Step 2 The change in temperature ATC (degrees Celsius) of the gasoline as it is poured into the underground tank is ATC = Tf₁c-T₁,c 1.03 X = Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. -10.6 °C. Your response differs from the correct answer by more than 10%. Double check your calculations.)°C = -25

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**Problem Statement:**
An underground gasoline tank can hold 1.19 × 10³ gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outdoor temperature (and the gasoline in a tanker truck) is 97.0°F, how many gallons from the truck can be poured into the tank? Assume the temperature of the gasoline quickly cools from 97.0°F to 52.0°F upon entering the tank.

**Step 1: Temperature Conversion**
We will convert the initial and final Fahrenheit temperatures of the gasoline to degrees Celsius using the formula:

\[ T_C = \frac{5}{9} (T_F - 32.0) \]

- **For the initial temperature (97°F) in degrees Celsius:**

  \[
  T_{i,C} = \frac{5}{9} (97 - 32.0) = 36.11^\circ C \approx 36.1^\circ C
  \]

- **For the final temperature (52°F) in degrees Celsius:**

  \[
  T_{f,C} = \frac{5}{9} (52 - 32.0) = 11.11^\circ C \approx 11.1^\circ C
  \]

**Step 2: Calculate Temperature Change**
The change in temperature \( \Delta T_C \) (degrees Celsius) of the gasoline as it is poured into the underground tank is:

\[ \Delta T_C = T_{f,C} - T_{i,C} \]

The correct calculation is:

\[ \Delta T_C = 11.1 - 36.1 = -25^\circ C \]

**Note:**
An error initially occurred in calculating \(\Delta T_C\). It was input as \(1.03\)°C and \(-10.6\)°C, which are incorrect. The correct answer is \(-25\)°C. Ensure all steps are thoroughly checked to avoid such mistakes.
Transcribed Image Text:**Problem Statement:** An underground gasoline tank can hold 1.19 × 10³ gallons of gasoline at 52.0°F. If the tank is being filled on a day when the outdoor temperature (and the gasoline in a tanker truck) is 97.0°F, how many gallons from the truck can be poured into the tank? Assume the temperature of the gasoline quickly cools from 97.0°F to 52.0°F upon entering the tank. **Step 1: Temperature Conversion** We will convert the initial and final Fahrenheit temperatures of the gasoline to degrees Celsius using the formula: \[ T_C = \frac{5}{9} (T_F - 32.0) \] - **For the initial temperature (97°F) in degrees Celsius:** \[ T_{i,C} = \frac{5}{9} (97 - 32.0) = 36.11^\circ C \approx 36.1^\circ C \] - **For the final temperature (52°F) in degrees Celsius:** \[ T_{f,C} = \frac{5}{9} (52 - 32.0) = 11.11^\circ C \approx 11.1^\circ C \] **Step 2: Calculate Temperature Change** The change in temperature \( \Delta T_C \) (degrees Celsius) of the gasoline as it is poured into the underground tank is: \[ \Delta T_C = T_{f,C} - T_{i,C} \] The correct calculation is: \[ \Delta T_C = 11.1 - 36.1 = -25^\circ C \] **Note:** An error initially occurred in calculating \(\Delta T_C\). It was input as \(1.03\)°C and \(-10.6\)°C, which are incorrect. The correct answer is \(-25\)°C. Ensure all steps are thoroughly checked to avoid such mistakes.
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