Answer:
The model for the temperature of the drink can be written as
[tex]T=72-20e^{-0.08t}[/tex]
Step-by-step explanation:
For a cold drink in a hotter room, we can say that the rate of change of temperature of the drink is proportional to the difference of temperature between the drink and the room.
We can model that in this way
[tex]\frac{dT}{dt}=k*(T_r-T)[/tex]
If we rearrange and integrate
[tex]\int\frac{dT}{(T-Tr)} =-k*\int dt\\\\ln(T-T_r)=-kt+C1\\\\T-T_r=Ce^{-kt}\\\\T=T_r+Ce^{-kt}[/tex]
We know that at time 0, the temperature of the drink was 52°F. Then we have:
[tex]T=T_r+Ce^{-kt}\\\\52=72+Ce^0=72+C\\\\C=-20[/tex]
We also know that at t=2, T=55°F
[tex]T=T_r+Ce^{-kt}\\\\55=72-20e^{-k*2}\\\\e^{-k*2}=(72-55)/20=0.85\\\\-2k=ln(0.85)=-0.1625\\\\k=0.08[/tex]
The model for the temperature of the drink can be written as
[tex]T=72-20e^{-0.08t}[/tex]
