The questions says that at 110 degrees the steam condenses (becmes liquid), and then the temperature is decrased from 110 to 80 degrees
This give you two important facts:
1. during the cooling the liquid doesn't have state changes, this means that all the energy is in the change of temperature
2. there is no information about the mass and the answer is asked interms of it, this simplify the calculation
[tex]\begin{gathered} Q=m\cdot c\cdot\Delta T \\ \frac{Q}{m}=c\cdot T \end{gathered}[/tex]Q/T is the heat over gram asked, c is the specific heat that we would use. As the problem doesn't says anything about the liquid we arsume that is water.
[tex]\frac{Q}{m}=30\degree\cdot4.186=125.58J/g[/tex]
