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Expert-verified Found in: Page 359 ### Physics Principles with Applications

Book edition 7th
Author(s) Douglas C. Giancoli
Pages 978 pages
ISBN 978-0321625922 # Question: (II) An ordinary glass is filled to the brim with 450.0 mL of water at 100.0°C. If the temperature of glass and water is decreased to 20.0°C, how much water could be added to the glass?

The volume of water that can be added to the glass container is $$6.59{\rm{ mL}}$$.

See the step by step solution

## Step 1: Identification of given data

• The coefficient of volumetric expansion of glass is $${\beta _{{\rm{Glass}}}} = 27 \times {10^{ - 6}}\;{\rm{/^\circ C}}$$.
• The coefficient of volumetric expansion of water is $${\beta _{{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = 210 \times {10^{ - 6}}\;{\rm{/^\circ C}}$$.
• The initial volume of both water and container is $${V_{{\rm{iGlass}}}} = {V_{{\rm{i}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}} = {V_{\rm{i}}}$$.
• The initial volume of glass and water is $${V_{\rm{i}}} = 450{\rm{ mL}}$$.
• The final decreased temperature is $${T_f} = 20^\circ {\rm{C}}$$.
• The initial temperature of glass and water is $${T_i} = 100^\circ {\rm{C}}$$.

## Step 2: Understanding the variation in the volume of water

The initial volume of water is equal to that of the glass container.

The volume of water added to the glass container is obtained by subtracting the final volume of water from the final volume of the glass container.

## Step 3: Determination of the volume of water added to the glass container

The volume of water that can be added to the glass container can be expressed as shown below:

\begin{aligned}{c}{V_{{\rm{add}}}} &= {\left( {{V_i} + \Delta V} \right)_{{\rm{Glass}}}} - {\left( {{V_i} + \Delta V} \right)_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\\ &= \Delta {V_{{\rm{Glass}}}} - \Delta {V_{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\\ &= {\beta _{{\rm{Glass}}}}{V_{{\rm{iGlass}}}}\Delta T - {\beta _{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}{V_{{\rm{i}}{{\rm{H}}_{\rm{2}}}{\rm{O}}}}\Delta T\\ &= \left( {{\beta _{{\rm{Glass}}}} - {\beta _{{{\rm{H}}_{\rm{2}}}{\rm{O}}}}} \right){V_{\rm{i}}}\left( {{T_f} - {T_i}} \right)\end{aligned}

Substitute the values in the above equation.

\begin{aligned}{c}{V_{{\rm{add}}}} &= \left( {27 \times {{10}^{ - 6}}\;{\rm{/^\circ C}} - 210 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right) \times 450{\rm{ mL}} \times \left( {20^\circ {\rm{C}} - 100^\circ {\rm{C}}} \right)\\ &= \left( { - 183 \times {{10}^{ - 6}}\;{\rm{/^\circ C}}} \right) \times 450{\rm{ mL}} \times \left( { - 80^\circ {\rm{C}}} \right)\\ &= 6.59{\rm{ mL}}\end{aligned}

Thus, the volume of water that can be added to the glass container is $$6.59{\rm{ mL}}$$. ### Want to see more solutions like these? 