Question

• A certain substance can cool itself from 110°C to 70°C in 15minutes. If the air’s temperature is 30°C, how long will it take to cool from 110°C to 50°C?

Answer 1

Newton's Law of Cooling 

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

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

$$T=T_s+Ce^{kt}$$

$T_s=30\degree C, T(0)=110\degree C$

$$110=30+C=>C=80\degree C$$

$$T=30+80e^{kt}$$

$T(0.25)=70\degree C$

$$70=30+80e^{0.25k}$$

$$e^{0.25k}=\dfrac{1}{2}$$

$$k=-4\ln 2$$

$$T=30+80e^{(-4\ln2) t}$$

If $T=50\degree C$

$$50=30+80e^{(-4\ln2) t}$$

$$e^{(-4\ln2) t}=\dfrac{1}{4}$$

$$(-4\ln2) t=-2\ln2$$

$$t=0.5\ hr$$

30 minutes.