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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 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

1 Answer

1 vote
0.0439 this is the answer

User Chris Kerekes
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