# Effect of ice on ambient temperature in closed space

Premise:

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.

Question:

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.

Premise:

First, as regards to your premise.

It really doesn't matter what kind of electrical load (inductive, resistive, or capacitive) that is connected to the source. If the source delivers a kilowatt of power for an hour, it will have delivered one kilowatt-hour of energy to the room. When the power source stops after an hour, the internal energy of a perfectly insulated room with rigid walls that cannot expand or contract will have increased by one kilowatt hour. Since the "room" does no boundary work (expand or contract) on the surroundings, and transfers no heat with the surroundings (being perfectly insulated), by the first law for a closed system (which is what we have) the change in internal energy will be one kilowatt hour.

Now the internal energy of the room consists of the sum of its microscopic internal kinetic and potential energy. If none of the materials of the room undergo a phase change, then we can say that the change in internal energy is primarily kinetic. And since a change in the internal kinetic energy manifests itself primarily as a change in temperature, we can say there would be an equal increase in the temperature of all the materials in the room once we have thermal equilibrium.

Question:

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.

First, you need to realize that each of the materials in the room that are now theoretically at the same temperature, including the air in the room, have different masses and specific heats. Those masses and specific heats will determine how much heat each material in the room can contribute to the melting of the ice introduced into the room. @Gert has already adequately addressed the heat requirements for melting ice.

Bottom line: In order to answer what a quantity of ice introduced to the same closed system would do, you would need to know all of the materials in the room, their masses, and their specific heats in order to do the necessary calculations.

A fair, helpful point. I take it that the floor, for example, would transfer its heat to the ice that sits upon it more readily than the air in the room, depending what material it is. The main question was whether the end result would be a lower-temperature room, and I think it's fair to say 'yes' and omit exact calculations for now.

I think it is fair to say that, yes, the end result "would be a lower temperature room". I agree with that because the heat transferred to the ice, be it from the floor or from any where else, would not raise the temperature of the ice (during the phase change) but would lower the temperature of what is transferring heat to the ice if it were not also undergoing a phase change.

Hope this helps.

• A fair, helpful point. I take it that the floor, for example, would transfer its heat to the ice that sits upon it more readily than the air in the room, depending what material it is. The main question was whether the end result would be a lower-temperature room, and I think it's fair to say 'yes' and omit exact calculations for now. Sep 28 '19 at 23:53
• @JYelton Good points. See my update answer in response to your follow up comment. Hope it helps. Sep 29 '19 at 1:12

Am I starting with correct fundamentals and is my explanation about the ice mostly correct?

Yes it is. But you are lacking a few pieces to complete the puzzle.

Firstly, the Laws of Thermodynamics dictate that, given enough time, both air and the molten ice (water, of course) will be at the same equilibrium temperature ($$T_e$$). This is true because as long as the two objects (air and ice/water) are not at the same temperature they keep exchanging heat energy.

Secondly, to melt ice costs heat energy, the so-called Latent Heat of Fusion (for water ice $$L_{ice}=333.55\text{kJ/kg}$$ at $$0$$ $$\text{Celsius}$$)

Thirdly, a simply model of the change heat content of an object when it changes temperature $$\Delta T$$ is given by:

$$\Delta H=mc_v\Delta T$$

where $$c_v$$ is the specific heat capacity and $$m$$ the mass.

You've indicated the system is isolated from the rest of the universe. Mathematically this means that the sum of all changes of heat content of all the objects in the system must be $$\text{zero}$$:

$$\displaystyle\sum_{n=1}^{n}\Delta H_i=0\tag{1}$$

From this so-called heat balance, the equilibrium temperature $$T_e$$ can be calculated.

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.

You're overthinking, I think. Lets just look at what $$(1)$$ turns out to be, shall we?

For simplicity's sake we'll assume the walls don't take part in the heat balance and that the ice is initially at $$0$$ $$\text{Celsius}$$.

Firstly, the air in the room will cool down so that:

$$\Delta H_1=m_{air}c_v(T_e-T_i)$$

where $$T_i$$ is the initial temperature of the air.

Secondly the ice has to melt:

$$\Delta H_2=-m_wL_{ice}$$

where $$m_w$$ is the mass of ice.

Finally the molten ice has to heat up from $$0$$ $$\text{Celsius}$$ to $$T_e$$:

$$\Delta H_3=m_{w}c_v(T_e-0)=m_{w}c_vT_e$$

Inserting into $$(1)$$ we get:

$$\boxed{m_{air}c_v(T_e-T_i)-m_wL_{ice}+m_{w}c_vT_e=0}\tag{2}$$

$$(2)$$ is a linear equation with only $$T_e$$ as unknown. It can be easily solved.

• My apologies on not quite knowing how to use this information. (I'm, at best, a physics newbie.) I've edited my question to try to clarify what I am asking. Sep 29 '19 at 5:22
• I hope the edit is useful. Ta.
– Gert
Sep 29 '19 at 14:22