# Reversible heat exchange between water and ice

• Soren4
In summary: One thing, I would be careful about writing ##\Delta m_I## to mean the change in mass of the ice. It could be interpreted to mean the mass of ice that melts, and that's not the same thing.In summary, the conversation discusses a scenario where a mass of ice and a mass of water are placed in an adiabatic container at different temperatures. The goal is to determine the total mass of water present at equilibrium and the change in thermodynamic entropy of the universe in the process. The conversation also asks for the total mass of water present at equilibrium in a case where the process is reversible and the equilibrium temperature is 0°C. The solution involves using the equations ΔUsystem
Soren4

## Homework Statement

In an adiabatic container are placed , in rapid succession , a mass of ice , ##m_I= 2 kg## , at temperature ##T_I = -10 ◦C## and a mass of water , ##m_W = 1 kg## , at the temperature ##T_W = + 20 ◦C## . Determine :
a) the total mass of water present in the container at equilibrium ;
b) the variation of thermodynamic entropy of the universe in the process .

Also determine :

c ) the total mass of water present at equilibrium in the case in which the equilibrium condition is achieved through a series of infinitesimal and reversible heat exchanges, knowing that the equilibrium temperature is ##T_0 = 0 ◦C## .

## Homework Equations

##\Delta U_{system}=0## for isolated system, ##\Delta S_{system}=0## for reversible adiabatic

## The Attempt at a Solution

I'm ok with point a) and b), I just wanted to be sure about one thing. In this case the heat given by water to go to ##0 °C## is bigger than the heat absorbed by ice to go to the same temperature, but not enough to change all the ice in water. Is it correct to think that, once water reaches ##0 °C## (end some ice melted) then there would be no heat exchange any more?

The main doubt is about point c). I intrerpreted the question in terms of entropy, that is, the entropy of the system (which is isolated) does not change

$$\Delta S= m_I c_I ln(\frac{T_0}{T_I}) +m_W c_W ln(\frac{T_0}{T_W}) + \frac{\Delta m_{I} \lambda_I}{T_{0}}$$

Where ##\Delta m_{I}## is the mass of ice melted and ## \lambda_I## is the entalpy of melting ice. The only thing I'm not ok with is that, with this method I assumed (again) that water do not freeze, so that the situation is the same as the one of point a) and b) but the condision is not ##\Delta U_{system}=0## but ##\Delta S_{system}=0##.

Is it correct to do so and is there a way to be sure that water cannot freeze in this situation?

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I don't understand what is being asked in part c, but please show us your work for parts a and b.

Chet

Soren4
Chestermiller said:
I don't understand what is being asked in part c, but please show us your work for parts a and b.

Chet
Thanks for the answer, part c) is the same of a) (find the mass of the water present in Equilibrium conditions) but in the hypotesis of reversible process.

For part a) I did the following
Since the ##|m_W c_W (0°C-T_W)|> m_I c_I (0°-T_I)## ice melts but, since ##|m_W c_W (0°C-T_W)| - m_I c_I (0°-T_I) < m_I \lambda_I##, just some part of the ice ##\Delta m_I## becomes water and final temperature must be ##0°C##. So I wrote ##\Delta U_{system}=0## to calculate ##\Delta m_I##

$$m_W c_W (0°C-T_W) + m_I c_I (0°C-T_I)+\Delta m_I \lambda_I=0$$

For part b) I calculated entropy as

$$\Delta S_{system}= m_I c_I ln(\frac{0°C+273.15}{T_I}) +m_W c_W ln(\frac{0°C+273.15}{T_W}) + \frac{\Delta m_{I} \lambda_I}{0°C+273.15}$$

I wonder if water really do not freeze and if, for point c), is correct to impose ##\Delta S_{system}=0##

Soren4 said:
Thanks for the answer, part c) is the same of a) (find the mass of the water present in Equilibrium conditions) but in the hypotesis of reversible process.

For part a) I did the following
Since the ##|m_W c_W (0°C-T_W)|> m_I c_I (0°-T_I)## ice melts but, since ##|m_W c_W (0°C-T_W)| - m_I c_I (0°-T_I) < m_I \lambda_I##, just some part of the ice ##\Delta m_I## becomes water and final temperature must be ##0°C##. So I wrote ##\Delta U_{system}=0## to calculate ##\Delta m_I##

$$m_W c_W (0°C-T_W) + m_I c_I (0°C-T_I)+\Delta m_I \lambda_I=0$$

For part b) I calculated entropy as

$$\Delta S_{system}= m_I c_I ln(\frac{0°C+273.15}{T_I}) +m_W c_W ln(\frac{0°C+273.15}{T_W}) + \frac{\Delta m_{I} \lambda_I}{0°C+273.15}$$

I wonder if water really do not freeze and if, for point c), is correct to impose ##\Delta S_{system}=0##
I agree with your analyses of parts a and b. Nice job.

