A) greater than T minutes
B) equal to T minutes
C) less than T minutes
D) equal to T/2 minutes
Correct Answer: B
Solution :
According to Newton's law of cooling rate of cooling \[\frac{\Delta \theta }{t}=K[{{\theta }_{av}}-\theta ]\] or \[\frac{({{\theta }_{av}}-\theta )}{\Delta \theta }=\frac{1}{K}\] \[\therefore \] \[\left( \frac{{{\theta }_{av}}-\theta }{\Delta {{\theta }_{1}}} \right){{t}_{1}}=\left( \frac{{{\theta }_{av}}-\theta }{\Delta {{\theta }_{2}}} \right){{t}_{2}}\] \[\because \] \[{{\theta }_{a{{v}_{1}}}}=\frac{62+61}{2}=61.5\] \[\Delta \theta =62-61={{1}^{o}}C,\] \[{{t}_{1}}=T\] \[{{\theta }_{a{{v}_{1}}}}=\frac{46+45.5}{2}=45.75,\] \[\because \] \[\Delta {{\theta }_{2}}=46-45.5={{0.5}^{o}}C\] \[\therefore \] \[\frac{(61.5-30)T}{1}=\frac{(45.75-30){{t}_{2}}}{0.5}\] or \[{{t}_{2}}=T\]s
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