Applications of Differential Equations
Temperature
An object which has a temperature of 100C is placed into air at 20C. Its temperature drops to 50C in 10 minutes. Express T(t).
Summarize the known information.
t = 0 T = 100C
t = 10 T = 50C
Tm= 20C
dT/dt = k (T – Tm)
What do you do next?
Plug Tm= 20C into the equation and separate the variables.
dT/dt = k (T – 20)
dT/(T – 20) = k dt
Now what?
Integrate and solve for constants.
ln(T – 20) = kt + C
T – 20 = Cekt
Using t = 0 T = 100C
C = 80
Using t = 10 T = 50C
k = – 0.981
The equation becomes:
T(t) = 80e-0.0981t + 20
