ORE HEAT PRACTICE The data below refers to an unknown substance, X. Melting point Boiling Heat of Heat of Point Fusion Vaporization 695 kJ/g 425 kJ/g 1. Draw a heating curve for substance X, going from 32.0°C 112.0°C 125 Call formulas used to calculate heat. 112°C 32 °C- E gomblik! F Em SS AT SH SH Capacity (solid) (liquid) 2.3 J/gºC 5.9 J/gºC 1.1 J/gºC 15°C to 125°C on the axis below. Write in A GEMSVAT C. qm Hvap e. Find the final temperature if 150g of X, at q=ms, AT 15°C- b. Determine the amount of heat released when 15.0g of gaseous X at 112.0°C changes to liquid at 112.0°C. 9-10425K) 9=15.09x695k/g q=mal vaporization AH = 695k/g c. Determine the amount of heat necessary to convert 15.0g of solid X at-32.0°C to liquid at 32.0° C. 95 ΔΗ = d. Determine the mass of X that can be heated from-38°C to 102°C using 4500J of heat. m= AH receives 650J of heat. SH (vapor) AH-
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
![## Core Heat Practice
The data below refers to an unknown substance, X:
- **Melting Point:** 32.0°C
- **Boiling Point:** 112.0°C
- **Heat of Fusion:** 425 J/g
- **Heat of Vaporization:** 695 J/g
- **Specific Heat (Solid):** 2.3 J/g°C
- **Specific Heat (Liquid):** 5.9 J/g°C
- **Specific Heat (Vapor):** 1.1 J/g°C
### Instructions:
1. **Draw a heating curve for substance X,** going from 15°C to 125°C on the axis below. Include all formulas used to calculate heat.
- The diagram illustrates various segments labeled A to F:
- **A:** Represents heating of the vapor phase using the formula \( q = m \cdot s_v \cdot \Delta T \).
- **B - C:** Represents condensation using the formula \( q = m \cdot \Delta H_{vap} \).
- **C - D:** Represents heating of the liquid phase using the formula \( q = m \cdot s_l \cdot \Delta T \).
- **D - E:** Represents fusion using the formula \( q = m \cdot \Delta H_{fus} \).
- **E - F:** Represents heating of the solid phase using the formula \( q = m \cdot s_s \cdot \Delta T \).
2. **Problem Solving:**
b. **Determine the amount of heat released** when 15.0g of gaseous X at 112.0°C changes to liquid at 112.0°C.
\[
q = 15.0 \, \text{g} \times 695 \, \text{J/g} = 10425 \, \text{J}
\]
- Formula: \( q = m \cdot \Delta H_{vaporization} \)
c. **Determine the amount of heat necessary** to convert 15.0g of solid X at 32.0°C to liquid at 32.0°C.
- Fill in the formula and solution.
d. **Determine the mass of X** that can be heated from -38°C to 102°C using 4500J of heat.
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