Newton's Law of Cooling

Question

A cup of coffee with temperature 120 degrees° F is placed in a freezer with temperature 0 degrees°F.

After 55 minutes, the temperature of the coffee is 80 degrees°F. Use​ Newton's Law of Cooling to find the​ coffee's temperature after 20 minutes.

Answer 1

Newton's Law of Cooling: T(t) = T_s + (T₀ - T_s) e^-kt

T(t) = temperature at time t

T₀ = initial temperature =120°F

T_s = surrounding temperature = 0°F

solve for k using 55 minutes and 80°F

80 = 0 + (120-0) * e^-k*55

80/120 = e^-k*55

ln(80/120) = ln (e^-k*55)

-0.4054651081 = -k*55

-0.00737209287 = -k

k = 0.00737209287

Now solve for T(t) for 20 minutes

T(t) = 0 + (120-0) * e^{-0.00737209287*20 =} 103.54°F