Question

Certain quantity of water cools from 70$^{\circ}$C to 60$^{\circ}$C in the first 5 minutes and to 54$^{\circ}$C in the next 5 minutes. The temperature of the surroundings is

• A. $ 45 ^\circ C $
• B. $ 20 ^\circ C $
• C. $ 42^ \circ C $
• D. $ 10 ^\circ C $

Answer 1

The Correct Option is A

Let $ T_s $ be the temperature of the
surroundings.
According to Newton's law of cooling
$\frac{T_1 - T_2}{t} = K \bigg( \frac{T_1+T_2}{2} - T_s \bigg)$
For first 5 minutes,
$T_1= 70 ^\circ C , T_2 = 60 ^\circ C , t=5 minutes $
$\therefore \, \, \, \frac{70 - 60 }{ 5} = K \bigg( \frac{70+60}{2} - T_s \bigg) $
$ \frac{10}{5} = K(65-T_s )$
For next 5 minutes,
$ T_1 =60 ^\circ C , T_2 =54 ^\circ C , t= 5 minutes $
$\therefore \, \, \, \frac{ 60-54}{5} = K \bigg( \frac{60+54} {2} - T_s \bigg) $
$\frac{6}{5}= K(57 - T_5)$
Divide eqn. (i) by eqn. (ii), we get
$ \frac{5}{3} = \frac{65 - T_s}{57- T_s}$
$ 285 -5T_s =195 -3 T_s$
$ 2T_s = 90 \, \, \, or \, \, \, \, T_s = 45 ^ \circ C$