Equilibrium Temperature and composition

[RedBarchetta]

Homework Statement

Describe the composition and temperature of the equilibrium mixture after 1.0 kg of ice at -40*C is added to 1.0 kg of water at 5.0*C.

Homework Equations

[tex]
\begin{gathered}
Q = mc\Delta T \hfill \\
Q = mL \hfill \\
\end{gathered}
[/tex]

Book answer:1.2kg ice, 0.80 kg water, all at 0*C

The Attempt at a Solution

I'm not quite sure how to set this one up. I've tried:

[tex]
m_i c_i \Delta T_i + m_w c_w \Delta T_w = 0
[/tex]

With this I found the equilibrium temperature to be -9.8*C...

 

[horatio89]

To begin writing down the equations to yield the equilibrium temperature, the first thing one has to do is to determine the phase of matter the equilibrium point.

So, to start, you need to quantitatively compare the amount of heat required to raise the temperature of ice to 0 degC, heat absorbed from the water to decrease the temperature to 0 degC, and the heat required to freeze/melt all the water/ice. At which phase does the heat removed from the water= heat absorbed by the ice?

 

[thomate1]

I would approach this problem by first identifying the key variables and equations that are relevant to the problem. In this case, the variables are mass, specific heat capacity, and temperature, and the equations are the heat capacity equation (Q=mcΔT) and the latent heat equation (Q=mL).

Next, I would consider the physical properties of the substances involved. Ice has a specific heat capacity of 2.09 J/g°C and a latent heat of fusion of 334 J/g, while water has a specific heat capacity of 4.18 J/g°C. I would also note that the heat capacities and latent heats are different for the solid and liquid phases of water.

Using this information, I would set up an energy balance equation to determine the equilibrium temperature and composition of the mixture. This equation would take into account the heat gained by the ice and the water as they reach equilibrium, as well as the heat lost by the ice as it melts and the water as it cools.

The final result would show that after 1.0 kg of ice at -40°C is added to 1.0 kg of water at 5°C, the resulting equilibrium mixture would consist of 1.2 kg of ice and 0.8 kg of water, all at a temperature of 0°C. This is because the heat lost by the ice as it melts (334 J/g x 1.0 kg = 334 kJ) is equal to the heat gained by the water as it cools (4.18 J/g°C x 1.0 kg x 5°C = 20.9 kJ) and by the ice and water as they reach equilibrium (2.09 J/g°C x 1.2 kg x 40°C = 100.8 kJ and 4.18 J/g°C x 0.8 kg x 5°C = 16.7 kJ, respectively).

In conclusion, the final equilibrium mixture would consist of 1.2 kg of ice and 0.8 kg of water, all at a temperature of 0°C. This result is in line with the book answer provided.