A) \[40{}^\circ C\]
B) \[80{}^\circ C\]
C) \[586{}^\circ C\]
D) \[313{}^\circ C\]
Correct Answer: D
Solution :
The average kinetic energy of molecules is \[KE=\frac{3}{2}kT\]where k is Boltzmann constant, T the absolute temperature. \[\therefore \] \[\frac{K{{E}_{1}}}{K{{E}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] \[\Rightarrow \] \[{{T}_{2}}=\frac{K{{E}_{2}}}{K{{E}_{1}}}{{T}_{1}}\] Given, \[K{{E}_{2}}=2K{{E}_{1}}\], \[{{T}_{1}}={{20}^{o}}C=273+20=293\,K\]. \[\Rightarrow \] \[{{T}_{2}}=2\times 293=586\ K\] In centigrade, \[{{T}_{2}}=586-273={{313}^{o}}C\]
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