Cooling Chart

Serving Valley, Elmore, Boise and Ada Counties

Main Office  707 N. Armstrong Pl.  Boise Id 83704-0825  (208) 375-7499  Fax 327-8553

Cooling Chart

Name: ________________________ Address: ___________________________

Food Product

Refrigeration or Room Temp

Food / Container

L x W x H / Material

Date

Time at 135°F

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:

:

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:

:

°F

°F

°F

°F

°F

°F

Temperature

After 1 Hour

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:

:

:

:

:

Time

°F

°F

°F

°F

°F

°F

Temperature

After 2 Hours

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:

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:

:

:

Time

(must be 70°F or below)

°F

°F

°F

°F

°F

°F

Temperature

After 3 Hours

Time

:

:

:

:

:

:

°F

°F

°F

°F

°F

°F

Temperature

After 4 Hours

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:

Time

Temperature

°F

°F

°F

°F

°F

°F

After 5 Hours

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:

:

:

Time

°F

°F

°F

°F

°F

°F

Temperature

After 6 Hours

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:

(must be 41°F or below)

Time

Corrective Action Taken

Manager/Employee Signature

For Successful cooling of food use the following formula to help you determine if your rate of cooling is fast enough:

1)

Cool from 135°F to 70° F within two hours, the rate of cooling must be approximately 0.54°F/minute (135-70=65°F ÷120 minutes (2 hours)=0.54°F) or ~32°F per hour

2)

Cool from 70°Fto 41°F within 4 hours, the rate of cooling must be approximately 0.12°F /minute (70-41=29°F ÷ 240min( 4 hours)=0.12°F) or ~7°F/hour.

Example: Initial temperature of soup is 135°F. One hour later the temperature is 129°F. 135-129=6 ÷60min=0.1°F/min. The rate of cooling is not fast enough. The rate needs to be ~.54°/min.

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