Hello, turtle!

The temperature of an ingot of silver is 60°C above room temperature right now.

Twenty minutes ago, it was 70°C above room temperature.

How far above room temperature will the silver be 15 minutes from now?

I believe the temperatures in coordinate points are (0,60) and (-20,70) .Right!

and that Newton's Law is used to solve this problem [Ce^(kt) + T]

I may be wrong, but I don't believe that "T" is required.

We have: .d .= .Ce^{kt}

From (0,60), we have: .Ce^{k·0} .= .60 . → . C = 60

. . Hence, the function (so far) is: .d .= .60e^{kt}

From (-20,70), we have: .60e^{-20k) .= .70

Then: .e^{-20k} .= .7/6

Take logs: . ln(e^{-20k}) .= .ln(7/6)

. . (-20k)ln(e) .= .ln(7/6)

. . k .= .(-1/20)ln(7/6) .≈ .-0.0077

Hence, the function is: .d .= .60e^{-0.0077t}

When t = 15: . d .= .60e^{(-.0077)(15)} .≈ .53.5°C