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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.







0.0687

0.0732

0.0813

0.0916

Verified Answer :

HomertheGenius

t = 12 min,  T o = 80° C, T s = 50 °C,  T ( 12 ) = 60° C
60 = 50 + ( 80 - 50 ) · e^(-12k)
10 = 30 · e^(-12k)
e^(-12k) = 0.3333...
ln ( 0.333.. ) = - 12 k
- 12 k = - 1.1
k = ( - 1.1 ) : ( - 12 ) = 0.0916
Answer: D ) 0.0916

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