Question

A $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{-}\mathbf{g}$sample of ice at $\mathbf{-}\mathbf{10}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$  is mixed with $\mathbf{100}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{g}$of water at $\mathbf{80}\mathbf{.}{\mathbf{0}}^o\mathbf{C}$. Calculate the final temperature of the mixture assuming no heat loss to the surroundings. The heat capacities of ${\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{s}\mathbf{\right)}$and ${\mathbf{H}}_2\mathbf{O}\mathbf{\left(}\mathbf{l}\mathbf{\right)}$ are $\mathbf{2}\mathbf{.}\mathbf{08}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}$ and $\mathbf{4}\mathbf{.}\mathbf{18}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}$ respectively, and the enthalpy of fusion for ice is $\mathbf{6}\mathbf{.}\mathbf{01}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.

Short answer

The final temperature of the mixture, assuming no heat is lost to the surroundings, is $\mathrm{T}={52.5}^{\circ }\mathrm{C}$.

Step-by-step solution

Determine the energy gained and released.

The energies gained by the mixture:

$\begin{array}{c}{\mathrm{Q}}_1=20\mathrm{g}\times 2.08{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}\times (0^{\circ }\mathrm{C}-(-{10.0}^{\circ }\mathrm{C}))\\ =416\mathrm{J}\\ {\mathrm{Q}}_2={\mathrm{\Delta H}}_{\mathrm{fus}}\times {\mathrm{M}}_{{\mathrm{rH}}_2\mathrm{O}}\\ =6010{\mathrm{Jmol}}^{-1}\times \frac{20.0\mathrm{g}}{18.02{\mathrm{gmol}}^{-1}}\\ =6672.1\mathrm{J}\end{array}$

$\begin{array}{c}{\mathrm{Q}}_3=20\mathrm{g}\times 4.18{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}\times (\mathrm{T}-0^{\circ })\mathrm{C}\\ =(83.6\times \mathrm{T})\mathrm{J}\end{array}$

The energy released by the mixture:

$\begin{array}{l}{\mathrm{Q}}_4=100\mathrm{g}\times 4.18{{\mathrm{g}}^{-1}}^{\circ }{\mathrm{C}}^{-1}\times {(80-\mathrm{T})}^{\circ }\mathrm{C}\\ =(33440-418\times \mathrm{T})\mathrm{J}\end{array}$

Determine the final temperature of the mixture.

It is given that no heat is lostto the surroundings.

$\begin{array}{l}\Rightarrow {\mathrm{Q}}_1+{\mathrm{Q}}_2+{\mathrm{Q}}_3={\mathrm{Q}}_4\\ \Rightarrow 416+6672.1+83.6\mathrm{T}=33440-418\mathrm{T}\\ \Rightarrow 334.4\mathrm{T}=26351.9\\ \Rightarrow \mathrm{T}={52.4}^{\circ }\mathrm{C}\end{array}$

Hence, the final temperature of the mixture is $\mathrm{T}={52.5}^{\circ }\mathrm{C}$.