Is the answer to this question 221 J? If not, can someone explain it to me?
1 Answer
No, not even close.
Explanation:
!! VERY LONG ANSWER !!
The problem wants you to calculate the total change in enthalpy,
The first thing to do here is figure out the sign of
In other words, you're dealing with a series of endothermic processes. The change in enthalpy associated with an endothermic process is positive, so right from the start you know that the
Now, in order to be able to do the actual calculations, you need to know
#c_"ice" = "2.06 J g"^(-1)""^@"C"^(-1)#
#DeltaH_"fus" = "333.55 J g"^(-1)#
#c_"water" = "4.184 J g"^(-1)""^@"C"^(-1)#
#DeltaH_"vap" = "2259 J g"^(-1)#
#c_"steam" = "2.02 J g"^(-1)""^@"C"^(-1)#
So, break up your process into five parts
- Going from ice at
#-10.0^@"C"# to ice at#0^@"C"#
The heat required to heat your sample of ice from
#q_1 = m_"ice" * c_"ice" * DeltaT_"ice"#
Plug in your values to find
#q_1 = 21.5 color(red)(cancel(color(black)("g"))) * "2.06 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * [0 - (-10.0)]color(red)(cancel(color(black)(""^@"C")))#
#q_1 = "442.9 J"#
- Going from ice at
#0^@"C"# to liquid water at#0^@"C"#
This is where you encounter the first phase change. All the heat that you provide to the sample of ice at
Phase changes occur at constant temperature, which is why your phase diagram shows two horizontal lines
one is marked with
#DeltaH_"melt" -># here is where melting / freezing take placeone marked with
#DeltaH_"vap" -># here is where boiling / condensation take place
The heat required to promote a solid
#q_2 = m_"ice" * DeltaH_"fus"#
Plug in your values to find
#q_2 = 21.5 color(red)(cancel(color(black)("g"))) * "333.55 J" color(red)(cancel(color(black)("g"^(-1))))#
#q_2 = "7171.3 J"#
- Going from liquid water at
#0^@"C"# to liquid water at#100^@"C"#
The heat required to heat your sample from liquid water at
#q_3 = M_"water" * c_"water" * DeltaT_"water"#
Plug in your values to find
#q_3 = 21.5 color(red)(cancel(color(black)("g"))) * "4.184 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - 0)color(red)(cancel(color(black)(""^@"C")))#
#q_3 = "8995.6 J"#
- Going from liquid water at
#100^@"C"# to steam at#100^@"C"#
This is where you encounter the second phase change. All the heat you supply to the sample of liquid water will be used to help the water molecules completely overpower the hydrogen bonds and start going into the gaseous state.
The heat required to promote a liquid
#q_4 = m_"water" * DeltaH_"vap"#
Plug in your values to find
#q_4 = 21.5 color(red)(cancel(color(black)("g"))) * "2259 J" color(red)(cancel(color(black)("g"^(-1))))#
#q_4 = "48568.5 J"#
- Going from steam at
#100^@"C"# to steam at#115.0^@"C"#
Finally, you can calculate the heat required to heat your sample of steam from
#q_5 = m_"steam" * c_"steam" * DeltaT_"steam"#
Plug in your values to find
#q_5 = 21.5 color(red)(cancel(color(black)("g"))) * "2.02 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (115.0 - 100)color(red)(cancel(color(black)(""^@"C")))#
#q_5 = "651.5 J"#
Now, the total heat required to get your sample from ice at
#q_"total" = q_1 + q_2 + q_3 + q_4 + q_5#
In your case, you have
#q_"total" = "442.9 J" + "7171.3 J" + "8995.6 J" + "48568.5 J" + "651.5 J"#
#q_"total" = "65829.8 J"#
The problem wants you to find the change in enthalpy per mole, os use water's molar mass to figure out how many moles of water you have in your sample
#21.5 color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "1.1934 moles H"_2"O"#
You can thus say that the heat required to convert one mole of ice at
#1 color(red)(cancel(color(black)("mole H"_2"O"))) * "65829.8 J"/(1.1934color(red)(cancel(color(black)("moles H"_2"O")))) = "55161.6 J"#
Expressed in kilojoules and rounded to three sig figs, the amoiunt of heat needed will be
#55161.6 color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "55.2 kJ"#
Since one mole of water requires
#color(green)(|bar(ul(color(white)(a/a)color(black)(DeltaH = +"55.2 kJ mol"^(-1))color(white)(a/a)|)))#