Question 53 GLI01 - Master-Unlimited Tonnage
A cargo of 10,000 barrels of gasoline is loaded at a temperature of 90°F, and a cargo temperature of 55°F, is expected on this voyage. It has a coefficient of expansion of .0006. How many barrels would you expect to discharge at your destination?
The Correct Answer is A *** ### Explanation of why option A ("9790") is correct The problem asks for the expected discharge volume after the cargo contracts due to a temperature drop. This calculation uses the principles of thermal volume contraction. **1. Calculate the Change in Temperature ($\Delta T$):** $$\Delta T = \text{Loading Temperature} - \text{Expected Discharge Temperature}$$ $$\Delta T = 90^\circ\text{F} - 55^\circ\text{F} = 35^\circ\text{F}$$ **2. Calculate the Change in Volume ($\Delta V$):** The formula for volume change is: $$\Delta V = \text{Original Volume} \times \text{Coefficient of Expansion} \times \Delta T$$ $$\Delta V = 10,000 \text{ barrels} \times 0.0006 \text{ per }^\circ\text{F} \times 35^\circ\text{F}$$ $$\Delta V = 10,000 \times 0.021$$ $$\Delta V = 210 \text{ barrels}$$ **3. Calculate the Final (Discharge) Volume:** Since the temperature decreased, the volume contracted. We must subtract the change in volume from the original volume. $$\text{Discharge Volume} = \text{Original Volume} - \Delta V$$ $$\text{Discharge Volume} = 10,000 \text{ barrels} - 210 \text{ barrels}$$ $$\text{Discharge Volume} = 9,790 \text{ barrels}$$ *** ### Explanation of why the other options are incorrect **B) 9994** This value represents a volume loss of only 6 barrels, suggesting an extremely small temperature drop or a very low coefficient of expansion (e.g., if the coefficient was mistakenly calculated as $0.000006$ instead of $0.0006$). It does not reflect the significant $35^\circ\text{F}$ temperature change. **C) 10210** This result is calculated by *adding* the volume change instead of subtracting it ($10,000 + 210 = 10,210$). This would only be the correct answer if the cargo temperature had *increased* (thermal expansion), which contradicts the problem statement that the temperature dropped from $90^\circ\text{F}$ to $55^\circ\text{F}$. **D) 10410** This number implies a volume gain of 410 barrels. This is the result of using an incorrect coefficient or an incorrect temperature change calculation (e.g., if $\Delta T$ was mistakenly calculated as $68.3^\circ\text{F}$ and the result was added). This is inconsistent with the physics of contraction described in the problem.
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