Question

A cubic foot of water at 68°F = 20°C weighs 62.3 lbf on earth. (a) What is its density? (b) What does it weigh on the moon ( - 6 ft/s)? (c) What is its density on the moon?

Answer 1

Explanation:

(a)   The given data is as follows.

Water volume V = 1 $ft^{3}$

At $20^{o}C$ weight of water w = 62.3 lbf

It i s known that 1 lbf = 32.174 $lbm.ft.s^{-2}$

Then, weight w = $62.3 \times 32.174$ = 2004.4 $lbm.ft.s^{-2}$

                   Weight = $mass \times g$

where,        g = gravitational acceleration in earth

and,           Mass = $density \times volume$

            $Density \times volume \times g$

                       = 2004.4 lbm. ft. s-2

$Density \times 1 ft^{3} \times 32.174 ft/s^{2}$ = $2004.4 lbm. ft.s^{-2}$

                         Density = $62.3 lbm/ft^{3}$

Hence, density of the water will be $62.3 lbm/ft^{3}$.

(b)    When g = 6 $ft/s^{2}$ at moon then  density and volume are not variable due to gravity.

                     weight (w) = $Density \times volume \times g$  

               w = $(62.3 lbm/ft^{3}) \times 1 ft^{3} \times 6 ft/s^{2}$

                 = 373.8 $lbm.ft.s^{-2}$

Hence, weigh on the moon is 373.8 $lbm.ft.s^{-2}$.

(c)    Mass of any body is independent of the gravity, so density of the water is same as when it is at moon or earth.

Only weight of the body changes due to change in gravity.

It is known that 1 cubic miles = $1.101 \times E12$ us liquid gallon

and,                      1 bbl = 42 US gallon

Therefore,

               $30 \times 10^{9} bbl = 42 \times 30 \times 10^{9} US gallon = 1260 \times E9 \text{us gallon}$

Hence,

              total cubic miles = $1260 \times \frac{E9}{1.101 \times E12}$

                                          = 1.144 cubic miles

Thus, we can conclude that its density on the moon is 1.144 cubic miles.