Question

You have a 1.00-mole sample of water at-30.^oC, and you heat it until you have gaseous water at 140.^oC. Calculate qfor the entire process. Use the following data:

Specific heat capacity of ice=2.03 J ^oC^-1g^-1

Specific heat capacity of water=4.18 J ^oC^-1g^-1

Specific heat capacity of steam=2.02 J ^oC^-1g^-1

$$\begin{gathered} {\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}{\mathit{H}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{01}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{(}\mathit{a}\mathit{t}\mathbf{}{\mathbf{0}}^{\mathbf{\circ }}\mathit{C}\mathbf{)} \\ {\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{o}\mathbf{r}\mathbf{i}\mathbf{z}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathit{k}\mathit{J}\mathbf{/}\mathit{m}\mathit{o}\mathit{l}\mathbf{(}\mathit{a}\mathit{t}\mathbf{}{\mathbf{100}}^{\mathbf{\circ }}\mathit{C}\mathbf{)} \end{gathered}$$

Short answer

The total heat for the whole process is 56.8 kJ.

Step-by-step solution

Calculation of heat

Mass of ice = 1 mol = 18.0 g

The change in temperature can be broken into many steps,

Ice at -30^oC to ice at 0^oC

$$\begin{gathered} \mathit{q}\mathbf{=}\mathit{m}\mathit{C}\mathbf{∆}\mathit{T} \\ \mathit{q}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathit{g}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{303}\mathit{J}{\mathit{g}}^{\mathbf{-}\mathbf{1}}{}^{\mathbf{\circ }}\mathit{C}\mathbf{\times }\left(0^{\circ }C-\left({30}^{\circ }C\right)\right) \\ \mathit{q}\mathbf{=}\mathbf{1092}\mathbf{.}\mathbf{2}\mathit{J}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{10}\mathit{k}\mathit{J} \end{gathered}$$

Ice at 0^oC to water at 0^oC

$$\begin{gathered} \mathit{q}\mathbf{=}\mathbf{∆}{\mathit{H}}_{\mathbf{f}\mathbf{u}\mathbf{s}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{\times }\mathit{m}\mathit{o}\mathit{l} \\ \mathit{q}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathit{k}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{0}\mathit{m}\mathit{o}\mathit{l} \\ \mathit{q}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{02}\mathit{k}\mathit{J} \end{gathered}$$

Water at 0^oC to water at 100^oC

$$\begin{gathered} \begin{array}{l}\mathbf{q}\mathbf{=}\mathbf{m}\mathbf{\left(}\mathbf{0}\mathbf{\right|}\mathbf{T}\\ \mathbf{q}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathbf{g}\mathbf{\times }\mathbf{4}\mathbf{.}\mathbf{18}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}\mathbf{C}\mathbf{\times }\left(110C-0C\right)\end{array} \\ \mathit{q}\mathbf{=}\mathbf{7524}\mathit{J}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{52}\mathit{k}\mathit{J} \end{gathered}$$

Water at 100^oC to steam at 100^oC

$$\begin{gathered} \mathit{q}\mathbf{=}\mathbf{∆}{\mathit{H}}_{\mathbf{v}\mathbf{a}\mathbf{p}\mathbf{o}\mathbf{r}\mathbf{i}\mathbf{z}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\mathbf{\times }\mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s} \\ \mathit{q}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathit{k}\mathit{J}\mathit{m}\mathit{o}{\mathit{l}}^{\mathbf{-}\mathbf{1}}\mathbf{\times }\mathbf{1}\mathbf{.}\mathbf{0}\mathit{m}\mathit{o}\mathit{l} \\ \mathit{q}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathit{k}\mathit{J} \end{gathered}$$

Steam at 100oC to steam at 140oC

$$\begin{gathered} \mathit{q}\mathbf{=}\mathit{m}\mathit{C}\mathbf{∆}\mathit{T} \\ \mathit{q}\mathbf{=}\mathbf{18}\mathbf{.}\mathbf{0}\mathit{g}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{02}\mathit{J}{\mathit{g}}^{\mathbf{-}\mathbf{1}\mathbf{\circ }}\mathit{C}\mathbf{\times }\mathbf{(}{\mathbf{140}}^{\mathbf{\circ }}\mathit{C}\mathbf{-}{\mathbf{100}}^{\mathbf{\circ }}\mathit{C}\mathbf{)} \\ \mathit{q}\mathbf{=}\mathbf{1454}\mathbf{.}\mathbf{4}\mathit{J}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{4}\mathit{k}\mathit{J} \end{gathered}$$

Determination of q for whole processa

The q for the whole process is,

$$\begin{gathered} \mathit{q}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{10}\mathit{k}\mathit{J}\mathbf{+}\mathbf{6}\mathbf{.}\mathbf{02}\mathit{k}\mathit{J}\mathbf{+}\mathbf{7}\mathbf{.}\mathbf{52}\mathit{k}\mathit{J}\mathbf{+}\mathbf{40}\mathbf{.}\mathbf{7}\mathit{k}\mathit{J}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{45}\mathit{k}\mathit{J} \\ \mathit{q}\mathbf{=}\mathbf{56}\mathbf{.}\mathbf{8}\mathit{k}\mathit{J} \end{gathered}$$

Thus, the total heat for the whole process is 56.8 kJ.