# What total amount of heat input is required to heat "9 g" of water from a liquid at 92^@ "C" to steam at 103^@ "C"?

Jul 27, 2017

${q}_{t o t} = {2.}_{067} \times {10}^{1}$ $\text{kJ}$

where subscripts are past the last significant digit (how sad!).

Here, we heat hot water to boil, and then heat it a bit further.

• We assume that ${C}_{P \left(l\right)}$, the heat capacity at constant pressure for liquid water, is still $\text{4.184 J/g"^@ "C}$ at this $T$ and $P$.
• Since we are at constant atmospheric pressure, $q = \Delta H$. At constant temperature, the heat required is a constant dependent on the amount of substance available, and is related to the enthalpy of the vaporization process.

At the boiling point, ${q}_{v a p} = {n}_{w} \Delta {H}_{v a p}$, where ${n}_{w}$ is the mols of water at the boiling point and $\Delta {H}_{v a p}$ is $\text{40.67 kJ/mol}$.

• For steam, we have that ${C}_{P \left(g\right)} = \text{1.996 J/g"^@ "C}$ near ${100}^{\circ} \text{C}$.

The main steps are then:

$\underbrace{\text{H"_2"O"(l))_(92^@ "C") stackrel(("Part 1"))stackrel(q = m_wC_(P(l))DeltaT" ")(->) stackrel(("Part 2"))(overbrace("boiling point")^(q = n_wDeltaH_(vap))) stackrel(("Part 3"))stackrel(q = m_wC_(P(g))DeltaT" ")(->) underbrace("H"_2"O"(g))_(103^@ "C}}$

That is, one heats to the boiling point, then boils, then heats further. We assume constant mass throughout (i.e. closed system).

1) Heating the liquid

${q}_{1} = {m}_{w} {C}_{P \left(l\right)} \Delta T$

q_1 = "9 g" xx "4.184 J/g"^@ "C" xx (100^@ "C" - 92^@ "C")

$=$ $\text{301.248 J}$

2) Vaporizing the liquid

${q}_{2} = {n}_{w} \Delta {H}_{v a p}$

${q}_{2} = \left(9 \cancel{\text{g H"_2"O") xx cancel"1 mol"/(18.015 cancel("g"))) xx ((40.67 cancel("kJ"))/cancel"mol" xx "1000 J"/cancel("1 kJ}}\right)$

$=$ $\text{20318.07 J}$

3) Heating the steam

${q}_{3} = {m}_{w} {C}_{P \left(g\right)} \Delta T$

q_3 = "9 g" xx "1.996 J/g"^@ "C" xx (103^@ "C" - 100^@ "C")

$=$ $\text{53.892 J}$

Apparently, we only have one significant figure...?

$\textcolor{b l u e}{{q}_{t o t}} = {q}_{1} + {q}_{2} + {q}_{3}$

$= \text{301.248 J" + "20318.07 J" + "53.892 J}$

$=$ $\text{20673.21 J}$

or about $\textcolor{b l u e}{\text{20.67 kJ}}$. Keep in mind that as-written, you have only allowed yourself one significant figure...