Time & Temperature ControlQuestion 291 of 319

A cook places a large stock pot of hot chili (180°F) directly into the walk-in cooler at 6:00 PM. Under California Retail Food Code §114002, why is this practice INCORRECT, and what would be the right approach?

a.It is correct as long as the cooler is set below 41°F; no other action is needed
b.It is incorrect only because the walk-in needs to be raised to 45°F first to avoid 'shock'
c.It is incorrect because hot food should never be placed in a cooler; it should be left out at room temperature to cool
d.It is incorrect because a large, deep mass of hot food cannot pass through the 135°F to 70°F window in 2 hours by ambient cooling; the cook should divide the chili into shallow pans (less than 4 inches deep), use an ice bath, ice wand, or blast chiller, and leave the pan loosely covered to release steam

Explanation

California Retail Food Code HSC §114002 requires cooked TCS food to be cooled from 135°F to 70°F within 2 hours and from 70°F to 41°F within an additional 4 hours (6 hours total). A large, deep, dense mass of hot chili in a tall stockpot acts as a thermal reservoir — the geometric center may remain above 125°F for many hours, sailing past both checkpoints and giving Clostridium perfringens (a major chili/stew pathogen) ideal growth conditions. The correct method is to break the food into smaller masses: shallow stainless-steel pans no more than 4 inches deep (2 inches for very dense foods), or use an active cooling tool (ice bath in a prep sink with stirring, sealed plastic ice wand placed in the center of the pot, or a blast chiller). The pan should be loosely covered while still hot to allow steam to escape. Option A ignores the physical heat transfer problem. Option B is wrong because room-temperature ambient cooling is even slower and is non-compliant. Option D invents a 'thermal shock' concept that has no basis in the code and would itself warm the cooler and endanger other food.

Law Reference: HSC §114002

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