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 because hot food should never be placed in a cooler; it should be left out at room temperature to cool
c.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
d.It is incorrect only because the walk-in needs to be raised to 45°F first to avoid 'shock'

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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