How much energy (kJ) is needed to heat 25.0 g of isopropyl alcohol from -25.0°C to +90.0°C? Molar mass 60.1 g/mol Boiling point 82.5 °C Melting point -89.0 °C kJ Heat of vaporization, AHvap 39.85 kJ/mol Round your final answer to 2 Specific heat capacity, liquid 1.71 J/(g.K) significant digits. Specific heat capacity, gas 2.49 J/(g.K)

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### How much energy (kJ) is needed to heat 25.0 g of isopropyl alcohol from -25.0°C to +90.0°C?

**Table of Relevant Properties:**

| Property                                   | Value                 |
|--------------------------------------------|-----------------------|
| Molar mass                                 | 60.1 g/mol            |
| Boiling point                              | 82.5 °C               |
| Melting point                              | -89.0 °C              |
| Heat of vaporization, ΔH<sub>vap</sub>     | 39.85 kJ/mol          |
| Specific heat capacity, liquid             | 1.71 J/(g·K)          |
| Specific heat capacity, gas                | 2.49 J/(g·K)          |

**Calculation Steps:**

1. **Determine the temperature change required:**
   - Initial temperature: -25.0°C
   - Final temperature: 90.0°C

2. **Determine the phase of isopropyl alcohol at given temperatures:**
   - Melting point: -89.0 °C
   - Boiling point: 82.5 °C
  
3. **Calculate the energy needed for each phase and phase change:**

    1. *Heating from -25.0°C to 82.5°C (liquid phase):*
       - q<sub>1</sub> = m * c<sub>liquid</sub> * ΔT
       - Where:
         - m = 25.0 g
         - c<sub>liquid</sub> = 1.71 J/g·K
         - ΔT = 82.5°C - (-25°C) 
       
    2. *Phase transition from liquid to gas at boiling point:*
       - q<sub>2</sub> = n * ΔH<sub>vap</sub>
       - Where:
         - n = m / Molar mass
         - m = 25.0 g
         - Molar mass = 60.1 g/mol
         - ΔH<sub>vap</sub> = 39.85 kJ/mol
      
    3. *Heating from 82.5°C to 90.0°C (gas phase):*
       - q<sub>3</sub> = m * c<sub>gas</sub
Transcribed Image Text:### How much energy (kJ) is needed to heat 25.0 g of isopropyl alcohol from -25.0°C to +90.0°C? **Table of Relevant Properties:** | Property | Value | |--------------------------------------------|-----------------------| | Molar mass | 60.1 g/mol | | Boiling point | 82.5 °C | | Melting point | -89.0 °C | | Heat of vaporization, ΔH<sub>vap</sub> | 39.85 kJ/mol | | Specific heat capacity, liquid | 1.71 J/(g·K) | | Specific heat capacity, gas | 2.49 J/(g·K) | **Calculation Steps:** 1. **Determine the temperature change required:** - Initial temperature: -25.0°C - Final temperature: 90.0°C 2. **Determine the phase of isopropyl alcohol at given temperatures:** - Melting point: -89.0 °C - Boiling point: 82.5 °C 3. **Calculate the energy needed for each phase and phase change:** 1. *Heating from -25.0°C to 82.5°C (liquid phase):* - q<sub>1</sub> = m * c<sub>liquid</sub> * ΔT - Where: - m = 25.0 g - c<sub>liquid</sub> = 1.71 J/g·K - ΔT = 82.5°C - (-25°C) 2. *Phase transition from liquid to gas at boiling point:* - q<sub>2</sub> = n * ΔH<sub>vap</sub> - Where: - n = m / Molar mass - m = 25.0 g - Molar mass = 60.1 g/mol - ΔH<sub>vap</sub> = 39.85 kJ/mol 3. *Heating from 82.5°C to 90.0°C (gas phase):* - q<sub>3</sub> = m * c<sub>gas</sub
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