The temperature of a liquid after being placed in a refrigerator is given by #T=5+95e^(-0.123t)# degree Celsius where t is the time in minutes Show that #(dT)/dt=c(T-5)# for some constant c. Find the value of c?

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Answer 1 · 2018-01-27T10:05:46

The value of #c=-0.123#

The equation is

#T=5+95*e^(-0.123t)#

#T-5=95*e^(-0.123t)#

Differentiating with respect to #t#

#(dT)/(dt)=-0.123*95e^(-0.123t)#

Plugging in the value of #95*e^(-0.123t)=T-5#

#(dT)/(dt)=-0.123(T-5)#

Comparing this to #(dT)/(dt)=c(T-5)#

The value of the constant #c=-0.123#