Questions: A hot bar of iron at 1100 degrees Fahrenheit is quenched in oil at 350 degrees. After 2 minutes, the temperature of the iron is 750 degrees. (a) Find the value for k in Newton's Law of Cooling. Round your answer to the nearest thousandth. k = (b) What will the temperature of the iron be after 10 minutes? Round your answer to the nearest degree. degrees Fahrenheit (c) How long will it take for the iron to reach 400 degrees? Round your answer to the nearest tenth of a minute. minutes

I'll solve this physics problem about Newton's Law of Cooling step by step.

Find the value for $k$ in Newton's Law of Cooling.

Set up Newton's Law of Cooling equation

Newton's Law of Cooling states that the rate of change of temperature is proportional to the difference between the object's temperature and the surrounding temperature:

\(\frac{dT}{dt} = -k(T - T_s)\)

Where:

• $T$ is the temperature of the object
• $T_s$ is the temperature of the surroundings
• $k$ is the cooling constant we need to find

The solution to this differential equation is: \(T(t) = T_s + (T_0 - T_s)e^{-kt}\)

Where $T_0$ is the initial temperature of the object.

Substitute the known values

Given:

• Initial temperature $T_0 = 1100°F$
• Surrounding temperature $T_s = 350°F$
• After $t = 2$ minutes, $T(2) = 750°F$

Substituting into the equation: \(750 = 350 + (1100 - 350)e^{-k \cdot 2}\) \(750 = 350 + 750e^{-2k}\) \(400 = 750e^{-2k}\) \(\frac{400}{750} = e^{-2k}\) \(\frac{8}{15} = e^{-2k}\)

Solve for $k$

Taking the natural logarithm of both sides: \(\ln(\frac{8}{15}) = -2k\) \(-\ln(\frac{15}{8}) = -2k\) \(k = \frac{\ln(\frac{15}{8})}{2}\) \(k = \frac{\ln(15) - \ln(8)}{2}\) \(k = \frac{2.708 - 2.079}{2}\) \(k = \frac{0.629}{2}\) \(k = 0.3145\)

Rounding to the nearest thousandth: \(k = 0.315\)

\(\boxed{k = 0.315}\)

Find the temperature of the iron after 10 minutes.

Apply Newton's Law of Cooling with the calculated $k$ value

Using the cooling equation with our calculated $k = 0.315$: \(T(t) = T_s + (T_0 - T_s)e^{-kt}\)

Substituting the values: \(T(10) = 350 + (1100 - 350)e^{-0.315 \cdot 10}\) \(T(10) = 350 + 750e^{-3.15}\)

Calculate the final temperature

\(e^{-3.15} \approx 0.0429\) \(T(10) = 350 + 750 \cdot 0.0429\) \(T(10) = 350 + 32.175\) \(T(10) = 382.175\)

Rounding to the nearest degree: \(T(10) = 382°F\)

\(\boxed{382 \text{ degrees Fahrenheit}}\)

Find how long it will take for the iron to reach 400 degrees.

Set up the equation with the target temperature

We need to find the time $t$ when $T(t) = 400°F$.

Using Newton's Law of Cooling: \(T(t) = T_s + (T_0 - T_s)e^{-kt}\)

Substituting the values: \(400 = 350 + (1100 - 350)e^{-0.315t}\) \(400 = 350 + 750e^{-0.315t}\)

Solve for time $t$

\(400 - 350 = 750e^{-0.315t}\) \(50 = 750e^{-0.315t}\) \(\frac{50}{750} = e^{-0.315t}\) \(\frac{1}{15} = e^{-0.315t}\)

Taking the natural logarithm of both sides: \(\ln(\frac{1}{15}) = -0.315t\) \(-\ln(15) = -0.315t\) \(t = \frac{\ln(15)}{0.315}\) \(t = \frac{2.708}{0.315}\) \(t = 8.597\)

Rounding to the nearest tenth of a minute: \(t = 8.6\) minutes

\(\boxed{8.6 \text{ minutes}}\)

(a) \(\boxed{k = 0.315}\) (b) \(\boxed{382 \text{ degrees Fahrenheit}}\) (c) \(\boxed{8.6 \text{ minutes}}\)