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Question
A welder drops a piece of red-hot steel on the floor. The initial temperature of the steel is 2,500 degrees Fahrenheit. The ambient temperature is 80 degrees Fahrenheit. After 2 minutes the temperature of the steel is 1,500 degrees. The function $\mathrm{f}(\mathrm{t})=\mathrm{Ce}^{(-\mathrm{kt})}+80$ represents the situation, where $\mathrm{t}$ is time in minutes, $\mathrm{C}$ is a constant, and $\mathrm{k}$ is a constant.

After 2 minutes the temperature of the steel is 1,500 degrees. After how many minutes will the temperature of the steel be 100 degrees and therefore safe to pick up with bare hands? Round your answer to the nearest whole number, and do not include units.

Provide your answer below:

Step 1 ：Given the function \(f(t) = Ce^{-kt} + 80\), where \(t\) is the time in minutes, \(C\) and \(k\) are constants, and \(f(t)\) is the temperature in degrees Fahrenheit.

Step 2 ：We know that at \(t = 0\), the temperature is 2500 degrees Fahrenheit. So, we can substitute these values into the function: \(2500 = Ce^{-k*0} + 80\), which simplifies to \(2500 = C + 80\). Solving for \(C\), we get \(C = 2500 - 80 = 2420\).

Step 3 ：Next, we know that after 2 minutes (\(t = 2\)), the temperature is 1500 degrees Fahrenheit. We can substitute these values into the function to find \(k\): \(1500 = 2420e^{-2k} + 80\). Simplifying, we get \(1420 = 2420e^{-2k}\). Solving for \(e^{-2k}\), we get \(e^{-2k} = 1420/2420\). Taking the natural logarithm of both sides, we get \(-2k = \ln(1420/2420)\). Solving for \(k\), we get \(k = -\ln(1420/2420) / 2\), which approximates to \(k \approx 0.188\).

Step 4 ：Now we have the function \(f(t) = 2420e^{-0.188t} + 80\).

Step 5 ：We want to find the time \(t\) when the temperature will be 100 degrees Fahrenheit. So, we set \(f(t) = 100\) and solve for \(t\): \(100 = 2420e^{-0.188t} + 80\). Simplifying, we get \(20 = 2420e^{-0.188t}\). Solving for \(e^{-0.188t}\), we get \(e^{-0.188t} = 20/2420\). Taking the natural logarithm of both sides, we get \(-0.188t = \ln(20/2420)\). Solving for \(t\), we get \(t = -\ln(20/2420) / 0.188\), which approximates to \(t \approx 23.6\).

Step 6 ：So, it will take approximately 24 minutes (rounded to the nearest whole number) for the steel to cool to 100 degrees Fahrenheit and be safe to pick up with bare hands. Therefore, \(\boxed{t \approx 24}\) minutes.