Question

Newton’s law of cooling states that for a cooling substance with initial temperature T0, the temperature T(t) after t minutes can be modeled by the equation T(t)=Ts+(T0−Ts)e−kt, where Ts is the surrounding temperature and k is the substance’s cooling rate.

A liquid substance is heated to 80°C. Upon being removed from the heat, it cools to 60°C in 15 min.

What is the substance’s cooling rate when the surrounding air temperature is 50°C?

Round the answer to four decimal places.

0.0687
0.0732
0.0813
0.0872

Answer 1

I guessed 0.0813 which was wrong. The answer is 0.0732

Answer 2

so why did you guess? They gave you the formula and the data point.
T(15) = 60, so
50 + 30e^(-15k) = 60
k = ln3/15 = 0.0732

Answer 3

It's 0.0916