A) \[\frac{3}{2}\]minutes
B) 1 minutes
C) \[\frac{1}{2}\]minutes
D) \[\frac{1}{4}\]minutes
Correct Answer: A
Solution :
[a] : According to New tons law of cooling\[\frac{{{T}_{1}}-{{T}_{2}}}{t}=K\left( \frac{{{T}_{1}}+{{T}_{2}}}{2}-{{T}_{s}} \right)\]where \[{{T}_{s}}\] is the surrounding temperature. For the first case, \[{{T}_{1}}=91{}^\circ C,{{T}_{2}}=79{}^\circ C,{{\text{T}}_{s}}=25{}^\circ C,t=2\min \] \[\therefore \]\[\frac{{{91}^{o}}C-{{79}^{o}}C}{2\min }=K\left( \frac{{{91}^{o}}C+{{79}^{o}}C}{2}-{{25}^{o}}C \right)\]...(i) or \[\frac{{{12}^{o}}C}{2\min }=K({{60}^{o}}C)\] For the second case, \[{{T}_{1}}=91{}^\circ C,\text{ }{{T}_{2}}=79{}^\circ C,{{T}_{s}}=5{}^\circ C,t=?\] \[\therefore \]\[\frac{{{91}^{o}}C-{{79}^{o}}C}{t}=K\left( \frac{{{91}^{o}}C+{{79}^{o}}C}{2}-{{5}^{o}}C \right)\]...(ii) or\[\frac{{{12}^{o}}C}{t}=K({{80}^{o}}C)\] Dividing eqn. (i) by equation (ii), we get \[\frac{t}{2\min }=\frac{{{60}^{o}}C}{{{80}^{o}}C}\]or\[t=\frac{3}{4}(2\,min)=\frac{3}{2}\min \]
You need to login to perform this action.
You will be redirected in
3 sec
