A) \[35.5{}^\circ C\]
B) \[50{}^\circ C\]
C) \[{{42.85}^{o}}C\]
D) \[{{45}^{o}}C\]
Correct Answer: C
Solution :
According of Newton law of cooling \[\frac{{{60}^{o}}-{{50}^{o}}}{10}=K\left( \frac{{{60}^{o}}+{{50}^{o}}}{2}-{{25}^{o}} \right)\] ?..(1) \[\frac{{{50}^{o}}-\theta }{10}=K\left( \frac{{{50}^{o}}+\theta }{2}-{{25}^{o}} \right)\] ??(2) Dividing equation (1) by (2), we obtain \[\frac{10}{{{50}^{o}}-\theta }=\frac{{{60}^{o}}}{\theta }\]or \[10\theta =3000-60\theta \] So, \[70\theta =3000\] or \[\theta =\frac{3000}{70}={{42.85}^{o}}C\]
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