A body at a temperature of `50^0F` is placed outdoors where the temperature is `100^0F` . If the rate of change of the temperature of a body is proportional to the temperature difference between the body and its surrounding medium. If after 5 min the temperature of the body is `60^0F` , find (a) how long it will take the body to reach a temperature of 75 `^0` F and (b) the temperature of the body after 20 min.

Explanation

Let T be the temperature of the body at time t and `T_(m)=100` (the temperature of the surrounding medium). We have `(dT)/(dt)=-k(T-T_(m))`
`(dT)/(dt)+kT=kT_(m)`,
where constant of proportionally. Thus,
`(dT)/(dt)+kT=100k`
The solution of differential equation is
`T=ce^(-kt)+100`................(1)
Since T=50 when t=0
from equation (1), `50=ce^(-k(0))+100`, or `c=-50`.
Sustituting this value in equation (1), we obtain
`T=-50e^(-kt)+100`.....................(2)
At t=t, we are given that `T=60`, hence, from equation (2),
`60=-50e^(-5k)+100`.
Solving for k, we obtain `-40=-50e^(-5k)` or `k=1/5"ln"40/50`
Substituting this value in equation (2), we obtain the temperature of the body at any time t as
`T=-50e^((1//5)"In"(4//5)t)+100`..............(3)
a) We required t when T=75. Substituting T=75 in equation (3), we have
from which we get t=15.53 min
b) We require T when t=20, Subsituting t=20 in equation (3) and then solving for T, we find
`T=-50e^((1//5)"ln"(4//5)(20))+100=79.52^(@)`