Given a consistent source of heat, about how much longer would it take to boil 100 g of water at 100°C than to melt 100 g of ice at 0°C?
1 Answer
Here's what I got.
Explanation:
In order to be able to solve this problem, you need to know the values for water's enthalpy of fusion,
https://en.wikipedia.org/wiki/Enthalpy_of_fusion#Reference_values_of_common_substances
https://en.wikipedia.org/wiki/Enthalpy_of_vaporization#Other_common_substances
So, you now that
#color(purple)(|bar(ul(color(white)(a/a)color(black)(DeltaH_"fus" = "333.55 J g"^(-1))color(white)(a/a)|)))" "# and#" "color(purple)(|bar(ul(color(white)(a/a)color(black)(DeltaH_"vap" = "44.66 kJ g"^(-1))color(white)(a/a)|)))#
Now, the enthlapy of fusion tells you how much heat is required in order to convert
Similarly, the enthalpy of vaporization tells you how much heat is required in order to convert
Now, take a look at the two enthalpies. Notice that you need significantly more heat in order to vaporize
Right from the start, this should tell you that a consistent source of heat, i.e. a source of constant power, would require more time to vaporize the water than to melt the ice.
Use the two enthalpies as conversion factors to figure out how much heat would be needed for your two
#100color(red)(cancel(color(black)("g"))) * overbrace("333.55 J"/(1color(red)(cancel(color(black)("g")))))^(color(purple)(DeltaH_"fus")) = "33,355 J"#
#100color(red)(cancel(color(black)("g"))) * overbrace("44,660 J"/(1color(red)(cancel(color(black)("g")))))^(color(brown)(DeltaH_"vap")) = "4,466,000 J"#
I converted the enthalpy of vaporization from kilojoules to joules by using
#"1 kJ" = 10^3"J"#
Now, let's assume that you have two source of equal power. Power, which tells you the rate of energy transfer per unit of time, is measured in watts, which are equivalent to one joule per second
#"1 W" = "1 J s"^(-1)#
For
#"33,355" color(red)(cancel(color(black)("J"))) * "1 s"/(1color(red)(cancel(color(black)("J")))) = "33,355 s"#
Similarly, the time needed for the water to vaporize will be equal to
#"4,466,000" color(red)(cancel(color(black)("J"))) * "1 s"/(1color(red)(cancel(color(black)("J")))) = "4,466,000 s"#
You can thus say that
#"time vaporize"/"time melt" = ("4,466,000" color(red)(cancel(color(black)("s"))))/("33,355"color(red)(cancel(color(black)("s")))) = color(green)(|bar(ul(color(white)(a/a)"134 s"color(white)(a/a)|)))#
In other words, when sources of equal power are used, it would take approximately
I'll leave the answer rounded to three sig figs.