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This problem involves using Newton's Law of Cooling to find the time required for the beef roast to reach a specific temperature. Newton's Law of Cooling states that the rate of change of the temperature of an object is proportional to the difference between its temperature and the surrounding temperature. The formula for this is:

\[
T(t) = T_s + (T_0 - T_s) e^{-kt}
\]

where:
- \( T(t) \) is the temperature of the object at time \( t \),
- \( T_s \) is the surrounding temperature (oven temperature, in this case, \( 375^\circ F \)),
- \( T_0 \) is the initial temperature of the object (beef roast, \( 50^\circ F \)),
- \( k \) is a constant we need to find,
- \( t \) is the time in minutes.

### Step 1: Solve for \( k \)
We know that after 75 minutes, the temperature of the beef roast is \( 125^\circ F \). Plugging in this information:

\[
125 = 375 + (50 - 375) e^{-75k}
\]

Solving for \( k \):

1. Rearrange the equation:
   \[
   125 - 375 = (50 - 375) e^{-75k}
   \]
   \[
   -250 = -325 e^{-75k}
   \]

2. Divide by \(-325\):
   \[
   \frac{250}{325} = e^{-75k}
   \]
   \[
   \frac{5}{6.5} = e^{-75k}
   \]

3. Take the natural logarithm of both sides:
   \[
   \ln\left(\frac{5}{6.5}\right) = -75k
   \]

4. Solve for \( k \):
   \[
   k = -\frac{\ln\left(\frac{5}{6.5}\right)}{75}
   \]

### Step 2: Use \( k \) to find the time when the roast reaches \( 150^\circ F \)
Now that we have \( k \), we substitute back to solve for \( t \) when \( T(t) = 150^\circ F \):

\[
150 = 375 + (50 - 375

A four-pound beef roast, initially at 50°F, is placed in a 375°F oven. After 75 minutes, it is found that the temperature of the beef roast is 125°F. Determine the time when the temperature of the beef roast will be at 150°F.

Solve for the time required for a beef roast to reach 150°F in a 375°F oven, using Newton's Law of Cooling. This problem explores exponential functions and temperature change concepts, ideal for college or advanced high school students.

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