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.