Answer: Option B
:
B
As the temperature differences are small, we can use Newton's law of cooling.
$\frac{d\theta }{dt}=-kt(\theta -{\theta }_0)$
$\frac{d\theta }{\theta -{\theta }_0}=-kdt$
or, ...(i)
Where k is a constant, $\theta$ is the temperature of the body at time t and ${\theta }_0={16}^{\circ }C$ is the temperature of the surrounding. We have,
${\int }_{{40}^{\circ }C}^{{36}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt\left(5min\right)$
$ln\frac{{36}^{\circ }C-{16}^{\circ }C}{{40}^{\circ }C-{16}^{\circ }C}=-kt\left(5min\right)$
or,
$k=-\frac{ln\left(\frac{5}{6}\right)}{5min}$
or, .
If t be the time required for the temperature to fall from 36${}^{\circ }$C to 32${}^{\circ }$C then by (i),
${\int }_{{36}^{\circ }C}^{{32}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt$
or, $ln\frac{{32}^{\circ }C-{16}^{\circ }C}{{36}^{\circ }C-{16}^{\circ }C}=-\frac{ln\left(\frac{5}{6}\right)}{5min}$
or, t=$\frac{ln\left(\frac{4}{5}\right)}{\left(ln\right(\frac{5}{6})}\times 5min$
= 6.1 min.
Alternative method
The mean temperature of the body as it cools from 40${}^{\circ }$C to 36${}^{\circ }$C is $\frac{{40}^{\circ }C+{36}^{\circ }C}{2}={38}^{\circ }C$ The rate of decrease of temperature is
$\frac{{40}^{\circ }C+{36}^{\circ }C}{5min}={0.80}^{\circ }C/min$ .
Newton's law of cooling is
$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$
or, -0.8${}^{\circ }$C/min = - k(38${}^{\circ }$C - 16${}^{\circ }$C)=-k(22${}^{\circ }$C)
or, $k=\frac{0.8}{22}mi{n}^{-1}$.
Let the time taken for the temperature to become 32${}^{\circ }$C be t.
During this period,.
The mean temperature is
Now,