If a 100 mL glass beaker is filled to the brim with ethyl alcohol at a temperature of 5.20°C, how much will overflow (in mL) when the alcohol's temperature reaches the room temperature of 20.0°C? (Assume the beaker is initially at room temperature.) mL How much less water (in mL) would overflow under the same conditions? mL

Physics for Scientists and Engineers
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ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter18: Temperature
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### Thermodynamics and Fluid Expansion Problem

This section involves understanding the thermal expansion of liquids when subjected to temperature changes. Below are two related problems involving a glass beaker and two different liquids: ethyl alcohol and water.

#### Problem (a)
**Statement:**  
If a 100 mL glass beaker is filled to the brim with ethyl alcohol at a temperature of 5.20°C, how much will overflow (in mL) when the alcohol's temperature reaches the room temperature of 20.0°C? (Assume the beaker is initially at room temperature.)

**Answer:**
```
______ mL
```

#### Problem (b)
**Statement:**  
How much less water (in mL) would overflow under the same conditions?

**Answer:**
```
______ mL
```

---

### Explanation

In this exercise, we need to calculate the volume overflow when a liquid expands due to heating. When a liquid is heated, its volume increases due to thermal expansion. The amount of overflow can be determined using the formula for volumetric thermal expansion:

\[ \Delta V = V_0 \beta \Delta T \]

where:
- \(\Delta V\) is the change in volume.
- \(V_0\) is the initial volume.
- \(\beta\) is the coefficient of volumetric expansion of the liquid.
- \(\Delta T\) is the change in temperature.

**Given:**
- \(V_0\) for the beaker = 100 mL.
- Initial temperature \(T_i\) = 5.20°C.
- Final temperature \(T_f\) = 20.0°C.
- Change in temperature \(\Delta T = T_f - T_i = 20.0 - 5.20 = 14.80\)°C.

To solve each part, find the \(\beta\) value for ethyl alcohol and water, then calculate their respective \(\Delta V\). Finally, the difference between the volumes of overflow will answer part (b).

**Note:** The values for \(\beta\) are:
- Ethyl alcohol: \(\beta \approx 1.12 \times 10^{-3} \text{°C}^{-1}\)
- Water: \(\beta \approx 0.21 \times 10^{-3} \text{°C}^{-1}\)

These calculations would elucidate
Transcribed Image Text:### Thermodynamics and Fluid Expansion Problem This section involves understanding the thermal expansion of liquids when subjected to temperature changes. Below are two related problems involving a glass beaker and two different liquids: ethyl alcohol and water. #### Problem (a) **Statement:** If a 100 mL glass beaker is filled to the brim with ethyl alcohol at a temperature of 5.20°C, how much will overflow (in mL) when the alcohol's temperature reaches the room temperature of 20.0°C? (Assume the beaker is initially at room temperature.) **Answer:** ``` ______ mL ``` #### Problem (b) **Statement:** How much less water (in mL) would overflow under the same conditions? **Answer:** ``` ______ mL ``` --- ### Explanation In this exercise, we need to calculate the volume overflow when a liquid expands due to heating. When a liquid is heated, its volume increases due to thermal expansion. The amount of overflow can be determined using the formula for volumetric thermal expansion: \[ \Delta V = V_0 \beta \Delta T \] where: - \(\Delta V\) is the change in volume. - \(V_0\) is the initial volume. - \(\beta\) is the coefficient of volumetric expansion of the liquid. - \(\Delta T\) is the change in temperature. **Given:** - \(V_0\) for the beaker = 100 mL. - Initial temperature \(T_i\) = 5.20°C. - Final temperature \(T_f\) = 20.0°C. - Change in temperature \(\Delta T = T_f - T_i = 20.0 - 5.20 = 14.80\)°C. To solve each part, find the \(\beta\) value for ethyl alcohol and water, then calculate their respective \(\Delta V\). Finally, the difference between the volumes of overflow will answer part (b). **Note:** The values for \(\beta\) are: - Ethyl alcohol: \(\beta \approx 1.12 \times 10^{-3} \text{°C}^{-1}\) - Water: \(\beta \approx 0.21 \times 10^{-3} \text{°C}^{-1}\) These calculations would elucidate
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