Question

A glass flask of volume 200 cm3 is just filled with mercury at 20°C. The amount of mercury that will overflow
when the temperature of the system is raised to 100°C is (Yglass = 1.2 * 10%C”, Ymercun = 1.8 x 10*-4/C)
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Answer 1

Answer:

Explanation:

Delta V(glass)=beta(glass)×Vo×delta T

=1.2×10^-5×200×(100-20)

=0.19

Delta V (mercury)=beta (mercury)×Vo×delta T

=18×10^-5×200×(100-20)

=2.9

Delta V(mercury)- Delta V (glass)=2.9-0.19=2.69

=approx 2.7cm^3

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Answer 2

Given :

Initial volume of glass , $V=200\ cm^3$ .

$Y_{glass}=1.2\times 10^{-4}\ C^{-1}$

$Y_{mercury}=1.8\times 10^{-4}\ C^{-1}$

Initial temperature , $T_i=20^oC$ .

Final temperature , $T_f=100^oC$ .

To Find :

Amount of overflow of mercury .

Solution :

The amount of mercury that will overflow is proportional to the difference between them is :

$Y=Y_m-Y_g\\\\Y=0.6\times 10^{-4}\ C^{-1}$

Therefore , amount of mercury that will overflow is :

$V=200\times 0.6\times 10^{-4}\times (100-20)\\\\V=0.96\ cm^3$

The amount of mercury that will overflow is $0.96\ cm^3$ .