Question

A glass flask has volume 1x10^-4 m3.It is filled with a liquid at 30 °C.If the temperature of the system is raised to100 °C, how much of the liquid will overflow. (Coefficient ofvolume expansion of glass is 1.2 x 10^-5(°C)^-1 while that of the liquid is 75 x 10^-5(°C)^-1.​

Answer 1

Answer:

Volume of liquid that overflow is given as $\Delta V = 5.17 \times 10^{-6} m^3$

Explanation:

When liquid is filled in the glass container then we know that liquid and glass both expands due to which the apparent expansion coefficient of liquid is given as

$\gamma_{app} = \gamma_{real} - \gamma_{glass}$

$\gamma_{app} = 75 \times 10^{-5} - 1.2 \times 10^{-5}$

$\gamma_{app} = 73.8 \times 10^{-5}$

now the volume of liquid that overflows is given as

$\Delta V = V_o \gamma_{app} \Delta T$

$\Delta V = (1\times 10^{-4})(73.8 \times 10^{-5})(100 - 30)$

$\Delta V = 5.17 \times 10^{-6} m^3$

#Learn

Topic : Thermal expansion

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Answer 2

Answer:

Volume of liquid that overflow is given as \Delta V = 5.17 \times 10^{-6} m^3ΔV=5.17×10

−6

m

3

Explanation:

When liquid is filled in the glass container then we know that liquid and glass both expands due to which the apparent expansion coefficient of liquid is given as

\gamma_{app} = \gamma_{real} - \gamma_{glass}γ

app

=γ

real

−γ

glass

\gamma_{app} = 75 \times 10^{-5} - 1.2 \times 10^{-5}γ

app

=75×10

−5

−1.2×10

−5

\gamma_{app} = 73.8 \times 10^{-5}γ

app

=73.8×10

−5

now the volume of liquid that overflows is given as

\Delta V = V_o \gamma_{app} \Delta TΔV=V

o

γ

app

ΔT

\Delta V = (1\times 10^{-4})(73.8 \times 10^{-5})(100 - 30)ΔV=(1×10

−4

)(73.8×10

−5

)(100−30)

\Delta V = 5.17 \times 10^{-6} m^3ΔV=5.17×10

−6

m

3

#Learn

Topic : Thermal expansion

Hope it was helpful

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