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 25 min. What is the substance’s cooling rate when the surrounding air temperature is 50°C ? Round the answer to four decimal places.

Answer 1

T(t) = Ts + (T₀ - Ts)exp(-kt), substituting the values:
60 = 50 + (80 - 50)exp(-25k)
10/30 = exp(-25k)
k = 0.0439 °C/min