A hollow glass sphere has a density of 1.30g/cc at 200C. Glycerin has a density of

Question

1.26g/cc at 200C. At what temperature would the sphere begin to float in Glycerine? Given
coefficient of volume expansion of glycerine is 53 × 10-5°C-1.

Answer 1

To determine the temperature at which the hollow glass sphere will begin to float in glycerin, we need to understand the relationship between density and temperature for both the glass sphere and the glycerin.

Given data:
- Density of hollow glass sphere (\(\rho_{\text{glass}}\)) at 20°C = 1.30 g/cm\(^3\)
- Density of glycerin (\(\rho_{\text{glycerin}}\)) at 20°C = 1.26 g/cm\(^3\)
- Coefficient of volume expansion of glycerin (\(\beta_{\text{glycerin}}\)) = 53 × 10\(^{-5}\) °C\(^{-1}\)

First, we need to determine how the density of glycerin changes with temperature. The density of glycerin will decrease as the temperature increases because of thermal expansion.

The volume expansion formula is:
\[ V_T = V_0 (1 + \beta \Delta T) \]

Where:
- \(V_T\) is the volume at temperature \(T\)
- \(V_0\) is the initial volume
- \(\beta\) is the coefficient of volume expansion
- \(\Delta T\) is the change in temperature (i.e., \(T - T_0\))

Since \(\rho = \frac{m}{V}\) and the mass remains constant, the density \(\rho\) at a new temperature \(T\) can be expressed as:
\[ \rho_T = \frac{\rho_0}{1 + \beta \Delta T} \]

For glycerin at temperature \(T\):
\[ \rho_{\text{glycerin}, T} = \frac{\rho_{\text{glycerin}, 20}}{1 + \beta_{\text{glycerin}} (T - 20)} \]

We are interested in finding the temperature \(T\) at which the density of glycerin equals the density of the hollow glass sphere (1.30 g/cm\(^3\)). Thus, we set up the equation:
\[ \frac{1.26 \, \text{g/cm}^3}{1 + 53 \times 10^{-5} (T - 20)} = 1.30 \, \text{g/cm}^3 \]

Now, solve for \(T\):

1.30 (1 + 53 × 10\(^{-5}\) (T - 20)) = 1.26

Expand the expression:
1.30 + 1.30 × 53 × 10\(^{-5}\) (T - 20) = 1.26

Divide both sides by 1.30:
1 + 53 × 10\(^{-5}\) (T - 20) = \(\frac{1.26}{1.30}\)

1 + 53 × 10\(^{-5}\) (T - 20) = 0.9692

Isolate the temperature term:
53 × 10\(^{-5}\) (T - 20) = 0.9692 - 1

53 × 10\(^{-5}\) (T - 20) = -0.0308

Solve for \(T\):
(T - 20) = \(\frac{-0.0308}{53 × 10^{-5}}\)

(T - 20) ≈ -581.13

T ≈ 20 - 581.13

T ≈ -561.13°C

Since temperature cannot logically drop to such a low value for practical purposes, it seems more plausible that the assumption used in the equation or the data set provided needs review. Nonetheless, under the given premises, this physical state can't be achieved realistically under normal conditions for usual material behavior, pointing perhaps towards needing reevaluation of inputs or a constraint in problem scenario.