Answer:
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
Explanation:
We start from:
[tex]\frac{dT}{dt}=\frac{-1}{50}(T-21)[/tex]
Separating variables:
[tex]-50dT=(T-21)dt[/tex]
[tex]-50\frac{dT}{T-21}=dt[/tex]
Integrating with initial conditions:
[tex]-50\int\limits^{T}_{91} {\frac{1}{T-21} } \, dT= \int\limits^t_{0} {} \, dt[/tex]
[tex]-50ln(\frac{T-21}{91-21})=t[/tex]
[tex]ln(\frac{T-21}{71})=\frac{-t}{50}[/tex]
Isolating T:
[tex]\frac{T-21}{70} =e^\frac{-t}{50} }[/tex]
[tex]T=21+70e^{\frac{-t}{50} }[/tex]
You may note that when t is zero the temperature is 91 ºC, as is specified by the problem. As well, when t is bigger (close to infinite), the temperature tends to be the room temperature (21 ºC)
