Answer to Question #104958 in Differential Equations for RJ

Question

According to Newton’s law of cooling, the rate at which a substance cools in moving
air is proportional to the difference between the temperature of the substance and that
of the air. If the temperature of the air is C
o
290 and the substance cools from
C
o
370 to C
o
330 in 10 minutes, find when the temperature will be C
o
295

Answer 1

According to Newton’s law of cooling, the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is 290°C and the substance cools from 370°C to 330°C in 10 minutes, find when the temperature will be 295°C.

Newton's law of cooling is $Q/t∝(T _ s ​ −T _ f ​ )$

Solving for it, we get;

$T(t)=T_ S ​ +(T_ 0 ​ –T_ S ​ )e –kt ​$

where; t is the time ;

$T_o= ​$  initial temperature of substance

$T_s=$ temperature of surrounding fluid

In this case

$T _ s ​ =290°C$ $;T _ o ​ =370°C$

For t=10 minutes and T=330°C---(given)

$To find: t for T=295°CT=295°C$

$330=290+80e^{-10k}330=290+80e −10k$

$\implies k =ln2/10⟹k=ln2/10$

$295=290+80e −kt$

$⟹e ^{kt} =16⟹t=4ln2/k$

$⟹t=40min. (Answer)$