Question

The temperature of a body falls from $46{}^{o}C$ to $28{}^{o}C$ in 8 minutes, when the body is placed in a surrounding of constant temperature of $10{}^{o}C$. Find the time taken by the body to reach $19{}^{o}C.$

• A. 10 min
• B. 6 min
• C. 8 min
• D. 12 min

Answer 1

The correct option is C 8 min

Using Newton's law of cooling,
$\frac{dT}{dt}=-k(T_t-T_s)$
For, first decrease from $46{}^{o}C$ to $28{}^{o}C$:
$\frac{18}{8}=-k(46-10)\Rightarrow k=-1/16$
For, second decrease from $28{}^{o}C$ to $19{}^{o}C$:
$\frac{9}{t}=-\left(\frac{-1}{16}\right)(28-10)\Rightarrow t=$ 8 min