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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 removing it from the heat it cools to 60°C in 12 minutes. What is the substance’s cooling rate when the surrounding air temperature is 50°C?
Round the answer to four decimal places.

Respuesta :

nobillionaireNobley
T(t) = Ts + (T0 − Ts)e^−kt
60 = 50 + (80 - 50)e^-12k
30e^-12k = 60 - 50 = 10
e^-12k = 10/30 = 1/3
-12k = ln(1/3)
k = ln(1/3)/-12 = 0.0916

k = 0.0916

Answer:

0.0916

Step-by-step explanation:

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