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Step-by-Step Solution
Step 1: Identify the Relevant Physical Property
The problem involves the thermal expansion of a liquid (glycerin). Specifically, we need the fractional change in its density resulting from a temperature increase.
Step 2: Recall the Formula for Density Change
The density of a substance at temperature $T$, denoted $ \rho_T $, can be expressed in terms of its reference density $ \rho_0 $ (at some reference temperature) and its coefficient of volume expansion $ \gamma $ as:
$ \rho_T = \rho_0 \left(1 - \gamma \, \Delta T\right)
\quad$
(for small temperature rises assuming volume expansion dominates).
This relationship shows how density decreases when the volume expansion grows.
Step 3: Express the Fractional Change in Density
The fractional change in density from the reference temperature to the new temperature is given by:
$
\frac{\rho_0 - \rho_T}{\rho_0}.
$
Using the formula for $ \rho_T $ gives:
$
\frac{\rho_0 - \rho_T}{\rho_0}
= \frac{\rho_0 - \rho_0(1 - \gamma \Delta T)}{\rho_0}
= \gamma \,\Delta T.
$
Step 4: Substitute the Given Values
Coefficient of volume expansion, $ \gamma = 5 \times 10^{-4}\,\mathrm{K}^{-1} $.
Rise in temperature, $ \Delta T = 40^\circ \mathrm{C} = 40\,\mathrm{K} $ (for changes in temperature, the difference in Celsius is numerically the same as in Kelvin).
Thus,
$
\gamma \Delta T
= \left(5 \times 10^{-4} \,\mathrm{K}^{-1}\right) \times (40\,\mathrm{K})
= 0.020.
$
Step 5: State the Final Conclusion
The fractional change in the density of glycerin for a 40°C rise in temperature is:
$
\boxed{0.020.}
$
