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

Question

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

1 Answer

Impact of this question

1978 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-27T01:11:13

$q_{t o t} = 2 ._{067} \times 10^{1}$ $q_(t ot) = 2._(067) xx 10^1$ $kJ$ $"kJ"$

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 $C_{P (l)}$ $C_(P(l))$ , the heat capacity at constant pressure for liquid water, is still ${4.184 J/g}^{\circ} C$ $"4.184 J/g"^@ "C"$ at this $T$ $T$ and $P$ $P$ .
Since we are at constant atmospheric pressure, $q = DeltaH$ . 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, $q_(vap) = n_w DeltaH_(vap)$ , where $n_w$ is the mols of water at the boiling point and $DeltaH_(vap)$ is $"40.67 kJ/mol"$ .

For steam, we have that $C_(P(g)) = "1.996 J/g"^@ "C"$ near $100^@ "C"$ .

The main steps are then:

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

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

$q_1 = m_wC_(P(l))DeltaT$

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

$=$ $"301.248 J"$

$2)$ Vaporizing the liquid

$q_2 = n_wDeltaH_(vap)$

$q_2 = (9 cancel("g H"_2"O") xx cancel"1 mol"/(18.015 cancel("g"))) xx ((40.67 cancel("kJ"))/cancel"mol" xx "1000 J"/cancel("1 kJ"))$

$=$ $"20318.07 J"$

$3)$ Heating the steam

$q_3 = m_wC_(P(g))DeltaT$

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

$=$ $"53.892 J"$

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

$color(blue)(q_(t ot)) = q_1 + q_2 + q_3$

$= "301.248 J" + "20318.07 J" + "53.892 J"$

$=$ $"20673.21 J"$

or about $color(blue)("20.67 kJ")$ . Keep in mind that as-written, you have only allowed yourself one significant figure...

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What total amount of heat input is required to heat "9 g"9 g"9 g" of water from a liquid at 92^@ "C"92∘C92^@ "C" to steam at 103^@ "C"103∘C103^@ "C"?

1 Answer

Related questions

Impact of this question

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