# Question 4fbf8

Sep 29, 2015

Sorry...I cancelled my answer by mistake!
Ok, I found $10 g$ of steam.

#### Explanation:

Water uses the heat (=energy) to change its phase from liquid to gas maintaining the temperature constant at ${100}^{\circ} C$.

So basically you give heat $Q$ to your water and change a mass $m$ of it into steam, or:
$Q = m L$ where $L = 539 \frac{K c a l}{K g}$ is called Latent Heat of Vaporization and represents the heat needed to vaporize $1 k g$ of water.

Rearranging the above relationship you get:
$m = \frac{Q}{L} = \frac{5.4}{539} = 0.01 k g = 10 g$ of steam

Sep 29, 2015

$\text{10.0 g}$

#### Explanation:

So, you know that you're dealing with a mass of $\text{1 kg}$ of water at $100 \text{^@"C}$.

In order to be able to calculate how much steam is produced by adding $\text{5.40 kcal}$ of heat to the water, you need to know the enthalpy of vaporization, $\Delta {H}_{\text{vap}}$, of water.

To make the calculations a little more interesting, I'll use the value of $\Delta {H}_{\text{vap}}$ in kilojoules per mole

$\Delta {H}_{\text{vap" = "40.657 kJ/mol}}$

So, use the equation

$q = {n}_{\text{water" * DeltaH_"vap}}$

to figure out how much water can be converted to steam by the addition of $\text{5.40 kcal}$ of heat.

Use the conversion factor

$\text{1 kcal" = "4.184 kJ}$

to get

5.40color(red)(cancel(color(black)("kcal"))) * "4.184 kJ"/(1color(red)(cancel(color(black)("kcal")))) = "22.6 kJ"

This means that you have

$\text{22.6 kJ" = n_"water" * DeltaH_"vap}$

${m}_{\text{water" = (22.6color(red)(cancel(color(black)("kJ"))))/(40.657color(red)(cancel(color(black)("kJ")))/"mol") = "0.556 moles water}}$

Use water's molar mass to figure out how many grams would contain this many moles

0.556color(red)(cancel(color(black)("moles"))) * "18.02 g"/(1color(red)(cancel(color(black)("mole")))) = "10.01 g"

Therefore, out of your initial $\text{1.00 kg}$ of water at ${100}^{\circ} \text{C}$, only $\text{10.0 g}$ of water will be converted to steam at $100 \text{^@"C}$ by the addition of $\text{5.40 kJ}$ of heat.

If you want, you can express that in kilograms to get

10.0color(red)(cancel(color(black)("g"))) * "1 kg"/(1000color(red)(cancel(color(black)("g")))) = color(green)(1.00 * 10^(-2)"kg")

SIDE NOTE So, how much heat would be needed to convert all the water to steam?

1.00color(red)(cancel(color(black)("kg"))) * (1000color(red)(cancel(color(black)("g"))))/(1color(red)(cancel(color(black)("kg")))) * "1 mole"/(18.02color(red)(cancel(color(black)("g")))) = "55.5 moles"

q = 55.5color(red)(cancel(color(black)("moles"))) * 40.657"kJ"/color(red)(cancel(color(black)("mole"))) = "2256 kJ"

Finally, this is equivalent to

2256color(red)(cancel(color(black)("kJ"))) * "1 kcal"/(4.184color(red)(cancel(color(black)("kJ")))) = "539 kcal"#