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Certain quantity of water cools from $70^o C$ to $60^o C$ in the first $5$ minutes and to $54^o C$ in the next $5$ minutes. The temperature of the surroundings is ..... $^oC$
$45$
$20$
$42$
$10$
Solution
Let $T_s$ be the temperature of the surroundings.
According to $Newton's$ law of cooling
$\frac{{{T_1} – {T_2}}}{t} = K\left( {\frac{{{T_1} + {T_2}}}{2} – {T_s}} \right)$
For first $5\,minutes,$
${T_1} = {70^ \circ }C,{T_2} = {60^ \circ }C,t = 5\,minutes$
$\therefore \frac{{70 – 60}}{5} = K\left( {\frac{{70 + 60}}{2} – {T_s}} \right)$
$\frac{{10}}{5} = K\left( {65 – {T_s}} \right)$ $…(i)$
For next $5\,minutes$
${T_1} = {60^ \circ }C,{T_2} = {54^ \circ }C,t = 5 minutes$
$\therefore \frac{{60 – 54}}{5} = K\left( {\frac{{60 + 54}}{2} – {T_s}} \right)$
$\frac{6}{5} = K\left( {57 – {T_s}} \right)$ $…(ii)$
Divide eqn. $(i)$ by eqn. $(ii)$ , we get
$\frac{5}{3} = \frac{{65 – {T_s}}}{{57 – {T_s}}}$
$285 – 5{T_s} = 195 – 3{T_s}$
$2{T_s} = 90\,\,or\,\,{T_s} = {45^ \circ }C$
