Question

How much energy is required to convert 429 g of water to steam at 156 °C?

`C_text(water) = "4.184 J/(°C·mol)"; C_text(vapour) = "2.078 J/(°C·mol)"; Δ_text(vap)H = "2257 J/g"`

Answer 1

`~~1155(kJ)/(g*^oC)`.

Explanation:

We will be using the formula: `mcDeltaT`, where:
=> `m` is the mass of the object in grams.
=> `c` is the specific heat capacity of the object.
=> `DeltaT` is the change in temperature found by subtracting the initial temperature from the final.

Let's go over what we are given:

`m = 429 "g"`

`c_"liquid water" = 4.184 J/(g*^oC)`

`c_"water vapour" = 2.078 J/(g*^oC)`

`DeltaT = T_"final" - T_"initial"`

Now there's something we have to consider, we can't just get from liquid at `24^oC` to steam at `156^oC`. What we have to do is convert from liquid to it's boiling point of `100^oC`, and then from `100^oC` to `156^oC`.

So let's do the liquid to gas first.

`q_"liquid water" =mcDeltaT`

`=(429)(4.184)(100-24)`

`=136415.136 J/(g*^oC)`

Next, we convert the liquid at `100 °"C"` to vapour at `100 °"C"`. For this, we

`q_"conversion"=mDeltaH`

`= 429(2257)`

`=968253`

Now let's do the gas to hotter gas.

`q_"water vapour"=mcDeltaT`

`=(429)(2.078)(156-100)`

`=49921.87J/(g*^oC)`

Now all we have to do is add the energies.

`q_"total"=q_"liquid water" + q_"conversion" + q_"water vapour"`

`=136415.136+968253+49921.87`

`=1154590.01J/(g*^oC)`

`=1154.59001(kJ)/(g*^oC)`

`~~1155(kJ)/(g*^oC)`

Hope this helps :)