# Question cf9b5

Mar 4, 2015

You'd need $\text{286 kJ}$ to get from ice at $\text{-12"^@"C}$ to steam at $\text{121"^@"C}$.

Here are the constants that you must use

Heat of fusion of water: $\Delta {H}_{f}$ = $334$ $\text{J/g}$
Heat of vaporization of water: $\Delta {H}_{v}$ = $2257$ $\text{J/g}$
Specific heat of ice: $c$ = $2.09$ $\text{J}$/$\text{g"^@"C}$;
Specific heat of water: $c$ = $4.18$ $\text{J}$/$\text{g"^@"C}$;
Specific heat of steam: $c$ = $2.09$ $\text{J}$/$\text{g"^@"C}$;

Notice that you must go from ice to vapor, which means that you must go through two phase changes - solid to liquid and liquid to vapor.

The steps that describe your specific transition are

1. Going from ice at $\text{-12"^@"C}$ to ice at $\text{0"^@"C}$

q_1 = m * c_("ice") * DeltaT_1 = "93.0 g" * 2.09"J"/("g" * ^@"C") * (0 - (-12))^@"C"

${q}_{1} = \text{2332.44 J}$

2. Going from ice at $\text{0"^@"C}$ to water at $\text{0"^@"C}$

${q}_{2} = m \cdot \Delta {H}_{f} = \text{93.0 g" * 334"J"/"g" = "31062 J}$

3. Going from water at $\text{0"^@"C}$ to water at $\text{100"^@"C}$

${q}_{3} = m \cdot {c}_{\text{water}} \cdot \Delta {T}_{3}$

${q}_{3} = \text{93.0 g" * 4.18"J"/("g" * ^@"C") * (100 - 0)^@"C" = "38874 J}$

4. Going from water at $\text{100"^@"C}$ to steam at $\text{100"^@"C}$

${q}_{4} = m \cdot \Delta {H}_{v} = \text{93.0 g" * 2257"J"/"g" = "209901 J}$

5. Going from steam at $\text{100"^@"C}$ to steam at $\text{121"^@"C}$

${q}_{5} = m \cdot {c}_{\text{steam}} \cdot \Delta {T}_{5}$

${q}_{5} = \text{93.0 g" * 2.09"J"/("g" * ^@"C") * (121 - 100)^@"C" = "4081.77 J}$

The total amount of energy required to go from ice at $\text{-12"^@"C}$ to steam at $\text{121"^@"C}$ will be the sum of all the calculated energies

${q}_{\text{total}} = {q}_{1} + {q}_{2} + {q}_{3} + {q}_{4} + {q}_{5}$

q_("total") = "286251.21 J"

In kJ and rounded to three sig figs, the answer will be

q_("total") = "286.25121 kJ" = "286 kJ"#