A body cools from 80 ^ o C to 60 ^ o in 5 minutes. The temperature...

The formula to calculate the rate of cooling of the object is given by

$\dfrac{dT}{dt} = K\left[\dfrac{{T}_{f} + {T}_{i}}{2} - {T}_{0}\right] . . . \left(1\right)$

For the first case, it can be written, using equation (1) that

$\begin{cases}\dfrac{80 - 60}{5} & = & K\left[\dfrac{80 + 60}{2} - 20\right] \\ 4 & = & 50K . . . \left(2\right)\end{cases}$

If $t$ is the required time for the second case, from equation (1), it can be written that

$\begin{cases}\dfrac{60 - 40}{t} & = & K\left[\dfrac{60 + 40}{2} - 20\right] \\ \dfrac{20}{t} & = & 30K . . . \left(3\right)\end{cases}$

Divide equation (2) by equation (3) and solve to calculate the required time.

$\begin{cases}\dfrac{4}{\dfrac{20}{t}} & = & \dfrac{50K}{30K} \\ & ⇒ & \dfrac{t}{5} = \dfrac{5}{3} \\ & ⇒ & t = \dfrac{25}{3}min = 500s\end{cases}$