- #1

Soren4

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## 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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