Within 10 minute time a body cooled from 100 degrees to 60 degrees. The temperature of the outside environment is a constant $A=20$ degrees.When will the body cool to 25 degrees?
Basically I only know that $P(0)=A+C,$ where A is 20 degrees and because at $t=0$ the temperature is 100 degrees, it follows that $C=80.$ Then at what time $P(t)=20+80 \cdot e^{-kt}=25.$ It is known that at time $t=0 \ \ P(10)=20+80 \cdot e^{-10k}=60 \Rightarrow e^{-10k}=\frac{1}{2} \Rightarrow -10k = -ln(2).$ Natural logarithm of 2 is approximately $0,7$ so $k=0,07.$ Then at what time $P(t)=20+80 \cdot e^{-0,07t}=25? \ \ e^{-0,07t}=\frac{1}{16} \Rightarrow -0,07t=-ln(16) \Rightarrow t=\frac{2,77}{0,07}=39,57.$ Is it correct?
Well, using the exact value of $k=\frac{1}{10}\ln2$, we get $$t=\ln 16 \div (\frac{1}{10}\ln 2) = 40 .$$