Question

select the correct answer. a technician is placing a temperature monitoring device in the warmest part of a cooler. within what temperature should the device be accurate? a. within 1.5°f (±3.5°c) b. within 2.2°f (±1°c) c. within 4.4°f (±2.5°c) d. within 3.3°f (±1.5°c) e. within 1.1°f (±0.6°c)

Explanation:

To solve this, we analyze the temperature ranges. The warmest part of a cooker (like an oven) typically has higher temperatures. Let's convert the Celsius to Fahrenheit for clarity (formula: \( F = \frac{9}{5}C + 32 \)) or just compare the Fahrenheit values directly.

• Option A: within \( 15^\circ F \) (\( 35^\circ C \)) – \( 35^\circ C \) is \( \frac{9}{5}(35)+32 = 95^\circ F \), but the Fahrenheit range here is \( 15^\circ F \), which is inconsistent (likely a typo, maybe \( 150^\circ F \)? Wait, no, re - examining the options:

Wait, the options are likely about the temperature accuracy or the temperature range of the warmest part. In cooking, the warmest part (like the cooking zone) of a cooker (oven) for accurate temperature monitoring (e.g., for baking) often has a temperature around \( 35^\circ C \) ( \( 95^\circ F \)) but the options seem to have a typo. Wait, no, looking at the options again:

Option A: within \( 15^\circ F \) (\( 35^\circ C \)) – maybe it's a range of accuracy. But in standard cooking temperature monitoring, the warmest part (e.g., where the heating element is or the core cooking area) for a cooker (oven) when placing a temperature - monitoring device, the accurate temperature range (for the warmest part) is often around \( 35^\circ C \) ( \( 95^\circ F \)) with an accuracy of around \( 15^\circ F \) (but this is a bit confusing). Wait, maybe the question is about the temperature range of the warmest part. Let's check the Fahrenheit and Celsius conversions:

For a cooker (oven), the warmest part (cooking chamber) can have a temperature. Let's convert the Celsius in each option to Fahrenheit:

• Option A: \( 35^\circ C=\frac{9}{5}\times35 + 32=63 + 32 = 95^\circ F \), range \( 15^\circ F \) (so 95 - 7.5 to 95+7.5? No, range of \( 15^\circ F \) means from \( 95 - 7.5 = 87.5^\circ F \) to \( 95+7.5 = 102.5^\circ F \))
• Option B: \( 1^\circ C = 33.8^\circ F \), range \( 22^\circ F \) (33.8 - 11 to 33.8+11 = 22.8 to 44.8 \( ^\circ F \)) – too cold for warmest part.
• Option C: \( 25^\circ C=\frac{9}{5}\times25 + 32 = 45+32 = 77^\circ F \), range \( 24^\circ F \) (77 - 12 to 77+12 = 65 to 89 \( ^\circ F \)) – still cold.
• Option D: \( 15^\circ C=\frac{9}{5}\times15+32 = 27+32 = 59^\circ F \), range \( 23^\circ F \) (59 - 11.5 to 59+11.5 = 47.5 to 70.5 \( ^\circ F \)) – cold.
• Option E: \( 10^\circ C=\frac{9}{5}\times10 + 32=18 + 32 = 50^\circ F \), range \( 11^\circ F \) (50 - 5.5 to 50+5.5 = 44.5 to 55.5 \( ^\circ F \)) – cold.

The warmest part of a cooker (oven) should be the hottest, so the highest Celsius is \( 35^\circ C \) (Option A). So the correct answer is A. within \( 15^\circ F \) (\( 35^\circ C \))

Answer:

A. within \( 15^\circ F \) (\( 35^\circ C \))