A) \[40{}^\circ C\]
B) \[80{}^\circ C\]
C) \[586{}^\circ C\]
D) \[313{}^\circ C\]
Correct Answer: D
Solution :
We know that kinetic energy is given by \[E=\frac{3}{2}kT\] or \[E\propto T\] Hence, \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{T}_{1}}}{{{T}_{2}}}\] or \[\frac{{{E}_{1}}}{2{{E}_{1}}}=\frac{273+20}{{{T}_{2}}}\] or \[{{T}_{2}}=2\times 293=586K\] \[=586-273={{313}^{o}}C\]
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