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)
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)
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F38a9992d-9c05-40da-9474-aaafff450dc4%2F722c5d17-6f76-4115-ae2d-86dc00423977%2F8d7et6_processed.png&w=3840&q=75)
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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