Question

1) a 1.000 m aluminum bar at 30.00 C cools to a temperature of -50.00 C. What is the final length of the bar?

2) a large container is holding 2.50 L of mercury. What is the change in volume of the mercury if it experiences a temperature change of 78.0 C (assuming the temperature of the container remains constant)?

Answer 1

1) We should calculate the linear expansion of the bar. The length of the bar is the function of temperature (see formula (3) in https://x-engineer.org/undergraduate-engineering/physics/thermodynamics/calculate-thermal-expansion/)

$L=L_0(1+\alpha\Delta T) \;\; \mathrm{or} \;\; \Delta L = \alpha L_0\Delta T.$

Coefficient of linear expansion of aluminium is (see table in https://x-engineer.org/undergraduate-engineering/physics/thermodynamics/calculate-thermal-expansion/ ) $\alpha = 25\cdot10^{-6},$ therefore the expansion will be

$\Delta L = \alpha L_0\Delta T = 25\cdot10^{-6}\cdot1.000\,\mathrm{m}\cdot(-50-30) = -0.002\,\mathrm{m}.$

Therefore, the length will be $1.000-0.002 = 0.998\,\mathrm{m}.$

2) The volume expansion can be calculated as (see formula (10) in https://x-engineer.org/undergraduate-engineering/physics/thermodynamics/calculate-thermal-expansion/)

$\Delta V = 3\alpha V_0\Delta T$ and the final volume will be $V=V_0(1+3\alpha\Delta T).$

We know that $\Delta T = 78^\circ C$ and $3\alpha = 0.00018$ (see https://www.engineeringtoolbox.com/cubical-expansion-coefficients-d_1262.html). Therefore, change of volume will be

$\Delta V=2.50\cdot0.00018\cdot78.0= 0.0351\,\mathrm{L} .$