# 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*, *enthalpy of vaporization*,

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 *solid* *liquid* **phase change**.

Similarly, the enthalpy of vaporization tells you how much heat is required in order to convert *liquid* *solid* phase change.

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 *equal masses*.

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