Question

A Pyrex measuring cup was calibrated at normal room temperature. How much error will be made in a recipe calling for 375 mL of cool water, if the water and the cup are hot, at 95°C, instead of at room temperature? Neglect the glass expansion.

Short answer

The volume of the cool water is 5.5 mL less than the required water.

Step-by-step solution

Given data

The higher temperature is \({T_1} = {95^ \circ }{\rm{C}}\).

The normal room temperature is \({T_2} = {25^ \circ }{\rm{C}}\). (approx.)

The volume of the cup is \({V_ \circ } = 375\;{\rm{mL}}\).

Assume that the size of the cap remains the same.

Concepts

The volume of the water is equal to the volume of the cap.

The change in the volume of the water is\(\Delta V = {V_ \circ }\beta \Delta T\).

Calculation

The volume expansion coefficient of the water is \(\beta = 210 \times {10^6}\;{{\rm{/}}^{\rm{o}}}{\rm{C}}\).

The volume of the hot water is equal to the volume of the cup.

The change in temperature is:

\(\begin{aligned}{c}\Delta T &= {T_2} - {T_1}\\ &= {25^ \circ }{\rm{C}} - {95^ \circ }{\rm{C}}\\ &= - {70^ \circ }{\rm{C}}\end{aligned}\)

Now, the change in the volume of the water when it comes to room temperature is:

\(\begin{aligned}{c}\Delta V &= \left( {375\;{\rm{mL}}} \right) \times \left( {210 \times {{10}^6}\;{{\rm{/}}^{\rm{o}}}{\rm{C}}} \right) \times \left( { - {{70}^ \circ }{\rm{C}}} \right)\\ &= - 5.5\;{\rm{mL}}\end{aligned}\)

Hence, the volume of the cool water is 5.5 mL less than the required water.