibrium is a state in which two objects or systems are at the same temperature, meaning there is no net flow of heat between them. This is achieved when the rate of heat transfer between the objects is equal, and the temperatures of both objects remain constant.

## 2. How does thermal equilibrium apply to ice being added to liquid water?

When ice is added to liquid water, the two substances will exchange heat until they reach thermal equilibrium. This means that the ice will melt and the water will cool down until they are both at the same temperature, which is 0 degrees Celsius for pure water and ice.

## 3. What factors affect the rate of thermal equilibrium between ice and liquid water?

The rate of thermal equilibrium between ice and liquid water is affected by factors such as the initial temperatures of the two substances, the amount of ice and liquid water present, and any external influences such as the surrounding temperature or pressure.

## 4. How does thermal equilibrium impact the properties of ice and liquid water?

During thermal equilibrium, the properties of ice and liquid water will be the same as they are at their respective temperatures. This means that the ice will have a solid state and the liquid water will have a liquid state, and their densities and other physical properties will remain constant.

## 5. Can thermal equilibrium be achieved between ice and liquid water at any temperature?

No, thermal equilibrium between ice and liquid water can only be achieved at 0 degrees Celsius. This is because this is the temperature at which the two substances can coexist in their solid and liquid states and still maintain the same temperature without any net flow of heat.

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