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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 12 min.What is the substance’s cooling rate when the surrounding air temperature is 50°C ?The substances cooling rate when the surrounding air temperature is 50C is 0.0916.0.06870.07320.08130.0916

Respuesta :

Answer:

k  = 0.0916

Step-by-step explanation:

T(t) = [tex]T_{s} + ( T_{o} - T_{s} )e^{-kt}[/tex]

from question; t = 12 mins , [tex]T_{s}[/tex] = 50 C , [tex]T_{o}[/tex] = 80 C , T = 60 C

60 = 50 + (80 - 50) [tex]e^{-12k}[/tex]

60-50 = 30 [tex]e^{-12k}[/tex]

10/30 =  [tex]e^{-12k}[/tex] (Taking natural Log of both sides)

In(0.3333) = In [tex]e^{-12k}[/tex]

-1.0986 = -12k

k = 0.0916

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