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"#

1 Answer
Sep 12, 2017

Answer:

#~~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 :)