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Both teams concentrate their respective energy blast towards each other. Whichever beam gets through wins. No external powers or effects can be use to weaken an opposing team. They all focus on their energy blast.

Iceman
Killer Frost
Icicle (Junior)
Captain Cold
Ice

vs

Pyro
Human Torch
Sunspot
Starfire
Fire

Who wins ?

https://jordanmac1.files.wordpress.com/2015/01/ice-vs-pyro.gif

Kinda feel like Captain Cold wins it for his team.

Killer Frost soloes.

Team Ice wins most likely. Firestar is a weird one though.

Would it be a different story if I put in Firestorm or Phoenix ?

Just with their heat blasts, no other powers? Tricky. Because Captain Cold:


https://i.imgur.com/9on3ILN.png

Stupid, I know.

Negative absolute temperatures are not cold but, in fact, ridiculously hot.

A system with T < 0 is actually hotter than a system with any positive absolute temperature. If you heat an object to, say, +1000 000 000 kelvins and place it next to an object at -1 kelvin, heat energy will flow from the LATTER to the FORMER. Since heat energy flows from "hotter" to "colder" (by definition) we conclude that the -1 K object is the hotter one.

Some technical stuff follows. In thermodynamic theories where entropy S is defined independently from temperature, thermodynamic (absolute) temperature T is defined as:

https://wikimedia.org/api/rest_v1/media/math/render/svg/f9ae0da709f678a64f96d0165d084a535eb877e9

For our purposes, in the above equation dE may be considered as the amount of heat energy absorbed by the system and dS the resulting increase in system's entropy. We may solve for dS:

dS = dE/T (eqn 1)

The Second Law of Thermodynamics states that the total entropy increases in all natural, spontaneous processes. Based on this, and the equation above, we can see that heat energy always flows from an object with higher T to an object with lower T, as long as both Ts are positive.

For example, lets consider objects A and B with T(A) = 10 and T(B) = 100 (in some absolute temperature units). We can calculate the total entropy chance for a transfer of 1 joule of heat energy from B to A (i.e. dE(A) = +1 and dE(B) = -1 since B loses and A gains energy). The entropy chance of A, according to (eqn 1), is:
dS(A) = +1/10 = +0.1
Similarly, the entropy change of B is:
dS(B) = -1/100 = -0.01
The total entropy chance is dS(A) + dS(B) = +0.09 > 0.
--> In case of positive temperatures, heat flows from the higher T object (i.e. B) to lower T object (i.e. A) since this increases the total entropy.

What if one of the objects has negative absolute temperature? Lets repeat the above calculations, but now T(A) = -10 and T(B) = 100. Lets now consider a flow of 1 J of heat energy from A to B (i.e. dE(A) = -1 and dE(B) = +1) :
dS(A) = (-1)/(-10) = +0.1
dS(B) = +1/100 = +0.01
dS(total) = +0.11 > 0.
--> Heat flows from the negative T object (i.e. A) to the positive T object (i.e. A) regardless of how high the positive T is, since this increases the total entropy. Negative absolute T objects are "infinitely" hotter than positive T objects.

Most thermodynamic systems cannot attain negative absolute temperatures. Only if the system's phase space is bounded can this be achieved. A typical example is a spin-system that can be sufficiently isolated from the surrounding "lattice". Such a system has a finite number of energy levels and thus a bounded phase space. Spin-systems with negative absolute Ts have indeed been prepared in laboratories.

Small typo in the above: "Heat flows from the negative T object (i.e. A) to the positive T object (i.e. A)" should read "Heat flows from the negative T object (i.e. A) to the positive T object (i.e. B)".