Calculate cooling capacity required to dissipate heat load from a server rack.
Nearly all electrical power consumed by IT equipment is ultimately dissipated as heat (a server doesn't store energy — it converts it to computation and, overwhelmingly, waste heat), so a rack's power draw in watts converts directly to a cooling load using the standard conversion of 3.412 BTU/hr per watt. Cooling capacity is often specified in 'tons of refrigeration' (a legacy unit equal to 12,000 BTU/hr, originally referencing the cooling effect of melting one ton of ice per day), so dividing total BTU/hr by 12,000 gives the tonnage of CRAC/CRAH capacity needed — with a safety margin added to avoid running cooling infrastructure at its absolute limit.
Heat load
BTU_hr = watts × 3.412
Tons of cooling
tons = BTU_hr / 12000
By the first law of thermodynamics, energy can't be destroyed — the electrical power consumed by servers, storage, and networking gear is converted into computation (a negligible fraction) and, almost entirely, waste heat radiated into the room, so the cooling system must remove heat at essentially the same rate the IT equipment consumes power.
It's the standard energy unit conversion: 1 watt = 1 joule/second, and 1 BTU = 1055.06 joules, so 1 watt sustained for one hour equals 3600 joules = 3600/1055.06 ≈ 3.412 BTU — this is a fixed physical conversion, not an approximation specific to data centers.
Running cooling infrastructure at exactly its rated capacity leaves no margin for equipment additions, higher ambient temperatures, partial CRAC/CRAH unit failures, or airflow inefficiencies (hot spots, bypass air) — a 15-20% margin is a common planning buffer to maintain safe operating temperatures under real-world, non-ideal conditions.