I still don't understand the question in part c. Is that the exact wording of the question?

Soren4
Chestermiller said:
I agree with your analyses of parts a and b. Nice job.

I still don't understand the question in part c. Is that the exact wording of the question?

Thanks for the reply! I'm a little in late! Yes I translated literaly.. I'm quite sure it is asking to consider the same process of point a) and b) but in the case that process is reversible (in fact the heat exchanges are infinitesimal) and calculate the mass of water present at equilibrium.

What I'm less sure about is if it is correct - under the hypotesis of reversibility of the process - to proceed as I did and impose ##\Delta S_{system}=0## to get ##\Delta m_I## (and therefore the mass of water present at equilibrium), that is

$$m_I c_I \mathrm{ln}(\frac{273.15 K}{T_I}) +m_W c_W \mathrm{ln}(\frac{273.15 K}{T_W}) + \frac{\Delta m_{I} \lambda_I}{273.15K}=0$$
Is this the correct approach?

Soren4 said:
Thanks for the reply! I'm a little in late! Yes I translated literaly.. I'm quite sure it is asking to consider the same process of point a) and b) but in the case that process is reversible (in fact the heat exchanges are infinitesimal) and calculate the mass of water present at equilibrium.

What I'm less sure about is if it is correct - under the hypotesis of reversibility of the process - to proceed as I did and impose ##\Delta S_{system}=0## to get ##\Delta m_I## (and therefore the mass of water present at equilibrium), that is

$$m_I c_I \mathrm{ln}(\frac{273.15 K}{T_I}) +m_W c_W \mathrm{ln}(\frac{273.15 K}{T_W}) + \frac{\Delta m_{I} \lambda_I}{273.15K}=0$$
Is this the correct approach?
I guess. It seems to be the only thing that kind of makes sense.

Soren4
For part c) one way to imagine how this reversible heat transfer might work would be to separate the water from the ice and run a small carnot engine between the water (as the hot reservoir) and the ice (as the cold reservoir) for an infinite number of cycles until the temperatures of water and ice were the same (each cycle drawing an arbitrarily small amount of heat from the water so as to create an arbitrarily small temperature change in the reservoirs each cycle). The work output would be kept separate and apart from the system (ie. separate from the reservoirs + engine).

Since the change in entopy of the system is 0, you can determine the heat flow into the ice by using an entropy approach.

You need to break it down into two parts. First determine the temperature of the water when the ice is brought up to 273K using entropy (##\Delta S = 0##).

$$\Delta S_{I} = \int_{263}^{273}m_Ic_IdT/T = m_Ic_I\ln\frac{273}{263} = - \Delta S_{w} = -\int_{293}^{T_{i0}}m_wc_wdT/T = m_wc_w\ln\frac{293}{T_{i0}}$$

Then use the same method to determine how much heat is delivered to the ice at 273K to bring the water down to 273K from ##T_{i0}##.

AM

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## What is reversible heat exchange between water and ice?

Reversible heat exchange between water and ice is a process in which heat energy is transferred between water and ice, causing them to change temperature without changing phase.

## What factors affect reversible heat exchange between water and ice?

The factors that affect reversible heat exchange between water and ice include the initial temperature of the water and ice, the heat transfer coefficient, and the amount of time allowed for the exchange.

## What is the purpose of studying reversible heat exchange between water and ice?

Studying reversible heat exchange between water and ice can help us understand the thermodynamics of heat transfer and phase changes, as well as the behaviors of water and ice in different environments.

## How is reversible heat exchange between water and ice different from irreversible heat exchange?

Reversible heat exchange between water and ice involves a transfer of heat energy in both directions, while irreversible heat exchange only involves a one-way transfer. Additionally, reversible heat exchange does not result in a change in phase, while irreversible heat exchange may cause a change in phase.

## What applications does reversible heat exchange between water and ice have?

Reversible heat exchange between water and ice has many practical applications, such as in refrigeration systems, thermal storage systems, and cryotherapy treatments. It is also important in understanding and predicting weather patterns and climate changes.

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