A) \[80{}^\circ C\]
B) \[444{}^\circ C\]
C) \[333{}^\circ C\]
D) \[171{}^\circ C\]
Correct Answer: D
Solution :
Since, vessel is open, therefore p remains constant so by Charles' law \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] \[\because \] \[V=\frac{m}{\rho }\] or \[V\propto \frac{1}{\rho }\] \[\Rightarrow \] \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{\rho }_{2}}}{{{\rho }_{1}}}\] \[\therefore \] \[\frac{{{\rho }_{2}}}{{{\rho }_{1}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] or \[{{\rho }_{1}}{{T}_{1}}={{\rho }_{2}}{{T}_{2}}\] \[\because \] \[{{\rho }_{1}}=\frac{m}{V},\] \[{{\rho }_{2}}=\frac{m-m/4}{V}=\frac{3m}{4V},\] \[{{T}_{1}}=60+273=333K\] \[{{T}_{2}}=\frac{{{\rho }_{1}}}{{{\rho }_{2}}}.{{T}_{1}}=\frac{4}{3}\times 333=444K\] or \[={{171}^{o}}C\]
You need to login to perform this action.
You will be redirected in
3 sec
