A) 8°C
B) 16°C
C) 32°C
D) 24°C
Correct Answer: D
Solution :
From Newton's law of cooling \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-\theta \right)\] \[{{\theta }_{1}}=\]initial temperature of the object \[{{\theta }_{2}}=\]final temperature of the object \[t=time\] \[K=\]coefficient of thermal conductivity \[\theta =\]temperature of the surrounding In the first case, \[\frac{{{80}^{o}}-{{64}^{o}}}{5}=K\left( \frac{{{80}^{o}}+{{64}^{o}}}{2}-\theta \right)\] ?..(1) In the second case, \[\frac{{{80}^{o}}-{{52}^{o}}}{10}=K\left( \frac{{{80}^{o}}+{{52}^{o}}}{2}-\theta \right)\] ?..(2) From Eqs. (1) and (2), we get \[\frac{16}{5}\times \frac{10}{28}=\frac{{{72}^{o}}-\theta }{{{66}^{o}}-\theta }\] By solving, we get \[\theta =24{}^\circ C\]
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