Question

Zirconium tungstate is an unusual material because its volume shrinks with an increase in temperature for the temperature range 0.3 K to 1050 K (where it decomposes). In fact, the volumetric coefficient of thermal expansion is –26.4 × 10–6/K. Determine the ratio ΔV/V0 for the above mentioned temperature range. Express your answer in percent.

Answer 1

Answer:

2.771208%

Explanation:

$\Delta V$ = Change of volume

$V_0$ = Initial volume

$\Delta T$ = Change in temperature = (0.3-1050)

$\beta$ = Volumetric coefficient of thermal expansion = $-26.4\times 10^{-6}\ /K$

Volumetric expansion of heat is given by

$\frac{\Delta V}{V_0}=\beta \Delta T\\\Rightarrow \frac{\Delta V}{V_0}=-26.4\times 10^{-6}\times (0.3-1050)\\\Rightarrow \frac{\Delta V}{V_0}=0.02771208$

Finding percentage

$\frac{\Delta V}{V_0}=0.02771208\times 100=2.771208\%$

The ratio of change of volume to initial volume is 2.771208%