A) \[27{}^\circ C\]
B) \[40.3{}^\circ C\]
C) \[23.3{}^\circ C\]
D) \[33.3{}^\circ C\]
Correct Answer: D
Solution :
According to Newtons law of cooling \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}\propto \left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] where \[{{\theta }_{0}}\] is the temperature of the surrounding. First case: \[\frac{50-45}{5}\propto \left( \frac{50+45}{2}-{{\theta }_{0}} \right)\] ?(i) Second case: \[\frac{45-41.5}{5}\propto \left( \frac{45+41.5}{2}-{{\theta }_{0}} \right)\] ?(ii) Dividing Eq.(i) by Eq. (ii), we get \[\frac{(50-45)}{5}\times \frac{5}{(45-41.5)}=\frac{\left( \frac{50+45}{2}-{{\theta }_{0}} \right)}{\left( \frac{45+41.5}{2}-{{\theta }_{0}} \right)}\] or \[\frac{5}{3.5}=\frac{(47.5-{{\theta }_{0}})}{(43.25-{{\theta }_{0}})}\] or \[216.25-5{{\theta }_{0}}=166.25-3.5{{\theta }_{0}}\] or \[1.5{{\theta }_{0}}=50\] or \[{{\theta }_{0}}=\frac{50}{1.5}=33.3{{\,}^{o}}C\]
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