• # question_answer A liquid cools from $50{}^\circ C$ to $45{}^\circ C$ in 5 minutes and from $45{}^\circ C$ to $41.5{}^\circ C$ in the next 5 minutes. The temperature of the surrounding is A) $27{}^\circ C$B) $40.3{}^\circ C$C)      $23.3{}^\circ C$D)      $33.3{}^\circ C$

According to Newton's law of cooling $ms\left( \frac{{{\theta }_{1}}-{{\theta }_{2}}}{t} \right)\propto \left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{t}-{{\theta }_{o}} \right]$ where ${{\theta }_{2}}=$ temperature of surrounding In first case,$ms\left( \frac{50-45}{50} \right)\propto \left[ \frac{50+45}{2}-{{\theta }_{o}} \right]$(1) In second case, $ms\left( \frac{45-41.5}{5} \right)\propto \left[ \frac{45+41.5}{2}-{{\theta }_{0}} \right]$           (2) Solving (1) and (2), we get${{\theta }_{o}}=33.3{{\,}^{o}}C$ Hence, the correction option is (d).