The temperature T t of a body at time t = 0 is 160 ° F and it decreases continuously as per the differential…

Given: $\frac{dT}{dt}=-K\left(T-80\right)$

$\Rightarrow \frac{dT}{\left(T-80\right)}=-Kdt$

$\Rightarrow \underset{160}{\overset{T}{\int }}\frac{dT}{\left(T-80\right)}=\underset{0}{\overset{t}{\int }}-Kdt$

$\Rightarrow {\left[\log \left|T-80\right|\right]}_{160}^T=-Kt$

$\Rightarrow \log \left|T-80\right|-\log 80=-Kt$

$\Rightarrow \log \left|\frac{T-80}{80}\right|=-Kt$

$\Rightarrow T=80+80e^{-Kt}$

Now, using the value $T\left(15\right)={120}^{°}$ we get,

$\Rightarrow 120=80+80e^{-K·15}$

$\Rightarrow \frac{40}{80}=e^{-15k}$

$\Rightarrow e^{-15k}=\frac{1}{2}$

$\therefore T\left(45\right)=80+80e^{-45k}$

$\Rightarrow T\left(45\right)=80+80{\left(e^{-15k}\right)}^3$

$\Rightarrow T\left(45\right)=80+80\times \frac{1}{8}$

$\Rightarrow T\left(45\right)={90}^{°} F$