Q:

# 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 f(t)=Ce(−kt)+80 represents the situation, where t is time in minutes, C is a constant, and 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?

Accepted Solution

A:
Answer: After 18.05 minutes, the temperature of steel becomes 100 degrees.Step-by-step explanation:Since we have given that Initial temperature = 2500At t = 0, we get that $$f(t)=Ce^{-kt}+80\\\\2500=C+80\\\\2500-80=C\\\\2420=C$$After 2 minutes, the temperature of the steel is 1500 degrees.so, it becomes,$$1500=2420e^{-2k}+80\\\\1500-80=2420e^{-2k}\\\\\dfrac{1420}{2420}=e^{-2k}\\\\0.587=e^{-2k}\\\\\ln 0.587=-2k\\\\-0.533=-2k\\\\k=\dfrac{0.533}{2}\\\\k=0.266$$So, We need to find the number of minutes when the temperature of steel would be 100 degrees.So, it becomes,$$100=2420e^{-0.266t}+80\\\\100-80=2420e^{-0.266t}\\\\20=2420e^{-0.266t}\\\\\dfrac{20}{2420}=e^{-0.266t}\\\\\ln \dfrac{20}{2420}=-0.266t\\\\-4.8=-0.266t\\\\t=\dfrac{4.8}{0.266}\\\\t=18.05$$Hence, after 18.05 minutes, the temperature of steel becomes 100 degrees.