The temperature of a glass vessel containing 100 cm\(^3\) of mercury is raised from 10°C to 100°C. Calculate the apparent cubic expansion of the mercury. (Real cubic expansivity of mercury = 1.8 × 10\(^{-4}\) K\(^{-1}\), Cubic expansivity of glass = 2.4 × 10\(^{-5}\) K\(^{-1}\))
The apparent cubic expansion of mercury is the increase in its volume relative to the glass vessel (i.e., the volume of mercury that would overflow).
Let: \( V_0 = 100 \, \mathrm{cm}^3 \) (initial volume of mercury), \( \gamma_m = 1.8 \times 10^{-4} \, \mathrm{K}^{-1} \) (real cubic expansivity of mercury),
\( \gamma_g = 2.4 \times 10^{-5} \, \mathrm{K}^{-1} \) (cubic expansivity of glass), \( \Delta \theta = 100^\circ\mathrm{C} - 10^\circ\mathrm{C} = 90 \, \mathrm{K} \) (temperature rise).
The coefficient of apparent expansion is \(\gamma_a = \gamma_m - \gamma_g = 1.8 \times 10^{-4} - 2.4 \times 10^{-5} = 1.56 \times 10^{-4} \, \mathrm{K}^{-1}. \)
The apparent cubic expansion (change in volume) is \(\Delta V_a = V_0 \cdot \gamma_a \cdot \Delta \theta = 100 \times (1.56 \times 10^{-4}) \times 90.\)
First compute \( 1.56 \times 10^{-4} \times 90 = 0.01404 \),
then \(\Delta V_a = 100 \times 0.01404 = 1.404 \, \mathrm{cm}^3.\)
(Alternatively:
Real expansion of mercury = \( 100 \times 1.8 \times 10^{-4} \times 90 = 1.62 \, \mathrm{cm}^3 \).
Expansion of glass vessel = \( 100 \times 2.4 \times 10^{-5} \times 90 = 0.216 \, \mathrm{cm}^3 \).
Apparent expansion = \( 1.62 - 0.216 = 1.404 \, \mathrm{cm}^3 \).)
Thus, the apparent cubic expansion of the mercury is \( 1.404 \, \mathrm{cm}^3 \).
Contributions ({{ comment_count }})
Please wait...
Modal title
Report
Block User
{{ feedback_modal_data.title }}