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.



0.0439

0.0532

0.0613

0.0872

Answer 1

0.0439 this is the answer