Answer to Question #280572 in Differential Equations for Muffin

Question

A thermometer reading of 100 degrees C was plunged into a tub of frozen water. If the thermometer reads 70C after 5 seconds, what will be the reading after 15 seconds?

Answer 1

Let $T(t)$ be the thermometer reading after $t$ seconds. The surrounding temperature is $T_s=0\degree C,$ so Newton’s Law of Cooling states that

$$\dfrac{dT}{dt}=k(T-T_s)$$

$$\dfrac{dT}{dt}=kT$$

$$\dfrac{dT}{T}=kdt$$

$$T(t)=Ce^{kt}$$

$$T(0)=C=100$$

$$T(t)=100e^{kt}$$

$$T(5)=100e^{k(5)}=70$$

$$e^{5k}=0.7$$

$$T(15)=100e^{k(15)}=100(e^{5k})^3$$

$$=100(0.7)^3=34.3(\degree C)$$

The reading will be $34.3\degree C$ after 15 seconds.