Differential Equations Application (Newton's law of cooling)

Question

Let T (t) denote the temperature of an object after t minutes and let Ts denote the temperature of its surroundings. Newton’s Law of Cooling states that dT/dt = k (T − Ts), where k is a constant. Suppose that you start serving −10 C ice cream at noon outside on a day when it is 25C. If the ice cream is −5C at 12:10, at what time will the ice cream be 0C?

Answer 1

dT/dt = -k(T-T_S) is separable (The sign is wrong if k is >0): dT/(T-T_S) = -kdt

I prefer do a definite integral, but you could do an indefinite integral an solve for the constant of integration using the initial condition, T = -10 degrees C at t = 0.

I get ln[(T-T_S)/(T₀-T_S)] = -kt

plug in T(10 min) = -5 degrees in order to find k in min^-1

Now solve for t with the k obtained and a T of 0 degrees.

Good luck!