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.

Δl=0.0710 in

change is not noticeable!

Explanation:

Quantity that have to stay the same in the tank, no matter of temperature change, is mass of the water. We can express that in the next way:  

m_1=m_2

V_1*p_1=V_2*p_2

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.  

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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