- #1

lorenz0

- 148

- 28

- Homework Statement
- In a box containing ##m_{ice}=0.42kg## of ice at a temperature ##T_{ice}=-15°C##, ##m_{w}=0.16kg## of water at a temperature ##T_w=12°C## are added.

Ignore all dispersions of heat in the environment.

Find the equilibrium temperature and how much ice and how much water there is in the equilibrium state.

- Relevant Equations
- ##\sum \Delta Q=0##

If there weren't phase changes occurring I know that the temperature equilibrium would be ##T_e=\frac{m_{ice}c_{ice}T_{ice}+m_{w}c_{w}T_{w}}{m_{ice}c_{ice}+m_{w}c_{w}}##.

Now, by repeating the reasoning to get the above formula (##\sum \Delta Q=0##) and adding the phase changes of the water freezing I get ##T_e=\frac{ m_{ice}c_{ice}T_{ice}+m_{w}c_{w}T_{w}-\delta m_w L_c }{m_{ice}c_{ice}+m_{w}c_{w}}##, where ##L_c## is the latent heat of fusion and condensation, respectively, but how do I find how much (##\delta m_w##) of the water freezes?

And, is my reasoning correct in general? I would like to understand this process in general. Thanks

Now, by repeating the reasoning to get the above formula (##\sum \Delta Q=0##) and adding the phase changes of the water freezing I get ##T_e=\frac{ m_{ice}c_{ice}T_{ice}+m_{w}c_{w}T_{w}-\delta m_w L_c }{m_{ice}c_{ice}+m_{w}c_{w}}##, where ##L_c## is the latent heat of fusion and condensation, respectively, but how do I find how much (##\delta m_w##) of the water freezes?

And, is my reasoning correct in general? I would like to understand this process in general. Thanks

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