Answer to Question #124017 in Differential Equations for carl

Question

A bowl of water has a temperature of 50◦C. It is put into a refrigerator where the temperature is 5◦C. After 0.5 hours, the water is stirred and its temperature is measured to be 20◦C. It is then left to cool further, take t in Newton’s law of cooling to be measured in
minutes. Use Newton’s law of cooling to predict when the temperature will be 10◦C.

Answer 1

Solution:

T(t)= T_s+(T₀-T_s)_e^-kt

T(t) =temperature of the object at time t

T_s= temperature of surrounding

T₀=temperature of the object at time

 t=0

k=constant

A bowl of water has a temperature of 50◦C. It is put into a refrigerator where the

 temperature is 5◦C.After 0.5 hours, 

the water is stirred and its temperature

is measured to be 20◦C. So we have:

20= 5+(50-5)_e^-0.5k

e^-0.5k=$\frac{1}{3}$

-(0.5)k= In$\frac{1}{3}$

k= In 9

It is then left to cool further. 

We should predict when the temperature will be 10◦C:

10= 5+(20-5)_e^{-(In 9)t}

e^{-(In 9)t} =$\frac{1}{3}$

t= 0.5

The temperature will be 10◦C in 30 minutes

(after we put a bowl of water into a

 refrigerator the second time).