Q20P - Fundamentals Of Physics

Question

Question: The record for the largest glass bottle was set in 1992 by a team in Millville, New Jersey—they blew a bottle with a volume of 193 U.S. fluid gallons. (a) How much short of 1.0 million cubic centimeters is that? (b) If the bottle were filled with water at the leisurely rate of 1.8 g/min, how long would the filling take? Water has a density of 1000 kg/m³_.

Step-by-Step Solution

Answer

Answer:

 

1. The volume is 269,417.8 cm³ short of 1 million cm³ .
2. It would take 405,879 minutes to fill the bottle with water.

 

Step 1: Given data

The volume of fluid that filled the bottle is 193 us gallons .

The rate of flow of water is 18 g/min.

Density of density of water is 1000 kg/m³.

Step 2: Understanding the density

factor for converting the US gallons to cubic centimeters. 

 

The expression for the density is given as follows: 

               $\rho =\frac{m}{v}$                                                                      … (i)

 

Here, $\rho$ is the density, m is the mass and v is the volume.

Step 3: (a) Conversion of US gallons into cubic centimeters

Using the converted values, find the shortfall.

 

First, 193 US Gallons convert  into cm³.

 1 US Gallons  = 3785.4 cm³ 

$$\begin{gathered} 93 US Gallons=93 US Gallons\times \left(\frac{3785.4 c{m}^3}{1 gallon}\right) \\ =730582.2 c{m}^2 \end{gathered}$$

Subtracting it from to get amount of water.

 1000,000-730582.2 =269,417.8 cm³ 

Thus, it is 269,417.8 cm³   short of 1 million cm³  .

Step 4: (b) Conversion into cubic meters from cubic centimeters

Now to find how long it takes to fill the bottle, convert the volume into m^3..

 

The conversion factor for

 $1 c{m}^3=1\times {10}^{-6}{m}^3$to m³.

 Use the conversion factor to convert 

  ^{$$\begin{gathered} 730,582.2c{m}^3 in to m^3 \\ 730,582.2c{m}^3 =730,582.2c{m}^3\frac{1\times {10}^{-6}{m}^3}{1 c{m}^3} \\ =0.7305822 m^3 \end{gathered}$$}

 

So, volume of water in cubic meter is $0.7305822 m^3$.

Step 5: Determination of the mass of water

$$\begin{gathered} \text{Using equation (i), the mass of water is calculated as follows: } \\ m=\rho \times v \\ = 1000 kg/m^3\times 0.7305822 m^3 \\ = 7305822 kg \\ \text{ So, the mass of water filled is }7305822 kg. \end{gathered}$$

Step 6: Determination of the time to fill the bottle

$$\begin{gathered} \text{Now, convert the 1.8 gm/min rate into kg/min } \\ 1 gm =0.001kg \\ \left(1.8\frac{g}{min}\right)\left(\frac{0.001kg}{1gm}\right)=0.0018\frac{kg}{min} \\ \text{ The time required to fill the bottle is calculated as} \\ t=\frac{730,5822kg}{0.0018{\displaystyle \frac{kg}{min}}} \\ =405,879 min \\ \text{: Thus, it will take }405,879 min\text{ to fill the bottle.} \end{gathered}$$