Chapter 13, Problem 13P

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?

To determine

The additional volume of water that can be added to the glass after the decrease in temperature

Answer to Problem 13P

Solution:

The additional volume of water that can be added to the glass is 6.588 ml

Explanation of Solution

Given:

Initial volume of glass, $V_{gi}=450ml$

Initial volume of water, $V_{wi}=450ml$

Initial temperature, $T_i=100°C$

Final temperature, $T_f=20°C$

Formula used:

$\Delta V=\beta V_i\Delta T$, where V_i is the initial volume and β is the coefficient of volume expansion

Calculation:

The relation between change in volume of a substance (ΔV)and change in temperature (ΔT)is given by,

$\Delta V=\beta V_i\Delta T$, where V_i is the initial volume and β is the coefficient of volume expansion

Here, $\Delta V=V_f–V_i$, where V_f is final volume

$\Delta T=T_f–T_i,$ where T_f is final temperature and T_i is initial temperature

For glass, ${\beta }_g=27\times {10}^{-6}/°C$

Therefore, $V_{gf}–V_{gi}=(27\times {10}^{-6}/°C)\times V_{gi}\times \left(T_f–T_i\right)$

$\begin{array}{l}or{V}_{gf}–450ml=(27\times {10}^{-6}/°C)\times \left(450ml\right)\times (20°C-100°C)\hfill \\ or{V}_{gf}–450ml=-0.972ml\hfill \\ or{V}_{gf} =449.028ml\hfill \end{array}$

For water, ${\beta }_w=210\times {10}^{-6}/°C$

Therefore, $V_{wf}–450ml=(210\times {10}^{-6}/°C)\times \left(450ml\right)\times (20°C-100°C)$

$\begin{array}{l}or{V}_{wf}–450ml=-7.56ml\hfill \\ or{V}_{wf} =442.44ml\hfill \end{array}$

This shows that while the volume of water has come down to 442.44 ml, the volume of glass is 449.028 ml

Therefore, additional water that can be added to the glass is,

$V_{gf} -V_{wf}=449.028ml–442.44ml=6.588ml$

(Note that while the above steps are for a clearer illustration of the problem, we can obtain this result by just subtracting ΔV_g from ΔV_w also)