A) \[{{5}^{o}}C\]
B) \[{{10}^{o}}C\]
C) \[{{15}^{o}}C\]
D) \[{{20}^{o}}C\]
Correct Answer: B
Solution :
According to Newton's law of cooling \[\frac{{{\theta }_{1}}-{{\theta }_{2}}}{t}=K\left[ \frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}-{{\theta }_{0}} \right]\] In the first case, \[\frac{(60-50)}{10}=K\,\left[ \frac{60+50}{2}-{{\theta }_{0}} \right]\] \[1=K\,(55-\theta )\] ?.(i) In the second case, \[\frac{(50-42)}{10}=K\,\left[ \frac{50+42}{2}-{{\theta }_{0}} \right]\] \[0.8=k\,(46-{{\theta }_{0}})\] ?.(ii) Dividing (i) by (ii), we get \[\frac{1}{0.8}=\frac{55-\theta }{46-\theta }\] or \[46-{{\theta }_{0}}=44-0.8\theta \] Þ \[{{\theta }_{0}}={{10}^{o}}C\]
You need to login to perform this action.
You will be redirected in
3 sec
