The following information is given for water at 1 atm: AHvap (100.00°C) = 2.259 × 10³ J/g AĦfus (0.00°C) = 333.5 J/g T, = 100.00°C Tm = 0.00°C Specific heat solid = 2.100 J/g °C Specific heat liquid = 4.184 J/g °C A 38.10 g sample of solid water is initially at -15.00°C. If the sample is heated at constant pressure (P = 1 atm), kJ of heat are needed to raise the temperature of the sample to 27.00°C.
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.
![**References**
Use the References to access important values if needed for this question.
The following information is given for water at 1 atm:
- **Boiling point (T_b)**: 100.00°C
- **Melting point (T_m)**: 0.00°C
- **Heat of vaporization (ΔH_vap at 100.00°C)**: 2.259 × 10³ J/g
- **Heat of fusion (ΔH_fus at 0.00°C)**: 333.5 J/g
- **Specific heat of solid water (ice)**: 2.100 J/g°C
- **Specific heat of liquid water**: 4.184 J/g°C
**Problem Statement:**
A 38.10 g sample of solid water (ice) is initially at -15.00°C. If the sample is heated at constant pressure (P = 1 atm), how many kJ of heat are needed to raise the temperature of the sample to 27.00°C?
To calculate the total heat required, you need to account for the following steps:
1. Heating the ice from -15.00°C to 0.00°C.
2. Melting the ice at 0.00°C.
3. Heating the liquid water from 0.00°C to 27.00°C.
You can use the following formulas for each step:
1. **Heating the ice:**
\[ q_1 = m \times c_{\text{solid}} \times \Delta T_1 \]
where:
- \( q_1 \) = heat required to raise the temperature of ice,
- \( m \) = mass of the sample,
- \( c_{\text{solid}} \) = specific heat of solid water,
- \( \Delta T_1 \) = change in temperature (final temperature - initial temperature).
2. **Melting the ice:**
\[ q_2 = m \times \Delta H_{\text{fus}} \]
where:
- \( q_2 \) = heat required to melt the ice,
- \( \Delta H_{\text{fus}} \) = heat of fusion.
3. **Heating the liquid water:**
\[ q_3 = m \times c_{\text{liquid}} \times \Delta T_2 \]
where:
- \( q_3 \) = heat required to raise the temperature](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24192f2d-e403-49d4-ab0d-d6896014fc0b%2F8b509e93-e125-402d-b89b-e33413bdf2a6%2Fio08d0p.png&w=3840&q=75)
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