A flask of volume 10^3 text{ cc} is completel | vedclass

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(B) The initial volume of the flask and mercury is $V_0 = 10^3 	ext{ cc}$.The change in temperature is $\Delta T = 100 \ ^\circ	ext{C} - 0 \ ^\circ\

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) The initial volume of the flask and mercury is $V_0 = 10^3 	ext{ cc}$.The change in temperature is $\Delta T = 100 \ ^\circ	ext{C} - 0 \ ^\circ	ext{C} = 100 \ ^\circ	ext{C}$.The increase in the volume of the glass flask is $\Delta V_g = V_0 \gamma_g \Delta T = 10^3 	imes 40 	imes 10^{-6} 	imes 100 = 4 	ext{ cc}$.The increase in the volume of the mercury is $\Delta V_m = V_0 \gamma_m \Delta T = 10^3 	imes 180 	imes 10^{-6} 	imes 100 = 18 	ext{ cc}$.The volume of mercury that overflows is the difference between the expansion of mercury and the expansion of the flask:$\Delta V_{	ext{overflow}} = \Delta V_m - \Delta V_g = 18 	ext{ cc} - 4 	ext{ cc} = 14 	ext{ cc}$.

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