Sample of water at 80.0°C to form a saturated solution. Using the information on Table B, determine the amount of heat absorbed by the E when the water is heated from 20.0°C to 50.0°C.

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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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**Question 36:**

**Context for Answer: Chemistry**

A 100-gram sample of liquid water is heated from 30.0°C to 80.0°C. Enough KClO₃(s) is dissolved in the sample of water at 80.0°C to form a saturated solution.

Using the information on Table B, determine the amount of heat absorbed by the water when the water is heated from 20.0°C to 50.0°C.

*Note: The text doesn't include any graphs or diagrams.*
Transcribed Image Text:**Question 36:** **Context for Answer: Chemistry** A 100-gram sample of liquid water is heated from 30.0°C to 80.0°C. Enough KClO₃(s) is dissolved in the sample of water at 80.0°C to form a saturated solution. Using the information on Table B, determine the amount of heat absorbed by the water when the water is heated from 20.0°C to 50.0°C. *Note: The text doesn't include any graphs or diagrams.*
**Table B: Physical Constants for Water**

| Property | Value |
|----------|-------|
| Heat of Fusion | 334 J/g |
| Heat of Vaporization | 2260 J/g |
| Specific Heat Capacity of H₂O(ℓ) | 4.18 J/g·K |

**Explanation:**

- **Heat of Fusion:** This is the amount of energy required to change 1 gram of solid water (ice) to liquid water at its melting point, without changing its temperature (334 Joules per gram).

- **Heat of Vaporization:** This is the energy needed to convert 1 gram of liquid water to vapor (gas) at its boiling point, without a temperature change (2260 Joules per gram).

- **Specific Heat Capacity of H₂O(ℓ):** It refers to the amount of heat required to raise the temperature of 1 gram of liquid water by 1 Kelvin (or 1 degree Celsius), which is 4.18 Joules per gram per Kelvin.
Transcribed Image Text:**Table B: Physical Constants for Water** | Property | Value | |----------|-------| | Heat of Fusion | 334 J/g | | Heat of Vaporization | 2260 J/g | | Specific Heat Capacity of H₂O(ℓ) | 4.18 J/g·K | **Explanation:** - **Heat of Fusion:** This is the amount of energy required to change 1 gram of solid water (ice) to liquid water at its melting point, without changing its temperature (334 Joules per gram). - **Heat of Vaporization:** This is the energy needed to convert 1 gram of liquid water to vapor (gas) at its boiling point, without a temperature change (2260 Joules per gram). - **Specific Heat Capacity of H₂O(ℓ):** It refers to the amount of heat required to raise the temperature of 1 gram of liquid water by 1 Kelvin (or 1 degree Celsius), which is 4.18 Joules per gram per Kelvin.
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