A) \[13{}^\circ C\]
B) \[15{}^\circ C\]
C) \[18{}^\circ C\]
D) \[20{}^\circ C\]
Correct Answer: D
Solution :
: Let the temperature of surrounding be\[\theta {}^\circ C\] \[\therefore \] In first case, \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left( \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-\theta \right)\] Or \[\frac{(100-80)}{5}=K\left( \frac{100+80}{2}-\theta \right)\] or\[4=K(90-\theta )\] ............... (i) In second case, temperature cools from\[80{}^\circ C\] to\[65{}^\circ C\]in\[(10-5)=5\]minutes. \[\therefore \]\[\frac{80-65}{5}=K\left( \frac{80+65}{2}-\theta \right)\] or\[3=K(72.5-\theta )\] ............... (ii) or \[\frac{4}{3}=\frac{90-\theta }{72.5-\theta }\] or \[270-3\theta =290-4\theta \] or\[\theta =290-270~~or\text{ }\theta =20{}^\circ C\].
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