# (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 ${H}_{2} O \left(l\right)$ 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 $\setminus \Delta {H}_{v a p} = 40.7 k J \setminus \times m o {l}^{-} 1 {H}_{2} O$ express your answer using two significant figures

Nov 29, 2016

$\text{0.13 g}$

#### 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 ${100}^{\circ} \text{C}$, its normal boiling temperature, then some of the water will evaporate.

Now, you can assume that the equilibrium temperature of the metal + liquid water system will be ${100}^{\circ} \text{C}$. In order to prove that this is the case, calculate how much heat is needed in order to increase the temperature of the water from ${90}^{\circ} \text{C}$ to ${100}^{\circ} \text{C}$.

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{q = m \cdot c \cdot \Delta T}}}$

Here

• $q$ is heat absorbed / given off
• $m$ is the mass of the sample
• $c$ is the specific heat of the substance
• $\Delta T$ is the change in temperature, defined as the difference between the final temperature and the initial 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")))

${q}_{\text{water" = "842 J}}$

This is how much heat is needed to get all the water to ${100}^{\circ} \text{C}$. Now calculate how much heat is given off when the metal cools to ${100}^{\circ} \text{C}$

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")))

${q}_{\text{metal" = -"1125 J}}$

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 ${100}^{\circ} \text{C}$, which means that some of the water will indeed evaporate.

The excess heat given off by the metal is

${q}_{\text{evaporation" = "1125 J" - "842 J" = "283 J}}$

This is the amount of heat that will be absorbed by the water at ${100}^{\circ} \text{C}$ to produce vapor.

Now, the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, is expressed in 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

$0.0069533 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles H"_2"O"))) * "18.015 g"/(1color(red)(cancel(color(black)("mole H"_2"O")))) = color(darkgreen)(ul(color(black)("0.13 g}}}}$

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