Calculate Final Temperature from Phase Change: Cooling Soda with Ice Cubes

The ice cubes are at the melting temperature of $\mathrm{0º}\text{C}$. Heat is transferred from the soda to the ice for melting. Melting of ice occurs in two steps: first the phase change occurs and solid (ice) transforms into liquid water at the melting temperature, then the temperature of this water rises. Melting yields water at $\mathrm{0º}\text{C}$, so more heat is transferred from the soda to this water until the water plus soda system reaches thermal equilibrium,

The heat transferred to the ice is $Q_{\text{ice}}=m_{\text{ice}}{L}_{\text{f}}+m_{\text{ice}}{c}_{\text{W}}(T_{\text{f}}-\mathrm{0º}\text{C})$. The heat given off by the soda is $Q_{\text{soda}}=m_{\text{soda}}{c}_{\text{W}}(T_{\text{f}}-\text{20º}\text{C})$. Since no heat is lost, $Q_{\text{ice}}=-Q_{\text{soda}}$, so that

Bring all terms involving $T_{\text{f}}$ on the left-hand-side and all other terms on the right-hand-side. Solve for the unknown quantity $T_{\text{f}}$:

1. Identify the known quantities. The mass of ice is and the mass of soda is $m_{\text{soda}}=0\text{.}\text{25 }\mathrm{<mtext>kg</mtext>}$.
2. Calculate the terms in the numerator:
   $$m_{\text{soda}}{c}_{\mathrm{W}}\left(\text{20º}\text{C}\right)=\left(0\text{.25 kg}\right)\left(\text{4186 J/kg}\cdot º\text{C}\right)\left(\text{20º}\text{C}\right)=\text{20,930 J}$$

   14.22

   and

   $$m_{\text{ice}}{L}_{\text{f}}=\left(0\text{.}\text{018 }\mathrm{<mtext>kg</mtext>}\right)\left(\text{334,000}\text{ J/kg}\right)\text{=6012 }\mathrm{<mtext>J.</mtext>}$$

   14.23
3. Calculate the denominator:
   $$\left(m_{\text{soda}}+m_{\text{ice}}\right)c_{\text{W}}=\left(0\text{.}\text{25 }\text{kg + 0}\text{.}\text{018 kg}\right)\left(\text{4186 K/}(\text{kg}\cdot º\text{C}\right)\text{=1122 J/º}\text{C}\text{.}$$

   14.24
4. Calculate the final temperature:
   $$T_{\text{f}}=\frac{\text{20},\text{930 J}-\text{6012 J}}{\text{1122 J/º}\text{C}}=\text{13º}\text{C.}$$

   14.25

This example illustrates the enormous energies involved during a phase change. The mass of ice is about 7 percent the mass of water but leads to a noticeable change in the temperature of soda. Although we assumed that the ice was at the freezing temperature, this is incorrect: the typical temperature is $-\mathrm{6º}\text{C}$. However, this correction gives a final temperature that is essentially identical to the result we found. Can you explain why?