Answer to Question #124022 in Differential Equations for NK

Question

2. (a) 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

Given, $T_{out}=5^\circ C=278K,T_0=50^\circ C=323K$

At t=0.5 hour $T=293K$ ,and at some time t, $T=283K$

Now, from Newton's law of cooling we get

$$T-T_{out}=(T_0-T_{out})e^{-kt}$$

Thus,

$$293-278=(323-278)e^{-30k}\implies e^{-30k}=\frac{1}{3}\implies k=3.66\times 10^{-2}$$

Also,

$$283-278=(323-278)e^{-kt}\implies t=\frac{1}{k}\ln(9)=60$$

Hence, required time is 60 mins.