The temperature T(t) of a body at time t = 0 is 160 ^ ° F and it...

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

$⇒ \dfrac{dT}{\left(T - 80\right)} = - Kdt$

$⇒ \mathop{∫}_{160}^{T}\dfrac{dT}{\left(T - 80\right)} = \mathop{∫}_{0}^{t} - Kdt$

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

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

$⇒ \log \left|\dfrac{T - 80}{80}\right| = - Kt$

$⇒ T = 80 + 80{e}^{- Kt}$

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

$⇒ 120 = 80 + 80{e}^{- K \cdot 15}$

$⇒ \dfrac{40}{80} = {e}^{- 15k}$

$⇒ {e}^{- 15k} = \dfrac{1}{2}$

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

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

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

$⇒ T\left(45\right) = {90}^{°}F$