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Expert-verifiedQuestion: (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 \(6.59{\rm{ mL}}\).
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.
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 \(6.59{\rm{ mL}}\).
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