A) \[{{45}^{o}}C\]
B) \[{{20}^{o}}C\]
C) \[{{42}^{o}}C\]
D) \[{{10}^{o}}C\]
Correct Answer: A
Solution :
Let the temperature of the surrounding is\[{{t}^{o}}C\]. For first case, \[\frac{(70-60)}{5\min }=k({{65}^{{}^\circ }}C-{{t}^{{}^\circ }}C)\] (\[{{65}^{o}}\] is average of \[{{70}^{o}}C\] and\[{{60}^{o}}C\]) \[\frac{10}{5\min }=K({{65}^{{}^\circ }}C-{{t}^{{}^\circ }}C)\,\,\,\,\,\,\,.....(i)\] For second case, \[\frac{(60-54)}{5\min }=k(57-t)\,\,\,\,\,\,\,\,\,\,.....(ii)\] (\[{{57}^{o}}C\] is average of \[{{60}^{o}}C\] and \[{{54}^{o}}C\]) From Eqs. (i)/(ii), \[\frac{10}{6}=\frac{(65-t)}{(57-t)}\] Solving, we get \[t={{45}^{o}}C\]
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