The amount of energy required to maintain water at some temperature over some time is simply equal to the amount of energy that the water loses to the environment over that time.
It depends on the conditions that the water is in, specifically how much heat the water is losing the environment. For example, it will take more energy to maintain water at 132 F when you place it in a cold windy environment because it's losing lots of heat in that environment.
So let's say you have a simple scenario of a pot of water sitting in room temperature. A simple energy balance on your water gives:
$$ q_{in} = hA_s(T-T_0) $$
Where $q_{in}$ is the input power, $h$ is an effective heat transfer coefficient telling how much heat the water loses to the ambient (typical value of 5-10 $W/m^2K$ for a 56 C object sitting in a room with no wind), $A_s$ is the surface area of your water container, $T$ is the temperature of the water (56 C), and $T_0$ is room temperature (25 C).
So, all you need is the surface area of your water container. Then solve for the required power $q_in$, and multiply by time (30 min) to get the required energy.
$$ Q_{in} = tq_{in} $$
Where $t$ is the time you desire (30 min). That's how you do it.
