Question

Question #4fbf8

Answer 1

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

Explanation:

Water uses the heat (=energy) to change its phase from liquid to gas maintaining the temperature constant at `100^@C`.

So basically you give heat `Q` to your water and change a mass `m` of it into steam, or:
`Q=mL` where `L=539(Kcal)/(Kg)` is called Latent Heat of Vaporization and represents the heat needed to vaporize `1kg` of water.

Rearranging the above relationship you get:
`m=Q/L=5.4/539=0.01kg=10g` of steam

Answer 2

`"10.0 g"`

Explanation:

So, you know that you're dealing with a mass of `"1 kg"` of water at `100""^@"C"`.

In order to be able to calculate how much steam is produced by adding `"5.40 kcal"` of heat to the water, you need to know the enthalpy of vaporization, `DeltaH_"vap"`, of water.

To make the calculations a little more interesting, I'll use the value of `DeltaH_"vap"` in kilojoules per mole

`DeltaH_"vap" = "40.657 kJ/mol"`

So, use the equation

`q = n_"water" * DeltaH_"vap"`

to figure out how much water can be converted to steam by the addition of `"5.40 kcal"` of heat.

Use the conversion factor

`"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

`"22.6 kJ" = n_"water" * DeltaH_"vap"`

`m_"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 `"1.00 kg"` of water at `100^@"C"`, only `"10.0 g"` of water will be converted to steam at `100""^@"C"` by the addition of `"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"`