Step 1 :First, we need to find the values of the constants C and k. We know that at t=0, the temperature of the object is 150 degrees. So we can substitute these values into the equation: \(150 = Ce^0 + 77\)
Step 2 :Solving for C, we get \(C = 150 - 77 = 73\)
Step 3 :Next, we know that after 10 minutes, the temperature of the object is 120 degrees. So we can substitute these values into the equation: \(120 = 73e^{-10k} + 77\)
Step 4 :Solving for e^{-10k}, we get \(e^{-10k} = \frac{43}{73}\)
Step 5 :Taking the natural logarithm of both sides, we get \(-10k = \ln\left(\frac{43}{73}\right)\)
Step 6 :Solving for k, we get \(k = -\frac{\ln\left(\frac{43}{73}\right)}{10}\)
Step 7 :Now that we have the values of C and k, we can find the time it takes for the temperature of the object to reach 80 degrees. We set the function equal to 80 and solve for t: \(80 = 73e^{-kt} + 77\)
Step 8 :Solving for e^{-kt}, we get \(e^{-kt} = \frac{3}{73}\)
Step 9 :Taking the natural logarithm of both sides, we get \(-kt = \ln\left(\frac{3}{73}\right)\)
Step 10 :Solving for t, we get \(t = -\frac{\ln\left(\frac{3}{73}\right)}{k}\)
Step 11 :Substituting the value of k we found earlier, we get \(t = 10 \frac{\ln\left(\frac{3}{73}\right)}{\ln\left(\frac{43}{73}\right)}\)
Step 12 :Using a calculator, we find that t is approximately 47. Therefore, it will take about 47 minutes for the temperature of the object to reach 80 degrees. So, \(\boxed{t \approx 47}\)