# (full question below) "How much water would you expect to vaporize, assuming no water splashes out?"

##
**full question**

*A 50.0g piece of iron at 150°C is dropped into 20.0g* #H_2O(l)# *at 90°C in an open, thermally insulated container.*

How much water would you expect to vaporize, assuming no water splashes out?

*specific heats of iron and water: 0.45 and 4.21* #Jg^(-1)°C^(-1)# *respectively, and* #\DeltaH_(vap)=40.7kJ\timesmol^-1H_2O#

**express your answer using two significant figures**

**full question**

*A 50.0g piece of iron at 150°C is dropped into 20.0g* *at 90°C in an open, thermally insulated container.
How much water would you expect to vaporize, assuming no water splashes out?*

*specific heats of iron and water: 0.45 and 4.21* *respectively, and*

**express your answer using two significant figures**

##### 1 Answer

#### Answer:

#### Explanation:

The idea here is that the heat given off by the metal as it cools will be absorbed by the water.

If this heat exceeds the amount needed to increase the temperature of the water to

Now, you can assume that the equilibrium temperature of the metal + *liquid water* system will be

Your tool of choice here will be the equation

#color(blue)(ul(color(black)(q = m * c * DeltaT)))#

Here

#q# is heat absorbed / given off#m# is the mass of the sample#c# is the specific heat of the substance#DeltaT# is the change in temperature, defined as the difference between thefinal temperatureand theinitial temperature

For the water sample, you will have

#q_"water" = 20.0 color(red)(cancel(color(black)("g"))) * "4.21 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - 90)color(red)(cancel(color(black)(""^@"C")))#

#q_"water" = "842 J"#

This is how much heat is needed to get all the water to

#q_"metal" = 50.0 color(red)(cancel(color(black)("g"))) * "0.45 J" color(red)(cancel(color(black)("g"^(-1)))) color(red)(cancel(color(black)(""^@"C"^(-1)))) * (100 - 150) color(red)(cancel(color(black)(""^@"C")))#

#q_"metal" = -"1125 J"#

Keep in mind that the minus sign is used here because heat is being **given off**, i.e. the metal is going from a higher temperature to a lower temperature.

So, the metal gives off *more heat* than is needed to bring all the water to

The excess heat given off by the metal is

#q_"evaporation" = "1125 J" - "842 J" = "283 J"#

This is the amount of heat that will be absorbed by the water at

Now, the enthalpy of vaporization, *kilojoules per mole*, so make sure that you convert the excess heat to *kilojoules*

#283 color(red)(cancel(color(black)("J"))) * "1 kJ"/(10^3color(red)(cancel(color(black)("J")))) = "0.283 kJ"#

Use the enthalpy of vaporization as a **conversion factor** to calculate how many moles of water would evaporate

#0.283 color(red)(cancel(color(black)("kJ"))) * ("1 mole H"_2"O")/(40.7 color(red)(cancel(color(black)("kJ")))) = "0.0069533 moles H"_2"O"#

To convert this to *grams*, use the **molar mass** of water

#0.0069533 color(red)(cancel(color(black)("moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("0.13 g")))#

The answer is rounded to two **sig figs**, as requested by the problem.