Question

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

• A. 16.$3^0$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 cooling $\left(R\right)$ of a body from Newton's Law of cooling is :-

$R=\frac{dT}{dt}=-k(T-T_0)$

T: Temp. of body at time t

and $T_0:$ surounding r temp.

Given, initialy, $T={30}^{\circ }C$ and rate is R.

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

so, we have :-

$R=-k({50}^{\circ }C-{30}^{\circ }C)=-20k$ ${}^{\circ }C/s$

and $\frac{R}{2}=-k(T^{\prime }-{30}^{\circ }C)$ ${}^{\circ }C/s$

from above,

$\Rightarrow -20k=-2\times k(T^{\prime }-{30}^{\circ }C)\Rightarrow T^{\prime }=10+30={40}^{\circ }C$ (Reqd)