When a chicken is removed from oven temperature measured at 300ffthree minutes later 200fhow long will it take for the to cool off room of 70f?

- Initial temperature of the chicken: \(T_i = 300^\circ F\)

- Final room temperature: \(T_r = 70^\circ F\)

- Temperature measured three minutes later: \(T_1 = 200^\circ F\)

Assumptions

- The chicken is assumed to cool according to Newton's law cooling, which states that the rate of cooling of an object is directly proportional to the difference in its temperature and the ambient temperature.

Solving for the Cooling Rate Constant (k)

Using the given data at three minutes, we can calculate the cooling rate constant \(k\) using the equation:

$$T_1 = T_i + (T_i-T_r)(e^{-kt})$$

Where:

- \(T_1\) is the temperature at time \(t\)

- \(T_i\) is the initial temperature

- \(T_r\) is the room temperature

- \(k\) is the cooling rate constant

Substituting the values into the equation, we get:

$$200 = 300+ (300-70)(e^{-3k})$$

Solving for \(k\), we find that:

$$k \approx 0.0693 \ \text{min}^{-1}$$

Finding the Time to Cool to Room Temperature

Now we want to find the time \(t\) it will take for the chicken to cool from \(T_1 = 200 ^\circ F\) to the room temperature \(T_r = 70^\circ F\).

We can rearrange the equation above to solve for \(t\):

$$t = \frac{1}{k} \ln \left(\frac{T_i-T_r}{T_1-T_r}\right)$$

Plugging in the values, we get:

$$t =\frac{1}{0.0693} \ln \left(\frac{300-70}{200-70}\right) \approx 4.6 \text{ minutes}$$

Therefore, it will take approximately 4.6 minutes for the chicken to cool from 200°F to room temperature (70°F) in the given scenario.