A cup of coffee is made with boiling water at 100 C and stands in a room where the temperature is 20 C. The change in temperature, H in degrees C , with respect to time, t in minutes, is given by the following differential equation.

Question

A cup of coffee is made with boiling water at 100 C and stands in a room where the temperature is  20 C. The change in temperature, H  in degrees C , with respect to time, t in minutes, is given by the following differential equation.

anonymous · Answer

the equation is $$\frac{ dH }{ dt }=-k(H-20)$$

dumbcow · Answer

separate variables
$$\frac{dH}{H-20} = -k dt$$
integrate both sides

anonymous · Answer

that's where i was having trouble. would the integral be ln(H-20)=-k^2/2?

dumbcow · Answer

ahh no right side you are integrating for variable "t"
k is a constant

--> ln(H-20) = -kt + C

dumbcow · Answer

always add the "+C" to right side

anonymous · Answer

oohhh okay. and the next part of the problem says "Solve this differential equation. If the coffee cools to 90  in 3  minutes, how long will it take to cool to 60 degrees?" would i be plugging something into the t and H to solve for C?

dumbcow · Answer

yes t = 3 , H = 90

you also know when t=0 H = 100

dumbcow · Answer

you have 2 constants to find ... both C and k
C from initial conditions
k from rate of cooling ... H = 90 after 3 min

anonymous · Answer

oh okay! i think i got what to do. thanks! :)

dumbcow · Answer

yw