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
Explanation:
We will be using the formula:
=>
=>
=>
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
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
#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 :)