Question

A body at ${50}^{0}C$ cools in a surrounding maintained at ${30}^{0}C$. The temperature at which the rate of cooling is half that of beginning is:

• A. ${16.32}^{0}C$
• B. ${26.3}^{0}C$
• C. ${40}^{0}C$
• D. ${46.3}^{0}C$

Answer 1

The correct option is C ${40}^{0}C$

The rate of cooting $\left(R\right)$ of a body from Newton's Law of cooting is:-

$R=\frac{dt}{dx}=-k(T-T_0)T:$ temp of body at time $t$

and $T_0:$ Surrounding $r$ temp

Given initially $T={30}^{o}C$ and rate in $R$

when temp drops to half $(R^{\prime }=R/2)$ then temp of body be $T^{\prime }$.

so, we have:-

$R=-4({50}^{o}C-{30}^{o}C)=-20k{}^{o}c/s$

and $\frac{R}{2}=-x(T^{\prime }-{30}^{o}C){}^{o}c/s$

from above

$\Rightarrow -2k=-1\times k(T^{\prime }-{30}^{o}C)\Rightarrow T^{\prime }=10+30={40}^{o}C$ (required)