A) \[20{}^\circ \text{ }C\]
B) \[30{}^\circ C\]
C) \[15{}^\circ C\]
D) \[10{}^\circ C\]
Correct Answer: D
Solution :
According to Newtons law of cooling \[\frac{{{\theta }_{2}}-{{\theta }_{1}}}{t}=K\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{s}} \right]\]where, \[{{\theta }_{s}}\] is the temperature of the surroundings. \[\frac{60-50}{10}=K\left[ \frac{60+50}{2}-{{\theta }_{s}} \right]\] \[1=K[55-{{\theta }_{s}}]\] ?(i) Similarly,\[\frac{50-42}{10}=K(46-{{\theta }_{s}})\] \[\frac{8}{10}=K(46-{{\theta }_{s}})\] ?(ii) Dividing Eq. (i) by Eq. (ii), we get \[\frac{10}{8}=\frac{K(55-{{\theta }_{s}})}{K(46-{{\theta }_{s}})}\]\[\Rightarrow \]\[{{\theta }_{s}}={{10}^{o}}C\]
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