# 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?

Mar 26, 2016

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, $\Delta {H}_{\text{fus}}$, and enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, which you can find here

https://en.wikipedia.org/wiki/Enthalpy_of_fusionReference_values_of_common_substances

https://en.wikipedia.org/wiki/Enthalpy_of_vaporizationOther_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 $\text{1 g}$ of ice at ${0}^{\circ} \text{C}$ to liquid water at ${0}^{\circ} \text{C}$, i.e. in order for $\text{1 g}$ of ice to undergo a solid $\to$ liquid phase change.

Similarly, the enthalpy of vaporization tells you how much heat is required in order to convert $\text{1 g}$ of liquid water at ${100}^{\circ} \text{C}$ to vapor at ${100}^{\circ} \text{C}$, i.e. in order for $\text{1 g}$ of water to undergo a liquid $\to$ solid phase change.

Now, take a look at the two enthalpies. Notice that you need significantly more heat in order to vaporize $\text{1 g}$ of water at its boiling point than to melt $\text{1 g}$ of ice at its melting point.

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 $\text{100-g}$ samples.

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

$\text{1 kJ" = 10^3"J}$

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

${\text{1 W" = "1 J s}}^{- 1}$

For $\text{1-W}$ power sources, the time needed for the ice to melt will be equal to

$\text{33,355" color(red)(cancel(color(black)("J"))) * "1 s"/(1color(red)(cancel(color(black)("J")))) = "33,355 s}$

Similarly, the time needed for the water to vaporize will be equal to

$\text{4,466,000" color(red)(cancel(color(black)("J"))) * "1 s"/(1color(red)(cancel(color(black)("J")))) = "4,466,000 s}$

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 $\text{134 s}$ longer for water to vaporize than for ice to melt, provided that the two samples have equal masses.

I'll leave the answer rounded to three sig figs.