Question

The density of water at $20^{\circ}C$ is $998\, kg/m^3 $ and at $40^{\circ}C$ $992\,kg/m^3 $. The coefficient of volume expansion of water is

• A. $ 3 \times 10^{-4} \, ^ \circ C$
• B. $ 2 \times 10^{-4} \, ^ \circ C$
• C. $ 6 \times 10^{-4} \, ^ \circ C$
• D. $ 10^{-4} \, ^ \circ C$

Answer 1

The Correct Option is A

To determine the coefficient of volume expansion \((\beta)\) of water, we can use the formula for volume expansion due to temperature change:

\[\beta = \frac{\Delta V}{V_0 \Delta T}\]

Where:

• \(\Delta V\) is the change in volume.
• \(V_0\) is the initial volume.
• \(\Delta T\) is the change in temperature.

In case of density, since density \((\rho)\) is inversely proportional to volume \((1/V)\), the formula for coefficient of volume expansion can be rewritten as:

\[\beta = -\frac{\Delta \rho}{\rho_0 \Delta T}\]

Where:

• \(\Delta \rho = \rho_1 - \rho_0\) is the change in density.
• \(\rho_0\) is the initial density.
• \(\Delta T\) is the change in temperature, given by \(40^\circ C - 20^\circ C = 20^\circ C\).

Substituting the values:

• Initial density \(\rho_0 = 998 \, \text{kg/m}^3\)
• Final density \(\rho_1 = 992 \, \text{kg/m}^3\)
• Change in density \(\Delta \rho = 992 - 998 = -6 \, \text{kg/m}^3\)

Now, calculate the coefficient of volume expansion:

\[\beta = -\frac{-6}{998 \times 20}\]

\[\beta = \frac{6}{19960}\]

Solving further:

\[\beta = 3 \times 10^{-4} \, ^\circ C^{-1}\]

Hence, the coefficient of volume expansion of water is 3 \times 10^{-4} \, ^\circ C^{-1}.