Question

A body at temperature 40C is kept in a surrounding of constant temperature 20C. It is observed that its temperature falls to 35C in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30C

• A. 14 min
• B. `28 min
• C. 56 min
• D. 112 min

Answer 1

Answer: (A)

14 min

from equation (14.3.)
$\Delta {\Theta }_1=\Delta {\Theta }_1{e}^{-kt}$
for the interval in which temperature falls from 40 to 35 C.
(35 - 20) = (40 - 20)$e^{-k10}$
$\Rightarrow e^{-10k}=\frac{3}{4}\Rightarrow k=\frac{ln\frac{4}{3}}{10}$
for the next interval
(30 - 20) = (35 - 20)$e^{-kt}$
$\Rightarrow$e^{-kt}\, =\, \displaystyle \frac{2}{3}\, \quad\, \quad\, \Rightarrow \, \quad\, kt\, =\, ln \displaystyle \frac{3}{2}\Rightarrow\, \quad\, \displaystyle \frac{\left ( ln \displaystyle \frac{4}{3}\right)t}{10}\, =\, \ln\, \displaystyle \frac{3}{2}\, \quad \quad\, \Rightarrow\, t\, =\, 10\, \displaystyle \frac{\left(ln\, \displaystyle \frac{3}{2})}{\left(ln\, \displaystyle \frac{4}{3} \right )}$ minute = 14.096 min
Aliter : (by approximate method)
for the interval in which temperature falls from 40 to ${35}^{\circ }$C
$<\Theta >=\frac{40+35}{2}={37.5}^{\circ }C$
from equation (14.4) $⟨\frac{d\Theta }{dt}⟩=-k(<\Theta >-{\Theta }_{\circ })$
$\Rightarrow \frac{({35}^{\circ }C-{40}^{\circ }C)}{10\left(min\right)}=-K({37.5}^{\circ }C-{20}^{\circ }C)$
$\Rightarrow K=\frac{1}{35}\left(mi{n}^{-1}\right)$
fortheinterval inwhich temperature. falls from ${35}^{\circ }C$ to ${30}^{\circ }C$
$<\Theta >=\frac{35+30}{2}={32.5}^{\circ }C$
from equation (14.4)
$\frac{{30}^{\circ }C-{35}^{\circ }C}{t}=-({32.5}^{\circ }C-{20}^{\circ }C)K$
$\Rightarrow$ required time, t = $\frac{5}{12.5}\times 35min=14min$