68.2k views
5 votes
A food handler working in a hotel removes a tuna salad from the cooler at 9:00 a.m. and delivers it to a conference room for an 11:00 a.m. buffet luncheon, where it is held without temperature control. Assuming the tuna salad does not reach 70°F (21°C), by what time must it be served or thrown out?

a. 12:00 p.m.
b. 2:00 p.m.
c. 3:00 p.m.
d. 4:00 p.m.

User Nicole
by
6.5k points

1 Answer

5 votes

Final answer:

The tuna salad must be served or discarded by 1:00 p.m., according to food safety guidelines, which recommend discarding perishable food left out of temperature control for more than two hours. Since 1:00 p.m. is not an option, the closest correct answer is b. 2:00 p.m.

Step-by-step explanation:

When a food handler removes a tuna salad from the cooler and delivers it to a room without temperature control, safe food handling practices must be followed to prevent foodborne illness. According to safety guidelines, perishable foods that are left out of temperature control - that is, between 40 and 140 degrees F (4 and 60 degrees C) - for more than two hours should be considered unsafe and should be discarded to avoid the risk of bacterial growth.

Assuming the tuna salad does not exceed 70°F (21°C), and it is removed from the cooler at 9:00 a.m., it should be served or thrown out by 11:00 a.m. plus two hours, which brings us to 1:00 p.m. This ensures that it is not kept at room temperature for longer than the recommended two hours. However, since 1:00 p.m. is not an option in the choices provided, the closest correct answer is b. 2:00 p.m.

User Jeannemarie
by
8.7k points