Question

How much energy does it take to convert $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{kg}$of ice at $\mathbf{-}\mathbf{20}\mathbf{°}\mathbf{C}$ to steam at $\mathbf{250}\mathbf{°}\mathbf{.}\mathbf{C}$? Specific heat capacities: ice, $\mathbf{2}\mathbf{.}\mathbf{1}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}$liquid, $\mathbf{4}\mathbf{.}\mathbf{2}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}$; steam, $\mathbf{2}\mathbf{.}\mathbf{0}{\mathbf{Jg}}^{-1}{}^o\mathbf{C}^{-1}$; ${\mathbf{\Delta H}}_{\mathrm{vap}}\mathbf{=}\mathbf{40}\mathbf{.}\mathbf{7}\mathbf{kJ}\mathbf{/}\mathbf{mol}\mathbf{;}{\mathbf{\Delta H}}_{\mathrm{fus}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{01}\mathbf{kJ}\mathbf{/}\mathbf{mol}$.

Short answer

The energy required to convert$0.500\mathrm{kg}$of ice at$-{20.}^{\circ }\mathrm{C}$ to steam at${250}^{\circ }\mathrm{C}$is $1680\mathrm{kJ}$.

Step-by-step solution

The formula for the heat

The formula used to determine theheatis:

$\mathrm{q}=\mathrm{Specific}\mathrm{Heat}\mathrm{Capacity}\mathrm{Mass}\mathrm{of}\mathrm{Sample}\mathrm{Change}\mathrm{in}\mathrm{Temperature}$

$\mathrm{Specific}\mathrm{heat}\mathrm{capacity}=\frac{\text{Heatabsorb}}{\text{temperaturechange}}$

Here we will divide the heat into ${\mathrm{q}}_1,{\mathrm{q}}_2,{\mathrm{q}}_3,{\mathrm{q}}_4,\&{\mathrm{q}}_5$, where:

${\mathrm{q}}_1=$Heat of ice calculated at$-20{.}^{\circ }\mathrm{C}$todata-custom-editor="chemistry" $0^{\circ }\mathrm{C}$

${\mathrm{q}}_2=$Heat of ice calculated atdata-custom-editor="chemistry" $0^{\circ }\mathrm{C}$to water atdata-custom-editor="chemistry" $0^{\circ }\mathrm{C}$

$q_3=$Heat of water calculated atdata-custom-editor="chemistry" $0^{\circ }\mathrm{C}$ todata-custom-editor="chemistry" ${100}^{\circ }\mathrm{C}$

$q_4=$Heat of water calculated atdata-custom-editor="chemistry" ${100}^{\circ }\mathrm{C}$to steam at${100}^{\circ }\mathrm{C}$

$q_5=$ Heat of water calculated at ${100}^{\circ }\mathrm{C}$ to ${250}^{\circ }\mathrm{C}$

Calculating the heat of water at -20oC to 250oC .

Calculating the total heat ${\mathrm{q}}_{\mathrm{Total}}$ or energy required:

The molecular mass of the water is $18.02\mathrm{g}$

role="math" localid="1663774746988" $\begin{array}{c}{q}_1=\left(\frac{2.0\times 5\times {10}^2\times \left(0-(-{20}^{\circ }\right)}{1000}\right)\\ =20\mathrm{kJ}\\ {\mathrm{q}}_2=\left(5\times {10}^2\times \frac{1}{18.02}\times 6.02\right)\\ =167\mathrm{kJ}\end{array}$

$\begin{array}{l}{\mathrm{q}}_3=\left(\frac{4.2\times 5.0\times {10}^2\times (100-0)}{1000}\right)\\ =210\mathrm{kJ}\\ {\mathrm{q}}_4=\left(5\times {10}^2\times \frac{1}{18.02}\times 40.7\right)\\ =1130\mathrm{kJ}\end{array}$

$\begin{array}{c}{\mathrm{q}}_5=\left(\frac{2.0\times 5.0\times {10}^2\times (250-100)}{1000}\right)\\ =150\mathrm{kJ}\\ {\mathrm{q}}_{\mathrm{Total}}=20+167+210+113+150\\ =1680\mathrm{kJ}\end{array}$

The energy required to convert $0.500\mathrm{kg}$of ice at $-{20}^{\circ }\mathrm{C}$ to steam at${250}^{\circ }\mathrm{C}$is $1680\mathrm{kJ}$.