A friend and I were discussing the concept of a heater operating in a closed space (by this I mean an isolated area with no thermal conduction in or out) and how a given wattage will (eventually) heat the space the same. For example, consider two 1kW heaters, one with a fan and one without. If both units consume exactly 1kW, my understanding is that they will both heat the space the same amount, provided that the space is a closed system.
In reality, the fan is an inductive load while a heating element will be more resistive, so there are some power consumption caveats which matter, but for the purpose of this question I am proposing that 1kW of power is transferred into a closed system for a given amount of time. The time for the closed space to reach equilibrium may differ, but in theory both scenarios will result in the same temperature rise.
After explaining that a given amount of power transferred into a closed system will result in the same (eventual) temperature rise, I was asked what a quantity of ice introduced to the same closed system would do.
Imagine a 10 $m^3$ room of air at 20°C. If 1kg of ice at 0°C was introduced, what happens to the ice and the room temperature?
My answer was that the heat energy of the air will be transferred to the ice, melting it to liquid water. The air temperature will decrease as the ice temperature increases until the room reaches equilibrium. I would expect that the the resulting temperature would certainly be less than 20°C, but I am not sure by how much or how to calculate it.
Am I starting with correct fundamentals and is my explanation about the ice mostly correct?
Put another way:
The answers I have so far are helpful but I still am confused about the final outcome. To restate my question: If an air-filled, isolated $10 m^3$ room is at 20°C and I introduce 1kg of ice at 0°C, will the room temperature be lower after it reaches equilibrium? It seems that it would, but...
If thermal energy is transferred from the room (air and surfaces) to the ice to change its state (from solid to liquid), it seems logical the room's air temperature would drop whilst the ice/water temperature increases. However, since it is a closed system, that energy never leaves the room.
I feel as though "beaming" ice into this hypothetical room is like adding a sort of heat sponge. If it absorbs energy, but is still contained in the room, all of the heat from the initial condition is still there, but now there's extra mass.
Is it at all accurate to think of this closed system as having some quantity of heat energy and mass at the start, and that adding mass at lower temperature will lower the $T_e$? It seems intuitive that introducing, say, hot coals, is adding both mass and energy to the system, and its $T_e$ will be higher.
My apologies for this question becoming quite long. Put a final way, if I think about the room and ice temperature in Kelvin, then it would seem that adding ice is, in fact, adding energy to the system. But that doesn't seem correct; I am just confusing myself more.