Question

Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 190 degrees Fahrenheit when freshly poured, and 2 minutes later has cooled to 170 degrees in a room at 66 degrees, determine when the coffee reaches a temperature of 145 degrees.

Answer 1

Answer:

t=5.18 minute

Step-by-step explanation:

Using Newton's law of cooling

$\frac{dT}{dt}=-k(T-T_{s})$

T is temperature of a cup of coffee at any time t $T_{s}$, is the temperature of the surrounding and k is a constant of proportionality in this negative because temperature is decreasing.

$T(0)=190\\T(1)=170\\T_{s}=66$

$\frac{dT}{dt}=-k(T-T_{s})\\\frac{dT}{(T-T_{s})}=-k*dt\\\int\limits {\frac{1}{T-T_{s} } } \, dT=-\int\limits {k} \, dt\\ln(T-T_{s})=k*t+C\\e*ln(T-T_{s})=e^{k*t+C} \\T-T_{s}=e^{k*t}*e^{C}\\e^{C}=C\\T-T_{s}=e^{k*t}*C\\T=T_{s}+e^{k*t}*C$

To find constant knowing the time and the temperature the first step of the change of energy in cup of coffee

$T=T_{s} +e^{-k*t}*C\\ C=190-66=124\\170=66+124*e^{-k*2}\\ 170-66=124*e^{-k*2}\\ln(\frac{104}{124})=ln*e^{-k*2}\\-0.175=-k*2\\k=0.08794$

Now using the constant of decreasing can find the time to be a 145 temperature the cup of coffee

$145=66+124*e^{-k*t} \\145-66=124*^{-0.087*t}$

$ln(0.637)=ln*e(-0.087*t)\\-0.45=-0.087*t\\t=5.18minutes$