Question

PLEASE HELP!! Thanks! How much heat (in kJ) is required to warm 13.0 g of ice, initially at -10.0 ∘C, to steam at 111.0 ∘C? The heat capacity of ice is 2.09 J/g⋅∘C and that of steam is 2.01 J/g⋅∘C, the heat of fusion for water is 6.02 kJ/mol, and the heat of vaporization for water is 40.7 kJ/mol.

Answer 1

Answer:

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is $\rm 4.182\; J \cdot mol^{-1}$, the melting point of water is $\rm 0\, ^{\circ} C$, and that the boiling point of water is $\rm 100 \,^{\circ} C$.

Explanation:

It takes five steps to convert 13.0 grams of $\rm \text{-}10.0\, ^{\circ}C$ ice to steam at $\rm 111.0\,^{\circ}C$.

• Step one: heat the 13.0 gram of ice from $\rm \text{-}10.0\, ^{\circ}C$ to $\rm 0\,^{\circ}C$. The change in temperature would be $\rm 10.0\,^{\circ}C$.
• Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.
• Step three: heat the 13.0 gram of water from$\rm 0\,^{\circ}C$ to $\rm 100\,^{\circ}C$. The change in temperature would be $\rm 100\,^{\circ}C$.
• Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.
• Step five: heat the 13.0 gram of steam from$\rm 100\,^{\circ}C$ to $\rm 111.0\,^{\circ}C$. The change in temperature would be $\rm 11.0\,^{\circ}C$.

Energy required for step one, three, and five

The following equation gives the amount of energy $Q$ required to raise the temperature of an object  by a $\Delta T$:

$Q = c \cdot m \cdot \Delta T$.

In this equation,

• $c$ is the specific heat of this substance,
• $m$ is the mass of the substance, and
• $\Delta T$ is the change in the temperature of the object.

Assume that there's no mass loss in this whole process. The value of $m$ would stay the same at $13.0\; \rm g$.

$\begin{aligned}& &&\text{Energy required for raising temperature} \cr &=&& c(\text{Ice}) \cdot m \cdot \Delta(\text{Ice}) \cr & && + c(\text{Water}) \cdot m \cdot \Delta(\text{Water})\cr & && + c(\text{Steam}) \cdot m \cdot \Delta(\text{Steam}) \cr & = && (2.09 \times 13.0 \times 10) \cr & && + (4.182 \times 13.0 \times 100) \cr & &&+ ( 2.01 \times 13.0 \times 10) \cr & = && 5969.6\;\rm J \cr & = && 5.969\; \rm kJ\end{aligned}$.

Energy required for step two and four

The equations for the energy of fusion and energy of vaporization are quite similar:

$E(\text{Fusion}) = n \cdot \Delta H_\text{Fusion}$.

$E(\text{Vaporization}) = n \cdot \Delta H_\text{Vaporization}$.

where $n$ is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

• H: 1.008,
• O: 15.999.

Hence the molar mass of water:

$M(\rm H_2O) = 2\times 1.008 + 15.999 = 18.015\; g \cdot mol^{-1}$.

Number of moles of $\rm H_2O$ molecules in $\rm 13.0\; g$:

$\displaystyle n = \frac{m}{M} \approx 0.721621\; \rm mol$.

$\begin{aligned}& &&\text{Energy required for phase changes} \cr &=&& n \cdot \Delta H_\text{Fusion} \cr & &&+n \cdot \Delta H_\text{Vaporization} \cr & = &&0.721621 \times 6.02 + 0.721621 \times 40.7 \cr & = &&33.7\; \rm kJ \end{aligned}$

Energy required for all five steps, combined

$5.969\; \rm kJ + 33.7\; \rm kJ \approx 39.7\; \rm kJ$.