question_answer

A body initially at 80o C cools to 64o C in 5 minutes and to 52o C in 10 minutes. The temperature of the body after 15 minutes will be                                      [UPSEAT 2000; Pb. PET 2004]

A)            42.7 o C                                    

B)            35 o C

C)            47 o C                                       

D)            40 o C

Correct Answer: A

Solution :

                   According to Newton law of cooling                    \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\,\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right]\]                    For first process : \[\frac{(80-64)}{5}=K\left[ \frac{80+64}{2}-{{\theta }_{0}} \right]\]     ...(i)                    For second process : \[\frac{(80-52)}{10}=K\left[ \frac{80+52}{2}-{{\theta }_{0}} \right]\]            ...(ii)                    For third process : \[\frac{(80-\theta )}{15}=K\left[ \frac{80+\theta }{2}-{{\theta }_{0}} \right]\]              ?(iii) On solving equation (i) and (ii) we get \[K=\frac{1}{15}\] and \[{{\theta }_{0}}=24{}^\circ C\]. Putting these values in equation (iii) we get  \[\theta =42.7{}^\circ C\]