Let's solve the problem step by step:
1. Initial Condition Understanding:
- The temperature was initially [tex]\(135^\circ F\)[/tex] and it was cooled to [tex]\(70^\circ F\)[/tex] in 2 hours.
- Now we need to find out how long it will take to cool the soup from [tex]\(70^\circ F\)[/tex] to [tex]\(41^\circ F\)[/tex].
2. Temperature Differences:
- The initial temperature drop from [tex]\(135^\circ F\)[/tex] to [tex]\(70^\circ F\)[/tex] is [tex]\(135 - 70 = 65^\circ F\)[/tex].
- The new temperature drop we need to work with is from [tex]\(70^\circ F\)[/tex] to [tex]\(41^\circ F\)[/tex], which is [tex]\(70 - 41 = 29^\circ F\)[/tex].
3. Exponential Cooling Behavior:
- The cooling process behaves exponentially, and the time required for cooling can be related to the temperature drop proportionally.
4. Proportional Calculation:
- Given that the initial cooling of [tex]\(65^\circ F\)[/tex] took 2 hours, we need to find the time [tex]\(t\)[/tex] for a cooling of [tex]\(29^\circ F\)[/tex].
- The time required for the second cooling is proportional to the ratio of the temperature drops.
- Therefore, the time [tex]\(t\)[/tex] can be calculated using the proportion:
[tex]\[
\frac{t}{2 \text{ hours}} = \frac{29^\circ F}{65^\circ F}
\][/tex]
5. Solve for [tex]\(t\)[/tex]:
[tex]\[
t = 2 \text{ hours} \times \frac{29^\circ F}{65^\circ F}
= 2 \times \frac{29}{65}
= 2 \times 0.4461538461538462
= 0.8923076923076924 \text{ hours}
\][/tex]
6. Rounding the Time:
- We need to round the time to the nearest whole number.
- [tex]\(0.8923076923076924\)[/tex] hours rounded to the nearest whole number is 1 hour.
Hence, the food worker will need approximately 1 hour to cool the soup from [tex]\(70^\circ F\)[/tex] to [tex]\(41^\circ F\)[/tex].
The correct answer is: 1 hour.
