A) 135°C
B) 125°C
C) 112°C
D) 100°C
Correct Answer: C
Solution :
Initial volume \[{{V}_{1}}=47.5\] units Temperature of ice cold water \[{{T}_{1}}=0{}^\circ C=273\,K\] Final volume of \[{{V}_{2}}=67\] units Applying Charle?s law, we have \[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\] (where temperature \[{{T}_{2}}\] is the boiling point) or \[{{T}_{2}}=\frac{{{V}_{2}}}{{{V}_{1}}}\times {{T}_{1}}=\frac{67\times 273}{47.5}=385\,K=112{}^\circ C\]
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