Question

A cup of coffee at 181 degrees is poured into a mug and left in a room at 66 degrees. After 6 minutes, the coffee is 139 degrees. Assume that the differential equation describing Newton's Law of Cooling is (in this case) dT/dt=k(T-66).
1) What is the temperature of the coffee after 16 minutes?
2) After how many minutes will the coffee be 100 degrees?

Answer 1

Answer:

Step-by-step explanation:

$\frac{dT}{dt} =k(t-66)\\T=\int\ {k(t-66)} \, dt=K(\frac{t^2}{2} -66)+c\\when t=0,T=181\\181=K(0-66)+c\\181=-66k+c\\when t=6,T=139\\139=k(\frac{6^2}{2} -66)+c\\139=-48k+c\\181-139=-66k+48k\\-42=-18k\\7=3k\\k=\frac{7}{3} \\139=-48*\frac{7}{3} +c\\c=139+112=251\\T=\frac{7}{3} (t-66)+251\\now complete the question$