An open, rigid-walled, cylindrical tank contains 4 ft3 of
water at 40∘F . Over a 24 hour period of time the water temperature varies from 40 to 90∘F.
Make use of the appropriate thermodynamic data, to determine how much the volume of the water will change. For a tank diameter of 2 ft, would the corresponding change in water depth be
very noticeable? Explain.

Where V_1, p_1 represent values at 40 degrees Fahrenheit, and V_2, p_2, represent values of volume and density on 90 degrees Fahrenheit. Next, from the table we can read that p_1(40°F) = 1.940 sl/m^3 and p_2(70°F) = 1.931 sl/m^3 Apply that data, on former equation:

V_1*1.940 sl/m^3 =V_2*1.931 sl/m^3

V_2=4ft^3*1.940 sl/m^3/1.931 sl/m^3

V_2=4.0186 ft^3

Knowing that tank is of cylindrical shape, we can determine depth change by dividing volume change of the water, with surface of the cylinder:

Δl=ΔV/S

Δl=(4.0186-4)/Π/4*(2)^2

Δl=0.0710 in

Note, we multiplied expression with an proper factor, to obtain result in inches. Since change of depth is in order of 0.0... inches, change would not be noticeable by naked eye observation.

priyankatutor priyankatutor · Answer 2 · 2022-01-28T08:19:26+01:00

The change in the volume of water is 0.0186 ft. The corresponding change in the depth of water is not noticeable.

What is Conservation of mass?

In a closed thermodynamic system, the initial mass is equal to the final mass which means mass does not lose over time.

[tex]m_1 =m_2[/tex]

Since,

Mass = Volume x Density

So,

[tex]V_1\times \rho_1 = V_2\times \rho_2[/tex]

Where,

[tex]V_1[/tex] - Initial volume = 4 ft³

[tex]V_2[/tex] - Final Volume = ?

[tex]\rho _1[/tex] - Dencity at 40°F = 1.940 [tex]\bold {sl/m^3 }[/tex]

[tex]\rho _2[/tex] - Dencity at 90°F = 1.931 [tex]\bold {sl/m^3 }[/tex]

Put the values in the formula,

[tex]V_1\times 1.940 =V_2\times 1.931 \\\\V_2 =\dfrac {4\times 1.940 }{1.931 }\\\\V_2=4.0186 \rm \ ft^3[/tex]

So the change in volume will be,

[tex]\Delta V = V_2 - V_1\\\\\Delta V = 4.186 - 4.000\\\\\Delta V = 0.0186 \rm \ ft^3[/tex]

From the volume of a cylinder,

[tex]V = \pi r^2 h[/tex]

So,

[tex]0.0186 = 3.14 \times 1 ^2\times h\\\\h = \dfrac {0.0186}{ 3.14 }\\\\h = 0.00592 \rm \ ft \\[/tex]

or

[tex]h = 0.071 \rm \ inch[/tex]

Since the height of water that has been lost is less than an inch,

Therefore, the corresponding change in the depth of water is not noticeable.

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