# Mercury fills 90% of the volume of a sealed glass container at 20 degrees Celsius. What...

## Question:

Mercury fills 90% of the volume of a sealed glass container at 20 degrees Celsius. What temperature does the glass/Mercury have to be heated so that the mercury completely fills the container? (Hg = {eq}6 \times 10^{-5} {/eq} degrees Celsius{eq}^{-1} {/eq}, and glass = {eq}0.4 \times 10^{-5} {/eq} degrees Celsius{eq}^{-1} {/eq}).

## Coefficient of Volume Expansion:

The property of the substances that describes the volume of the substances varies, followed by the fluctuation of the temperature is known as the coefficient of the volume expansion. It measures in the unit of the inverse of the degree Celsius.

## Answer and Explanation:

**Given Data:**

- The coefficient of the volume expansion of the mercury is: {eq}{\alpha _m} = 6 \times {10^{ - 5}}/^\circ {\rm{C}} {/eq}

- The coefficient of the volume expansion of the glass is: {eq}{\alpha _g} = 0.4 \times {10^{ - 5}}/^\circ {\rm{C}} {/eq}

- The temperature is: {eq}{T_a} = 20^\circ {\rm{C}} {/eq}

The temperature at which the mercury and glass heated is: {eq}{T_h} {/eq}

The volume of the mercury {eq}\left( {{V_{am}}} \right) {/eq} fills {eq}90\% {/eq} of the volume of the glass {eq}\left( {Vg} \right) {/eq}

{eq}{V_{am}} = 0.90{V_{ag}} {/eq}

The expression for the volume of the mercury after heating is

{eq}{V_{bm}} = {V_{am}}\left( {1 + 3{\alpha _m}\left( {{T_h} - {T_a}} \right)} \right) {/eq}

The expression for the volume of the glass after heating is

{eq}{V_{bg}} = {V_{ag}}\left( {1 + 3{\alpha _g}\left( {{T_h} - {T_a}} \right)} \right) {/eq}

The glass/Mercury heated so that the mercury completely fills the container, the final volume of the mercury and glass will be same.

{eq}{V_{bm}} = {V_{bg}} {/eq}

Substitute the values and solve the above expression

{eq}\begin{align*} 0.90{V_{ag}}\left( {1 + 3 \times 6 \times {{10}^{ - 5}} \times \left( {{T_h} - 20} \right)} \right) &= {V_{ag}}\left( {1 + 3 \times 0.4 \times {{10}^{ - 5}} \times \left( {{T_h} - 20} \right)} \right)\\ 0.90 + 16.2 \times {10^{ - 5}}\left( {{T_h}} \right) - 3.24 \times {10^{ - 3}} &= 1 + 1.2 \times {10^{ - 5}}\left( {{T_h}} \right) - 0.24 \times {10^{ - 3}}\\ 15 \times {10^{ - 5}}\left( {{T_h}} \right) &= {\rm{0}}{\rm{.103}}\\ {T_h} &= 686.66^\circ {\rm{C}} \end{align*} {/eq}

Thus the temperature at which the mercury and glass heated is {eq}686.66^\circ {\rm{C}} {/eq}.

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from High School Physics: Help and Review

Chapter 17 / Lesson 12