Will upvote please help Newton’s law of cooling states that for a cooling substance with initial temperature T0 , the temperature T(t) after t minutes can be
Mathematics
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Question
Will upvote please help
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
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
1 Answer

1. User Answers 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