Question

A rigid-walled cubical container is completely filled with water at 40 °F and sealed. The water is then heated to 100 °F. Determine the pressure that develops in the container when the water reaches this higher temperature. Assume that the volume of the container remains constant and the value of the bulk modulus of the water remains constant and equal to 300,000 psi.

Answer 1

When the water in the cubical container is heated, the pressure inside the container will increase due to the increase in temperature. The pressure can be determined using the formula P = Bulk modulus x ΔV / V, where P is the pressure, Bulk modulus is the constant value of 300,000 psi, ΔV is the change in volume (which is assumed to be zero in this case), and V is the initial volume.

Step-by-step explanation:

When the water in the rigid-walled cubical container is heated from 40°F to 100°F, the pressure inside the container will increase. This is because the increase in temperature causes the water molecules to gain more kinetic energy, resulting in higher pressure. The increase in pressure can be determined by using the formula:

P = Bulk modulus x ΔV / V

Where P is the pressure, Bulk modulus is the constant value of 300,000 psi, ΔV is the change in volume (which is assumed to be zero in this case as the volume of the container remains constant), and V is the initial volume.

By substituting the given values into the formula, we can calculate the pressure that develops in the container when the water reaches 100°F.

Answer 2

Step-by-step explanation:

The given data is as follows.

For the constant volume condition,

`T_(1)` = 40 F

`V_(1) = 0.01602 ft^(3)/lb`

`V_(2)` = 0.01613

Therefore, expression for the compressibility is as follows.

`(dP)/((dV)/(V))` = K

`(dP)/(((0.01602 - 0.01613))/(0.01602)) = 30 * 10^(4)`

dP =
`30 * 10^(4) ((0.01602 - 0.01613)/(0.01602))`

= 2059.925 psi

=
`2.059 * 10^(3)` psi

therefore, we can conclude that pressure that develops in the container when the water reaches this higher temperature is
`2.059 * 10^(3)` psi.