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"#?
1 Answer
#q_(t ot) = 2._(067) xx 10^1# #"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))# , the heat capacity at constant pressure for liquid water, is still#"4.184 J/g"^@ "C"# at this#T# and#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).
#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"#
#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"#
#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
