The unit weight of water at 50 °F is 62.4 pcf. If the volume of the vessel that contains water is 3.50 ft A. What will be the change in its volume when it is heated to 160°F (Unit weight of water is 61 pcf 160°F) B. Compute the percentage change of volume. C. What weight of water must be removed to maintain the original volume.
The correct answer and explanation is :
Let’s analyze the problem step by step.
Given:
- Unit weight of water at 50 °F = 62.4 pcf
- Unit weight of water at 160 °F = 61 pcf
- Initial volume V1=3.50 ft3V_1 = 3.50 \, \text{ft}^3
- Temperature change from 50 °F to 160 °F
A. What will be the change in volume when heated from 50°F to 160°F?
Concept:
The unit weight γ\gamma is defined as: γ=WV\gamma = \frac{W}{V}
where WW is the weight and VV is the volume.
If the weight remains constant and the unit weight changes (due to temperature), then the volume must change accordingly.
Since weight WW is constant: W=γ1V1=γ2V2W = \gamma_1 V_1 = \gamma_2 V_2
Rearranged for V2V_2: V2=γ1γ2V1V_2 = \frac{\gamma_1}{\gamma_2} V_1
Substitute values: V2=62.461×3.50=1.02295×3.50=3.5803 ft3V_2 = \frac{62.4}{61} \times 3.50 = 1.02295 \times 3.50 = 3.5803 \, \text{ft}^3
Change in volume, ΔV\Delta V: ΔV=V2−V1=3.5803−3.50=0.0803 ft3\Delta V = V_2 – V_1 = 3.5803 – 3.50 = 0.0803 \, \text{ft}^3
B. Compute the percentage change of volume.
% change=ΔVV1×100=0.08033.50×100=2.29%\%\text{ change} = \frac{\Delta V}{V_1} \times 100 = \frac{0.0803}{3.50} \times 100 = 2.29\%
C. What weight of water must be removed to maintain the original volume of 3.50 ft³ at 160°F?
At 160 °F, if the volume is to remain the same V=3.50 ft3V = 3.50 \, \text{ft}^3, but the unit weight decreases to 61 pcf, the weight must change accordingly. W2=γ2×V=61×3.50=213.5 lbW_2 = \gamma_2 \times V = 61 \times 3.50 = 213.5 \, \text{lb}
Original weight at 50 °F: W1=γ1×V1=62.4×3.50=218.4 lbW_1 = \gamma_1 \times V_1 = 62.4 \times 3.50 = 218.4 \, \text{lb}
Weight to be removed: Wremoved=W1−W2=218.4−213.5=4.9 lbW_{\text{removed}} = W_1 – W_2 = 218.4 – 213.5 = 4.9 \, \text{lb}
Summary of answers:
| Part | Result |
|---|---|
| A | Volume increases by 0.0803 ft³ |
| B | Percentage volume change = 2.29% |
| C | Weight to be removed = 4.9 lb |
Explanation (around 300 words):
When water is heated, its density decreases, causing the volume to expand if the weight remains constant. The unit weight of water is directly proportional to its density, and the decrease in unit weight from 62.4 pcf at 50 °F to 61 pcf at 160 °F indicates that water expands upon heating.
The problem gives the initial volume of water at 50 °F and asks for the volume change after heating to 160 °F. Since the weight stays the same, we use the relationship between weight, volume, and unit weight. At constant weight, the volume changes inversely with the unit weight. As the unit weight decreases, volume must increase to keep weight constant.
Calculating the new volume at 160 °F showed a 2.29% increase in volume (0.0803 ft³) compared to the original 3.50 ft³. This expansion is typical for liquids as they heat.
If it’s necessary to keep the volume fixed at 3.50 ft³ despite the temperature increase, some water must be removed because the water expands and would occupy more space otherwise. Calculating the weight of water for the fixed volume at 160 °F yields a smaller weight (213.5 lb) compared to the original (218.4 lb). The difference (4.9 lb) is the amount of water to remove to maintain the original volume.
This problem exemplifies thermal expansion of liquids and its impact on volume and weight relationships. It’s important in practical applications such as fluid storage and transportation, where temperature changes can cause volume changes that affect system design or operation. Understanding these principles helps in maintaining desired system parameters by compensating for expansion or contraction.