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Chapter 1: Problem 94

A rigid-walled cubical container is completely filled with water at \(40^{\circ} \mathrm{F}\) and sealed. The water is then heated to \(100^{\circ} \mathrm{F}\) 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.

Short Answer

Expert verified
The pressure that develops in the container when the water reaches the higher temperature is 16.46 psi.

Step by step solution

01

Analyze given values

Given values are: Initial temperature, \(T_1=40^{\circ}F\); Final temperature, \(T_2 = 100^{\circ}F\); Bulk modulus, \(B = 300,000 \, psi\); and the fact that the volume remains constant because the container is rigid.
02

Calculate the change in temperature

Change in temperature, ΔT will be final temperature minus initial temperature. This equates to \(ΔT = T_2 - T_1 = 100^{\circ}F - 40^{\circ}F = 60^{\circ}F\)
03

Convert the temperature change to Rankine for the formula application

To apply the formula correctly, we need to convert the temperature change to Rankine (absolute temperature scale). We can use the conversion factor: \(1^{\circ}F = 1 \, Rankine\). This gives us \(ΔT = 60 \, Rankine\)
04

Apply the pressure-volume-temperature change relation

In the case of constant volume, the thermodynamic process is isochoric. So, the pressure-volume-temperature relationship can be expressed as \(ΔP=P_1 * ΔT / T_1\), where ΔP represents the change in pressure, P1 is the initial pressure, and T1 is the initial temperature. Since the initial pressure of water, P1 is atmospheric and assuming this to be 14.7 psi, we can substitute these values to the above formula.
05

Calculate the change in pressure

By substituting \(P_1 = 14.7 \, psi, \, ΔT = 60 \, Rankine\) and \(T_1 = 40^{\circ}F + 460 = 500 \, Rankine\) into the pressure change formula, we get \(ΔP = 14.7 * 60 / 500 = 1.76 \, psi\)
06

Calculate the actual pressure

The final pressure inside the container when the water reaches the higher temperature will be the original pressure plus the change in pressure. So, the final pressure, \(P_2 = P_1 + ΔP = 14.7 \, psi + 1.76 \, psi = 16.46 \, psi\)

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