A) \[16{}^\circ C\]
B) \[26{}^\circ C\]
C) \[36{}^\circ C\]
D) \[21{}^\circ C\]
E) \[31{}^\circ C\]
Correct Answer: B
Solution :
According to Newtons law of cooling \[\frac{({{\theta }_{1}}-{{\theta }_{2}})}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right)\] \[\therefore \] \[\frac{(62-50)}{10}=K\left( \frac{62+50}{2}-{{\theta }_{0}} \right)\] \[\frac{12}{10}=K(56-{{\theta }_{0}})\] ?(i) For further cooling \[\frac{(50-42)}{10}=K\left( \frac{50+42}{2}-{{\theta }_{0}} \right)\] \[\frac{8}{10}=K(46-{{\theta }_{0}})\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{12}{8}=\frac{(56-{{\theta }_{0}})}{(46-{{\theta }_{0}})}\] \[3(46-{{\theta }_{0}})=2(56-{{\theta }_{0}})\] \[138-2{{\theta }_{0}}=112-2{{\theta }_{0}}\] \[\therefore \] \[{{\theta }_{0}}=26{}^\circ C\]
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