Q67PE - College Physics (Urone)

Question

Question: To get an idea of the small effect that temperature has on Archimedes’ principle, calculate the fraction of a copper block’s weight that is supported by the buoyant force in 0°C water and compare this fraction with the fraction supported in 95.0°C water.

Step-by-Step Solution

Answer

Answer

The fraction of weight of a copper block supported by buoyant force at 0°C is 0.980 times more than the copper in water at 95°C.

Step 1: Define the fraction submerged

The fraction submerged is equal to the ratio of the average density of the object to the density of the fluid in which the object is submerged. This can be mathematically expressed as;

$\mathbf{fractionsubmerged}\mathbf{=}\frac{{\mathbf{\rho }}_{\mathbf{obj}}}{{\mathbf{\rho }}_{\mathbf{fl}}}$ …………(1)

Here, ${\mathit{\rho }}_{\mathbf{o}\mathbf{b}\mathbf{j}}$ is the density of the object and  ${\mathit{\rho }}_{\mathbf{f}\mathbf{l}}$ is the density of the fluid. The density of the fluid varies with the temperature. So, the effect of temperature on Archimedes’ principle can be understood by comparing the fraction of submerged copper in 0°C of water with 0°C of water.

Step 2: Find the density of water at both temperature

The density of water at 0°C is 1000 Kg/m³. We need to calculate the density at 95°C. Density is the ratio of mass to volume. The mass of water will be the same as at 0°C, that is, 1000 Kg. But the volume changes due to thermal expansion. This change in volume can be calculated as;

 $\Delta V=\beta V\Delta T$……………(2)

Here,  is the volume change,   is the coefficient of volume expansion,   is the initial volume, and   is the temperature change.

In water,

 $$\begin{gathered} \beta =210\times {10}^{--6} °C^{-1} \\ V=1 m^3 \\ \Delta T=95°C-0°C \\ =95°C \end{gathered}$$

Substitute these values in equation (2).

 $$\begin{gathered} \Delta V=\left(210\times {{10}^{-}-}^6 °C^{-1}\right)\times 1 m^3\times 95°C \\ =0.01995 m^3 \end{gathered}$$

Therefore,

  $$\begin{gathered} \text{density of water at 95}°C=\frac{mass}{v+\Delta v} \\ =\frac{1000 Kg}{1.01995 m^3} \\ =980.440 Kg/m^3 \end{gathered}$$

Step 3: Compare the fraction submerged at both cases of temperature

Fraction submerged can be calculated in both situations by substituting the corresponding values in equation (1),

 $$\begin{gathered} \text{fraction submerged at }0°C=\frac{{\rho }_{copper}}{1000 Kg/m^3} \\ \text{fraction submerged at 95}°C=\frac{{\displaystyle {\rho }_{copper}}}{{\displaystyle 980.440 Kg/m^3}} \end{gathered}$$

Divide these two equations.

 $$\begin{gathered} \frac{\text{fraction submerged at }0°C}{\text{fraction submerged at 95}°C}=\frac{980.440 Kg/m^3}{1000 Kg/m^3} \\ =0.980 \end{gathered}$$

The fraction of copper submerged at 0°C of water is 0.980 times greater than the copper submerged at 95°C.

Therefore, there is a small variation in the fraction submerged with a temperature rise.