Question 1100283: According to Newton's Law of Cooling, if a body with temperature Upper T 1T1 is placed in surroundings with temperature Upper T 0T0, different from that of Upper T 1T1, the body will either cool or warm to temperature T(t) after t minutes, whereT(t)equals=Upper T 0T0plus+(Upper T 1T1minus−Upper T 0T0)e Superscript negative kte−kt.
A cup of coffee with temperature 160 degrees°F is placed in a freezer with temperature 0 degrees°F. After 15 minutes, the temperature of the coffee is 47 degrees°F. Use Newton's Law of Cooling to find the coffee's temperature after 20 minutes.
Answer by jorel1380(3719)   (Show Source):
You can put this solution on YOUR website! First, find t(15)
t(15)=0+(160-0)e^-15k
47=160e^-15k
0.29375=e^-15k
ln 0.29375=ln e^-15k=-15k ln e=-15k
k=0.081668
So, for t(20), we have:
t(20)=0+(160-0)e^-(20*.081668)
t(20)=31.2433 deg. F as the temperature after 20 mins.
☺☺☺☺
|
|
|
