1a. You pour a cup of coffee. When it is poured it is at D=130F above room temp. 3 mins later it has cooled to D=117F above room temp. As the coffee cools the instantaneous rate of change of D with respect to time t min is directly proportional to D. Find the constants in the equation.
b. The coffee in a. is drinkable if it is at least 50F above room temp. Will it still be drinkable if you let it sit for 15 min after you pour it?
So for a. I wrote an equation dD/dt=kD , where k is the constant
then when I integrate it I get ln (abs) D= kt+C <=> D=e^kt+C <=>D=Ce^kt
Since D=130 when t=0, I solved for C, which is 130.
Since D=117 when t=3, I solved for k, which is -0.0351
My equation is now D=130e^-0.0351t
Please help me with part b?
February 18th 2013, 05:09 AM
Re: Coffee Cup Problem
This is my solution for part b, when t=15 minutes, D=130e^-0.035*15=70.99F above room temp, which is drinkable. Is it correct?