Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be modeled by the equation T(t) = T, + (To – T,)e-kt, where T, 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 12 min. What is the substance's cooling rate when the surrounding air temperature is 50°C? Round the answer to four decimal places. O 0.0687 O 0.0732 O 0.0813 O 0.0916

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Chapter6: Exponential And Logarithmic Functions
Section6.1: Exponential Functions
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Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be
modeled by the equation T(t) =T, + (To – T)e-kt, where T, 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 12 min.
What is the substance's cooling rate when the surrounding air temperature is 50° C?
Round the answer to four decimal places.
O 0.0687
O 0.0732
O 0.0813
O 0.0916
Transcribed Image Text:Newton's law of cooling states that for a cooling substance with initial temperature To, the temperature T(t) after t minutes can be modeled by the equation T(t) =T, + (To – T)e-kt, where T, 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 12 min. What is the substance's cooling rate when the surrounding air temperature is 50° C? Round the answer to four decimal places. O 0.0687 O 0.0732 O 0.0813 O 0.0916
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