Which food is at a temperature that allows bacteria to grow well?
A roast at 125°F(52°C)
Hamburgers at 165°F(74°C)
Pork chops at 170°F(77°C)
A steak at 180°F(82°C)
The Correct Answer and Explanation is:
The correct answer is: A roast at 125°F (52°C)
Explanation:
Bacteria grow most rapidly in the “temperature danger zone,” which ranges from 40°F (4°C) to 140°F (60°C). Foods that are within this range for extended periods are at higher risk of developing harmful levels of bacteria, such as Salmonella, Escherichia coli (E. coli), and Listeria monocytogenes, all of which can cause foodborne illness. Among the temperature choices listed, 125°F (52°C) is squarely within this danger zone.
At 125°F (52°C), a roast is warm enough to be comfortable to the touch but not hot enough to kill bacteria. In fact, temperatures between 70°F and 125°F (21°C to 52°C) are particularly favorable for rapid bacterial multiplication. If food remains in this range for more than 2 hours (or even less in hot environments), it becomes increasingly unsafe to consume.
In contrast, the other food temperatures listed are well above the danger zone:
- Hamburgers at 165°F (74°C): This temperature is considered safe and is the minimum internal temperature recommended by the USDA for ground beef to ensure bacteria are killed.
- Pork chops at 170°F (77°C): This is also above the safe threshold; pork only needs to be cooked to 145°F (63°C) followed by a 3-minute rest to be safe.
- A steak at 180°F (82°C): This is well-cooked and far beyond the required temperature for safety. Most bacteria would be destroyed at this level.
To prevent foodborne illness, it’s essential to keep hot foods above 140°F (60°C) and cold foods below 40°F (4°C). Holding or serving food at 125°F creates an ideal environment for bacteria to grow, making it unsafe. Therefore, the roast at 125°F is the item most likely to support bacterial growth.